Sponsor Area
= 1.05 V – 0.0295 x log 80
= 1.05 V – 0.0295 x 1.9031
= 1.05 V – 0.056 = 0.99 V.
Answer:
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer:
A secondary cell after use can be recharged by passing current through it in the opposite direction so that it can be used again.
Recharging: During charging process, hydrogen ions moves to cathode and sulphate ions to anode and the following reactions take place
At cathode:
PbSO4 + H2 → Pb + H2SO4
At anode:
PbSO4 + SO4 + 2H2O → PbO2 + 2H2SO4
Thus, during charging active materials namely Pb cut the cathode and PbO2 at the anode are formed. Sulphuric acid is formed and water is iconsumed. Due to this, the specific gravity of sulphuric acid increases and emf of the cell goes up to 2.2 volt.
The fuel cell thermodynamic efficiency is given by the ratio of the Gibbs function change to the Enthalpy change in the overall cell reaction. The Gibbs function change measures the electrical work and the enthalpy change is a measure of the heating value of the fuel.
Efficiency = (dG/dH)
Sponsor Area
Galvanic cells that are designed to convert
the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells
The electrode
reactions are given below:
Cathode: O2(g) + 2H2O(l) + 4e- -->4OH-(aq)
Anode: 2H2 (g) + 4OH-(aq) ---> 4H2O(l) + 4e-
Overall reaction being:
2H2(g) + O2(g) ---> 2 H2O(l )
Kohlrausch examined Ëm° values for a number of strong electrolytes and observed certain regularities. He noted that the difference in Ëm° of the electrolytes NaX and KX for any X is nearly constant. For example
at 298 K:
Ëm°(KCl) – Ëm° (NaCl)= Ëm°(KBr) – Ëm°(NaBr)
= Ëm°(KI) – Ëm° (NaI) ≃ 23.4 S cm2 mol–1
and similarly it was found that
Ëm°(NaBr)– Ëm°(NaCl)= Ëm°(KBr) – Ëm°(KCl) ≃ 1.8 S cm2 mol–1
On the basis of the above observations he enunciated Kohlrausch
law of independent migration of ions
The electrolysis of an aqueous solution of copper sulphate using copper electrodes (i.e. using active electrodes) results in transfer of copper metal from the anode to the cathode during electrolysis. The copper sulphate is ionised in aqueous solution.
CuSO4 - -> Cu2+ + S
The positively charged copper ions migrate to the cathode, where each gains two electrons to become copper atoms that are deposited on the cathode.
Cathode: Cu2+ + 2e → Cu
At the anode, each copper atom loses two electrons to become copper ions, which go into solution.
Anode: Cu → Cu2+ + 2e
The sulphate ion does not take part in the reaction and the concentration of the copper sulphate in solution does not change. The reaction is completed when the anode is completely eaten away. This process is used in electroplating
Galvanic cells that are designed to convert
the energy of combustion of fuels like hydrogen, methane, methanol,
etc. directly into electrical energy are called fuel cells.
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Answer:
The electrolysis of an aqueous solution of sodium sulphate using inert electrodes produces hydrogen at the cathode and oxygen at the anode and a neutral solution of sodium sulphate remains unaltered by the electrolysis.
At the cathode:
2H+ + 2e– → H2
At the anode:
2H2O → O2 + 4H+ + 4e–
Answer:
Anode: Zn + Zn2+ → 2e
Cathode: MnO2(s) + NH4+ + e– →MnO(OH) (s) + NH3 (aq)
Answer:
Degree of dissociation:
Answer:
At anode: Sn(s) →Sn2+ (aq) + 2e–
At cathode: 2H+ (aq) + 2e– → H2(g)
Answer:
Specific conductivity =
The quantity l/A is called cell constant denoted by the symbol, G*.
It depends on the distance between the electrodes and their area of cross-section and has the dimension of length–1 and can be calculated
if we know l and A.
Sponsor Area
Cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells.
Combustion of fuels such as hydrogen, CO and CH4 is the basis of obtaining electrical energy in fuel cells.
Strong electrolyte | Weak electrolyte |
1. These have higher molar conductivities at all concentrations. 2. λ°m values increase very slightly with dilution. 3. Degree of ionisation is very high at all concentration i.e., almost fully ionized. 4. Most of the salts like NaCl, KCl, NaNO3, BaCl2 and mineral acids like HCl, H2SO4, HNO3 and NaOH, KOH etc are common examples of strong electrolytes | 1. These have much lower conductivities at high concentration. 2. λ°m values increase sharply with dilution. 3. Degree of ionisation is very low at high concentration and increases with dilution. 4. Salts like ammonium acetate, acetic acid, aq NH4OH, aqueous CO2 and organic acids and bases are common examples of weak electrolytes. |
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3 with platinum electrodes.
(iii) A dilute solution of H2SO4 with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
(ii) Electrons move from anode (zinc electrode) to cathode (silver electrode) in the external circuit. Zinc ions go into solution at anode and Ag+ ions get deposited at cathode. Thus electrons in the external and metal ions in the internal circuit act as carrier of current in.
(iii) Overall reaction is obtained by anode and cathode reactions.
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)
Answer:
(i) Mn2+ is more stable than Mn3+ because Mn+2 has exactly half-filled orbitals, while Fe+2 after losing one e– half-filled orbitals.
(ii) Ce+4 achieve inert gas structure of Xenon, to require extra stability.
Answer:
The advantage of water fuel cell over ordinary cell is,
(i) It has high-efficiency and eco-friendly.
(ii) The H2O produced can be used by astronauts for drinking purposes.
Answer:
Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal.
Standard half-cell potential E° : If the ionic species have concentration of mol dm–3and pressure of the gaseous species is 1 atm (101.325 KPa), than half-cell potential is called as standard half-cell potential. The temperature being 298 K.
Absolute value of half-cell potential cannot be determined experimentally. However, its value relative to reference electrode can be determined.
Reference electrode 11 given half-cell. E°cell can be measured experimentally using a potentiometer.
Knowing the standard reduction potential of the reference electrode E°half-cell can be found out.
At cathode:
At anode:
36 g of requires
3.6 of
R is gas constant (8.314 JK–1 mol–1),
F is Faraday constant (96487 C mol–1), T is temperature in kelvin and [Mn+] is the concentration of the species, Mn
(i) Weak electrolytes: When the concentration of weak electrolyte becomes very low, its degree of ionisation rises sharply. There is sharp increase in the number of ions in the solution. Hence, the molar conductivity of a weak electrolyte rises steeply at low concentration.
(ii) Strong electrolytes: The molar conductivity of a strong electrolyte decreases slightly with the increase in concentration. This decrease is due to the increase in interionic attractions as a result of greater numbr of ions per unit volume. With dilution, the ions are far apart, inter ionic attractions become weaker and conductance increases.
Fuel cell: A galvanic cell in which the reactants are continuously fed into the cell and the products are continuously removed is called a fuel cell.
The most important fuel cell is hydrogen oxygen fuel cell.
The reactions taking place in this fuel cell are
Cathode: O2(g) + 2H2O(l) + 4e–--->4OH–(aq)
Anode: 2H2 (g) + 4OH–(aq)---> 4H2O(l) + 4e–
Overall reaction being:
2H2(g) + O2(g) ---> 2H2O(l )
What is a mercury cell? Give the electrode reactions?
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Zinc is more electro-positive than iron. Therefore, as long as zinc is there on the iron pipe, zinc acts as anode and the iron as cathode. As a result, rusting of iron is prevented.
On the other hand, tin is less electro-positive than iron. Therefore, when tin coating over iron gets broken, iron acts as anode and gets oxidised. Thus even when tin is there, the exposed iron gets rusted.
Equivalent conductivity : The conductivity of a volume (V) of a solution containing one equivalent of electrolyte placed between two electrodes separated by unit distance apart and of large enough area of cross-section to hold the entire volume (V) is called equivalent conductance. It is denoted by Λeq.
Λeq = k x V
where k = Specific conductance
V = Volume of solution containing one equivalent of electrolyte.
Molar conductivity : It is the conductivity of volume (V) of a solution containing 1 mole of a dissolved electrolyte place between two electrodes separated by unit distance apart and of enough area of cross-section to hold the entire volume V. It is denoted by Λm.
Λm= k x V = k / V where V = Volume of solution Containing 1 mole of electrolyte C = molarity of solution k = Specific conductivity.
conductivity and molar conductivity change with the concentration of the electrolyte. Conductivity always decreases with decrease in concentration both, for weak and strong electrolytes.
Molar conductivity varies with the concentration of electolytes.
Here b = Experimental constant
Λ∞m = Molar conductivity at infinite solution.
The plot of Λ∞m versus would give a straight line.
Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal. The rusting of iron, tarnishing of silver, development of green coating on copper and bronze are some of the
examples of corrosion.
Salt bridge. A salt bridge consists of a saturated solution of NH4NO3 or KCl mixed with gelatin or agar jelly filled in a glass tube bent according to the requirement of the experiment.
Significance:
(i) Salt bridge prevents mixing of two electrolytes.
(ii) Prevents junction potential.
(iii) Maintains electrical neutraility.
What is corrosion? Describe the electrochemical phenomenon of rusting of iron.
Corrosion is the process of slowly eating away of the metal due to attack of the atmospheric gases on the surface of the metal resulting into the formation of compounds such as oxides, sulphides, carbonates, etc.
The corrosion of iron is called rusting.
According to theory of rusting, impure iron surface behaves as a small electrochemical cell in the presence of water containing dissolved oxygen or CO2.
The pure iron acts as anode and impure surface as cathode.
At Anode : Iron atom undergo oxidation spontaneously forming Fe2+ ion.
Fe → Fe2+ (aq) + 2e- E°cell = – 0.44 V
Fe2+ ions move into solution and electrons into cathodic area where they are picked up by H+ ions of the solution.
At cathode:
H+ ions are produced by secondary reaction either from H2O or from H2CO3 (CO2 + H2O)
H2O ---->H+ + OH–
H2CO3------> H+ + HCO3–
The overall reaction of the corrosion cell may be represented as:
The Fe2+ ions move through water and come at the surface where these are further oxidized into Fe3+ ions by atmospheric oxygen to form hydrate ferric oxide known as rust, Fe2O3.xH2O.
The following curve is obtained when molar conductivity λm (y-axis) is plotted against the square root of concentration C1/2 (x-axis) for two electrolytes A and B.
(a) What can you about the nature of the two electrolytes A and B.
(b) How do you account for the increase in molar conductivity λm for the electrolytes A and B on dilution.
(a) A is strong electrolyte and B is weak electrolyte.
(b) The molar conductivity of a strong electrolyte:
(A) increase slightly as the concentration increased. This is because greater inter ionic attractions retrad the motion of the ions therefore molar conductivity decreases.The molar conductivity of a weak electrolyte.
(B) increase with decrease in the concentration of the solute in solution. The increase is very rapid at lower concentrations. The increase in the molar conductance with dilution is due to an increase in the effective degree of ionisation (α) at lower concentration and release of more full ions in the solution.
(i) The relation between free energy change and emf of a galvanic cell of the reaction is given as
ΔG = – nFEcell
(ii) The relation between free energy change and equilibrium constant of the reaction is given by
ΔG = – 2.303 RT log Kc.
The change in free energy is equal to the useful work done in a reversible process at constant temperature and pressure
ΔG = Wnet ...(i)
In an electrochemical cell and work obtained is equal to the charge transferred multiplied by the potential difference:
Welectrical = – nFE ...(ii)
The negative sign appears because the work is done by the charge. In equation (ii) n is the moles of electrons gained or lost in redox reaction and F is the Faraday constant.
In a cell when only electric work is done, then
Wnet= WeIectrical = – nFE ...(iii)
From relation (i) and (iii), we have
ΔG = – nFE
or ΔG = – mFE°cell.
Cathode: O2(g) + 2H2O(l) + 4e–-----> 4OH–(aq)
Anode: 2H2 (g) + 4OH–(aq)-----> 4H2O(l) + 4e–
Overall reaction being:
2H2(g) + O2(g) ----->2H2O(l )
Advantages: (i) Fuel cells are efficient and free from pollution.
(ii) The only product in the reaction of fuel cell is water which can be removed and the astronauts of a spacecraft can drink it.
The cell reactions when the battery is in use are given below:
Anode:
Cathode:
i.e., overall cell reaction consisting of cathode and anode reactions is:
ii) Recharging a lead-acid cell:
How much charge is required for the following reductions:
1 mol of Al3+ to Al?
How much charge is required for the following reductions:
1 mol of Cu2+ to Cu?
How much charge is required for the following reductions:
1 mol of MnO4– to Mn2+?
Zn/Zn2+(0.1 M) || Cd2+ (0.01) | Cd
Cu2+ + 2e– → Cu E° = + 0.34 V
Ag+ + 1e– → Ag E° = + 0.80 V
(i) Construct a galvanic cell using the above data.
(ii) For what concentration of Ag+ ions will the emf of the cell be zero at 25°C, if the concentration of Cu2+ is 0.01 M ? [log 3.919 = 0.593]
(i) Cu(s) | Cu2+ (aq) || Ag+ (aq) | Ag(s)
(ii) Cu(s) | Cu2+ (aq) || Ag+ (aq) | Ag(s)
2H2O → 4H+ + O2 + 4e–
2 moles of H2O require 4 x 96500 C
1 mole of H2O will need
Concentration of
= 0.1 +
According to Nernst equation,
(i) Resistance (k) =
Conductance (C) =
=
Specific conductance (k)
= Conductance x cell constant
(ii) Molar conductance (C)
Molar conductance
For cell
The cell reaction
On applying Nernst equation
or
or
or
At equilibrium
∴
or
or
or
Resistance of KCl solution,
R = 85
Cell constant = K x R
Resistance of unknown electrolyte solution,
Specific conductance
Concentration,
Molar conductance,
The following electrochemical cell has been set up
Pt(1) | Fe3+, Fe2+ (a = 1) || Ce4+, Ce3+ (a = 1) | Pt (2)
E°(Fe2+) = 0.77 V, and E°(Ce4+,Ce3+) = 1.61 V
If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time?
For the electrochemical cell
Pt(1) | Fe3+, Fe2+ (a = 1) || Ce4+, Ce3+ (a = 1) | Pt (2)
the cell regions are
Right-half cell: reduction
Left-half cell: oxidation
______________________________
Add
The net cell potential is
E° Cell = E° R – E° L = 1.61 V – 0.77 V = 0.84 V.
Since E°cell is positive, the cell reaction will be spontaneous.
The current in the external circuit will flow from Pt (1) (which serves as anode to Pt(2) which serves as cathode.
With the passage of time, Ecell will decrease and so is the current in the external circuit.
Zn acts as anode, and SHE acts as cathode
At anode: Zn(s) → Zn2+ (aq) + 2e–
At cathode: 2H+ (aq) + 2e– H2(g)
The net cell reaction is
Zn(s) + 2H+ (aq) → Zn2+ + H2(g)
Can a nickel spatula be used to stir the solution of CuSO4? Support your answer with reason. (E°Ni2+/Ni = – 0.25 V, E°Cu2+/Cu = + 0.34 V)
No.nickel spatula cannot be used to stir the solution of CuSO4
As E° Ni2+/Ni < E°cu2+/cu
Nickel has more tendency to lose electrons than copper. If a nickel spoon is used to CuSO4solution, it will displace Cu from CuSO4 solution.
Ni(s) + Cu2+ (aq) → Ni2+ (aq) + Cu(s)
Calculate the e.m.f. of the cell in which the reaction is
Mg(s) | Mg2+ (0.01 M) || Sn2+ (0.1 M) | Sn(s)
Mg(s) → Mg2+ (aq) + 2e–
(oxidation at anode)
Sn2+(aq) + 2e– → Sn(s)
(reduction at cathode)
Calculate the cell emf at 25° C for the following cell:
Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1 M) | Cu(s)
[Given E°Ni2+/Ni = – 0 25 V, E° Cu = + 0.34 V, 1 F = 96500]
Calculate the maximum work that can be accomplished by operation of this cell.
Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1 M) | Cu(s)
At anode Ni(s) → Ni2+ (aq) + 2e
At cathode Cu2+ + 2e → Cu(s)
Net cell reaction
The emf of a cell corresponding to the reaction
Zn(s) + 2H+ (aq) → Zn2+ (0.1 M) + H2(g) (1 atm) is 0.28 V at 15° C.Write the half cell reactions and calculate the pH of the solution at the hydrogen electrode.
From the table, standard electrode potents at 298 k are:
(E°Fe3+/VF = 0.77 V, E°I2/I– = 0.54 V)
(E°Ag+/Ag = E°Cu2+/Cu = 0.34)
(E°Fe3+/Fe= 0.77 V, E°Br2/Br- = 1.08 V)
(E°Ag+/Ag= 0.8 V E°Fe3+/Fe2+ = 0.77 V)
(E°Fe3+/Fe2+= 0.77 V,E°Br2/Br- = 1.08 V)
(a)
In this reaction, Fe3+ is reduced to Fe2+ and I– is oxidised to I2. The cell giving above reaction will be
As E0 is positive, the reaction between Fe3+ (aq) and I– (aq) occurs as indicated by possible reaction given above.
(b)
Here, in this reaction, Ag+ is reduced to Ag (i.e., it should be cathode) and Cu(s) is oxidised to Cu2+(aq) (i.e., it should be anode).
The cell can be represented as
As E°cell is positive, the reaction between (Ag+ (aq) and Cu(s) occurs as indicated by possible reaction given above.
(c)
In this reaction Fe3+ is reduced to Fe2+ (i.e., Fe3/Fe2+ electrode should be cathode) and Br is oxidised to Br2 (i.e., Br2/Br– electrode should be anode.
The cell can be represented as:
As E°cell is negative, no reaction will occur between Fe3+ (aq) and Br–(aq).
(d)
Two half-cell reactions can be expressed as:
As E°cell is negative, no reaction occurs between Fe3+(aq) and Ag(s).
(e)
The two half-cell reactions are
As E°cell is positive, the reaction is feasible, i.e., reaction between Br2(aq) and Fe2+ (aq) occurs as indicated by possible reaction given above.
For an electrochemical cell reaction
aA + bB → cC + dD
The Nernst’s equation for cell reaction is
The values of a, b, c, d and n can be obtained from the balanced cell reaction. (i) Anode reaction:
Cathode reaction:
Overall cell reaction:
Here, n= 2,
The Nernst equation for Ecell at 298 can be written as
For the given cell Anode reaction:
Sn(s) → Sn2+ (aq) + 2e–
Cathode reaction:
2H+(aq) + 2e– → H2(g)
Overall cell reaction:
Sn(s) + 2H+ (aq) → Sn2+(aq) + H2(g)
Here, n = 2, E°cell = E°cathode – E°anode = 0 – (– 0.14 V) = + 0.14 V.
The Nernst equation for Ecell and 298 k can be written as:
Corrosion is the process of slowly eating away of the metal due to attack of the atmospheric gases on the surface of the metal resulting into the formation of compounds such as oxides, sulphides, carbonates, etc. The corrosion of iron is called rusting.
The phenomenon of corrosion involves the destruction of metal in which metal is generally converted into oxide. Common examples are rusting of rion, tarnishing of silver and deposition of green coating on copper and bronze.
Factors affecting corrosion:
(a) Presence of oxygen, sulphur etc. (elements which gain electrons).
(b) Presence of moisture.
(c) Presence of carbon dioxide.
Electrolysis of AgNO3 using Ag electrodes
At cathode:
At anode:
Time = 25 min = 25 x 60 sec
Current = 0.75 A
Electricity passed = 25 x 60 x 0.75 = 1125 C of electricity deposit copper = 0.369 g. 2 x 96500 C
of electricity will deposit copper.
M= ZxIxT
Atomic mass of copper = 63.3 u.
B.
6.28 x 1018 electronsThe resistance of any conductor varies directly as its length (l) and inversely as its cross-sectional area (A), i.e.,
Mathematically
Where ρ is called the specific resistance.
l/A is known as cell constant.
If l = 1 cm and A = 1 cm2, then
R = ρ
The specific resistance is, thus defined as the resistance of one-centimetre cube of a conductor.
The reciprocal of specific resistance is termed the specific conductance or it is the conductance of one-centimetre cube of a conductor.
It is denoted by the symbol κ.
Thus,
Where (κ) kappa → the specific conductance
Specific conductance is also called conductivity.
Further,
or Specific conductance = Conductance × cell constant
In the case of electrolytic solutions, the specific conductance is defined as the conductance of a solution of definite dilution enclosed in a cell having two electrodes of unit area separated by one centimetre apart.
Unit of specific conductance: Ω-1 cm-1
It is not possible to store copper sulphate solution in iron vessel.since ,iron is more reactive than copper,it displaces copper from any if its solution.the reaction takes place as ,
Fe(s)+CuSO4(aq)..............>FeSO4(aq)+Cu(s)
A salt bridge is a U-shaped device containing concentrated solution of an inert electrolyte like KCl, KNO3, etc.or a solidified solution of those electrolytes in agar-agar solution and gelatin. It connects the oxidation and reduction half-cells of a galvanic cell. The inert electrolytes present do not take part in redox reaction of the cell and dont react with the electrolyte that has been used.
As electrons leave one half of a galvanic cell and flow to the other, a difference in charge is built up. If no salt bridge were used, this increasing charge difference would eventually prevent further flow of electrons. The salt bridge solves these problems. It has two main functions:
1. To allow the flow of ions from one solution to another without mixing of the two solutions and completing the electrical circuit.
2. To maintain the electical neutrality of the solutions in the two half cells.
The value of n for electrode potential is 2.
Free energy change G is measure of the spontaneity of a chemical reaction or process.
Where
-n =number of moles of electron
F= the quantity of electrical charge that is contained in 1mole of electrons.This is the Faraday constant i.e. 1F= 96500 Coloum/mole of electron
F and n are positive value. therefore as positive value of E (which indicte spontaneity )and thus negtive value for G (which also indicate spontaneity)
Standard Oxidation Potentials. The standard oxidation potential is much like the standard reduction potential. It is the tendency for a species to be oxidized at standard conditions.
it can be raised up by reducing hydrogen value in cell reaction.
Gibbs Free Energy is defined as the thermodynamic potential that signifies the maximum or reversible work performed by a thermodynamic system at constant temperature and pressure. i.e.
Weak electrolytes: When the concentration of weak electrolyte becomes very low, its degree of ionisation rises sharply. There is sharp increase in the number of ions in the solution. Hence, the molar conductivity of a weak electrolyte rises steeply at low concentration.
Fluorine cannot be prepared from fluorides by chemical oxidation.
Fluorine is the strongest oxidizing agent as it has highest E0 value. Therefore fluorine has the highest tendency to get reduced to F-. As a result, F- ion has the least tendency to get oxidized, hence fluorine cannot be prepared by chemical oxidation of fluorides.
Secondary batteries are the rechargeable batteries. They have the advantage of being more cost-efficient over the long term, although individual batteries are more expensive. Generally, secondary batteries have a lower capacity and initial voltage, a flat discharge curve, higher self discharge rates and varying recharge life ratings.
The concentration of all species involved in the species involved in the electrode reaction is unity.This need not be always true.
Nernst shows that for the electrode reaction:
the electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by:
but concentration of solid M is taken as unity as we have
R is gas constant (8.314 JK–1 mol–1),
F is Faraday constant (96487 C mol–1), T is temperature in kelvin and [Mn+] is the concentration of the species, Mn
Let us take a electrode reaction
The Nernst equation of this electrode
Instead of activity, we can take molar concentration.
For pure solid and liquid molar concentration is taken as unity.
Nernst equation is given as:
Ecell =
Where Q reaction quotient
Ecell is the cell potential at the temperature of interest.
is the standard cell potential
R is the universal gas constant
T is the abolute temperture
F is the Farday constant
n i the number of electron
At equilibrium
Ecell =0 and Q=k
K isthe equilibrium constant
0=
Strong electrolytes: The molar conductivity of a strong electrolyte decreases slightly with the increase in concentration. This decrease is due to the increase in interionic attractions as a result of greater numbr of ions per unit volume. With dilution, the ions are far apart, inter ionic attractions become weaker and conductance increases
Fuel cell: A galvanic cell in which the reactants are continuously fed into the cell and the products are continuously removed is called a fuel cell.
The most important fuel cell is hydrogen oxygen fuel cell
A half-cell called standard hydrogen electrode
represented by Pt(s) l H2(g) l H+(aq), is assigned
a zero potential at all temperatures corresponding to the reaction
A galvanic cell (left) transforms the energy released by a spontaneous redox reaction into electrical energy that can be used to perform work. The oxidative and reductive half-reactions usually occur in separate compartments that are connected by an external electrical circuit.
Oxidation half reaction:
Y ----> Y+ +e-
Reduction half reaction:
Z + e- ----> Z-
overall cell reaction
Y+Z ------> Y+ + Z- (G<0)
A second connection that allows ions to flow between the compartments (shown here as a vertical dashed line to represent a porous barrier) is necessary to maintain electrical neutrality. The potential difference between the electrodes (voltage) causes electrons to flow from the reductant to the oxidant through the external circuit, generating an electric current.
In cell reaction :
Cu(s) +2Ag+(aq) ---->Cu2+(aq) +2Ag(s)
Half cell reaction
Cathode (reduction ):
2Ag+(aq) +2e- -----> 2Ag(s)
Anode (oxidation)
Cu(s) -----> Cu2+(aq) +2e-
In overall reaction of the cell, silver electrode act as cathode and copper electrode act as anode.
A salt bridge is a U-shaped device containing concentrated solution of an inert electrolyte like KCl, KNO3, etc.or a solidified solution of those electrolytes in agar-agar solution and gelatin. It connects the oxidation and reduction half-cells of a galvanic cell. The inert electrolytes present do not take part in redox reaction of the cell and dont react with the electrolyte that has been used.
As electrons leave one half of a galvanic cell and flow to the other, a difference in charge is built up. If no salt bridge were used, this increasing charge difference would eventually prevent further flow of electrons. The salt bridge solves these problems. It has two main functions:
1. To allow the flow of ions from one solution to another without mixing of the two solutions and completing the electrical circuit.
2. To maintain the electical neutrality of the solutions in the two half cells.
A half-cell called standard hydrogen electrode
represented by Pt(s)l H2(g)lH+(aq), is assigned
a zero potential at all temperatures corresponding to the reaction;
H+ (aq) + e- ---->1/2H2(g)
The standard hydrogen electrode consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it.
The electrode potential measures the tendency of electrons to flow away from or towards a redox equilibrium. They are always measuredwith respect to the standard hydrogen electrode (which is assigned a value of zero volts).
The concentration of all species involved in the species involved in the electrode reaction is unity.This need not be always true.
Nernst shows that for the electrode reaction:
the electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by:
but concentration of solid M is taken as unity as we have
R is gas constant (8.314 JK–1 mol–1),
F is Faraday constant (96487 C mol–1), T is temperature in kelvin and [Mn+] is the concentration of the species.
In Daniell cell, the electrode potential for any given concentration of
Cu2+ and Zn2+ ions, we write
For cathode:-
Electrolysis of molten NaCl gives sodium at cathode while aqueous NaCl gives H2 gas at cathode.
Molten NaCl dissociates to give Na+ and Cl-, Na+ then move towards cathode, picks up one electron and gets reduced to form Na.
Na+ +e- ---> Na
In aqueous NaCl, as , it is the hydrogen gas which liberates at cathode.
The cells from which electric energy is derived by irreversible chemical action are called primary cells. The primary cell is capable of providing an EMF when its constituent’s two electrodes and a suitable electrolyte are assembled together. The three main primary cells namely are the Daniel cell, the Leclanche cell, and the dry cell. None of these cells can be recharged electrically.
The dry cell consists of a zinc container that also acts as anode and the cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide and carbon. The space between the electrodes is filled by a moist paste of ammonium chloride (NH4Cl) and zinc chloride (ZnCl2). The electrode reactions are complex, but they can be written approximately as follows :
Anode: Zn(s)----> Zn2+ + 2e–
Cathode: MnO2+ NH4+ + e–----> MnO(OH) + NH3
In the reaction at cathode, manganese is reduced from the + 4 oxidation state to the +3 state. Ammonia produced in the reaction forms a complex with Zn2+ to give [Zn (NH3)4]2+. The cell has a potential of nearly 1.5 V.
Strong electrolytes: The molar conductivity of a strong electrolyte decreases slightly with the increase in concentration. This decrease is due to the increase in interionic attractions as a result of greater numbr of ions per unit volume. With dilution, the ions are far apart, inter ionic attractions become weaker and conductance increases.
Kohlrausch’s law of independent migration of ions states molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of the contribution of individual ions. If molar conductivity of cations and anions are represented by λ∞+ and λ∞– respectively.
where v+ and v– are number of cations and anions per formula of electrolyte e.g.,
Λ∞ CaCl2 = λ∞ (Ca2+) + 2 λ∞ (CI–)
Λ = KCl = λ∞ (K+) + λ∞ (CI–)
Uses 1. It is used to find molar conductivity of weak electrolyte at infinite dilution which
cannot be obtained by extrapolation.
2. It is used to calculate degree of dissociation of weak electrolyte at a particular concentration.
Degree of dissociation
where Λm is molar conductivity of weak electrolyte at a particular concentration and Λemis molar conductivity of weak electrolyte at infinite dilution.
Gibbs energy of the reaction given by:
= – nFE(cell)
thus the reaction
Zn(s) + Cu2+(aq)---> Zn2+(aq) + Cu(s)
= – 2FE(cell)
but when we write the reaction
2 Zn (s) + 2 Cu2+----->2 Zn2+(aq) + 2Cu(s)
= – 4FE(cell)
If the concentration of all the reacting species is unity, then
E(cell) =
and we have
= – nF
Thus, from the measurement of we can obtain an important thermodynamic quantity, , standard Gibbs energy of the reaction.
From the latter we can calculate equilibrium constant by the equation:
= –RT ln Kc
A secondary cell after use can be recharged by passing current through it in the opposite direction so that it can be used again. A good secondary cell can undergo a large number of discharging and charging cycles. The most important secondary cell is the lead storage battery commonly used in automobiles and invertors.
It consists of a lead anode and a grid of lead packed with lead dioxide (PbO2 ) as cathode. A 38% solution of sulphuric acid is used as an electrolyte.
The cell reactions when the battery is in use are given below:
Anode: Pb(s) + SO42–(aq)----> PbSO4(s) + 2e–
Cathode: PbO2(s) + SO42–(aq) + 4H+(aq) + 2e– ----> PbSO4 (s) + 2H2O (l )
i.e., overall cell reaction consisting of cathode and anode reactions is:
Pb(s) + PbO2(s) + 2H2SO4(aq)---> 2PbSO4(s) + 2H2O(l)
On charging the battery the reaction is reversed and PbSO4(s) on anode and cathode is converted into Pb and PbO2, respectively.
On the basis of the standard electrode potential values stated for acid solution. Predict whether Ti4+ species may be used to oxidise FeII to FeIII.
Reaction: E°/V TiIV + e– → Ti3+ + 0.01
Fe3+ + e– → Fe2+ + 0.77
Ti4+ would not oxidize Fe+2 . since the standard electrode potential of iron is high as compare to the titanuim standard electrode potential.
Predict the products of electrolysis obtained at the electrodes in each case when the electrodes used are of platinum.
(i) An aqueous solution of AgNO3.
(ii) An aqueous solution of H2SO4.
(i) Cathodic protection (CP) is a technique used to control the corrosion of a metal surface by making it the cathode of an electrochemical cell. A simple method of protection connects the metal to be protected to a more easily corroded "sacrificial metal" to act as the anode.
for example zinc is used to pervent iron
Zinc is more electro-positive than iron. Therefore, as long as zinc is there on the iron pipe, zinc acts as anode and the iron as cathode. As a result, rusting of iron is prevented.
(ii)Electrochemical series is a series of chemical elements arranged in order of their standard electrode potentials. The hydrogen electrode. H+(aq) + e- →← 1/2H2(g) is taken as having zero electrode potential. An electrode potential is, by definition, a reduction potential
(iii)The quanitty 1/A is called cell constant denoted by the symbol. G*. It depends on the distance between the electrodes and their area of cross -section and has the dimension of length-1 and can be calculated if l and A
G* =l/A =Rk
(iv) A strong electrolyte is a solute that completely, or almost completely, ionizes or dissociates in a solution. While the specificconductance of a solution increases with concentration, the equivalent conductance decreases as the concentration increases. unit of equivalent conductance
(v)electrolytes :A substance that when dissolved in water produced a solution that can conduct electric current.
there are two electrolytes
1. strong
2.weak
strong Electrolytes conduct current very efficiently.Completely ionized or dissociate when dissolved in water
a. Soluble Ionic compounds
b. Strong acids (HNO3(aq), H2SO4(aq), HCl(aq))
HNO3--> H+ + NO3- (100% ionization)
c. Strong bases (KOH and NaOH)
KOH -->K+ +OH - (100% dissociation)
Weak electrolytes conduct only a small current
Slightly ionized in solution
a. Weak acids (organic acids-->acetic, citric, butyric,malic, etc.)
HC2H3O2 <==> H+ + C2H3O2-
b. Weak bases (ammonia)
NH3 + H2O <==> NH4+ + OH-
If we connect the zinc and copper by means of a metallic conductor, the excess electrons that remain when Zn2+ ions emerge from the zinc in the left cell would be able to flow through the external circuit and into the right electrode, where they could be delivered to the Cu2+ ions which become "discharged", that is, converted into Cu atoms at the surface of the copper electrode. The net reaction is the oxidation of zinc by copper(II) ions:
but this time, the oxidation and reduction steps (half reactions) take place in separate locations:
left electrode : Zn----> Zn2+ +2e-
| galvanic cell | Electrolytic cell |
|---|---|
| A Galvanic cell converts chemical energy into electrical energy. | An electrolytic cell converts electrical energy into chemical energy. |
| Here, the redox reaction is spontaneous and is responsible for the production of electrical energy. | The redox reaction is not spontaneous and electrical energy has to be supplied to initiate the reaction. |
| The two half-cells are set up in different containers, being connected through the salt bridge or porous partition. | Both the electrodes are placed in a same container in the solution of molten electrolyte. |
| Here the anode is negative and cathode is the positive electrode. The reaction at the anode is oxidation and that at the cathode is reduction. | Here, the anode is positive and cathode is the negative electrode. The reaction at the anode is oxidation and that at the cathode is reduction. |
| The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit. | The external battery supplies the electrons. They enter through the cathode and come out through the anode. |
Definition: The standard hydrogen electrode is the standard measurement of electrode potential for the thermodynamic scale of redox potentials.
The standard is determined by the potential of a platinum electrode in the redox half reaction
2 H+(aq) + 2 e- → H2(g) at 25 °C.
The standard hydrogen electrode is often abbreviated SHE.
Also Known As: normal hydrogen electrode or NHE
Corrosion is the process of slowly eating away of the metal due to attack of the atmospheric gases on the surface of the metal resulting into the formation of compounds such as oxides, sulphides, carbonates, etc.
The rusting of iron, tarnishing of silver, development of green coating on copper and bronze are some of the examples of corrosion.
The main factors which affect corrosion are
Presence of water and the electrolytes present in it.
1. More the reactivity of metal, the more will be the possibility of the metal getting corroded.
2. The impurities help in setting up voltaic cells, which increase the speed of corrosion
3. Presence of electrolytes in water also increases the rate of corrosion
4. Presence of CO2 in natural water increase rusting of iron.
5. When the iron surface is coated with layers of metals more active than iron, then the rate of corrosion is retarded.
6. A rise in temperature (with in a reasonable limit) increases the rate of corrosion.
a)
|
electrolytes |
nonelectrolyte |
|
An electrolyte dissociates in solution and thus produce ion.
|
A nonelectrolyte does not dissociate at all in solution and therefore does not produce any ions. |
|
Electrolytes are ionic substance that dissolve in water |
Nonelectrolytes are typically polar covalent substances that do dissolve in water as molecules instead of ions. |
C)
|
Primary cell |
Secondary |
|
Lower initial cost. |
Higher Initial Cost |
|
Higher life-cycle cost ($/kWh). |
Lower life-cycle cost ($/kWh) if charging in convenient and inexpensive |
|
Disposable. |
Regular maintenance required. |
|
Typically lighter and smaller thus traditionally more suited for portable applications. |
Traditionally less suited for portable applications, although recent advances in Lithium battery technology have lead to the development of smaller/lighter secondary batteries. |
|
Molar conductivity |
Specific conductivity |
|
Molar Conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length. Therefore, Distance is unit so l = 1 Volume = area of base × length So V = A × 1 = A Λm =κA/l Λm = κV
|
Conductivity of a solution is equal to the conductance of a solution of 1 cm length and cross section area of 1 square cm. it may also be define as the conductance of ine centimeter cube of the conductor . It is represented by the symbol Kappa (κ). mathematically we can write κ = 1/ p here ρ is resistivity the unit of K is ohm –1 cm –1 or S cm–1 The conductivity, κ, of an electrolytic solution depends on the concentration of the electrolyte, nature of solvent and temperature.
|
The conductivity of 0.01 M solution of acetic acid at 25°C is 1.63 x 10–4 s cm–1. Given:
Λ°m (HCl) = 426 s cm2 mol–1, Δ°m (Na AC) = 91.5 cm2 mol–1
Λ°m (NaCl) = 126.5 cm2 mol–1 Calculate:
(a) the molar conductivity of acetic acid
(b) the degree of dissociation of acetic acid.
(c) the dissociation constant.
(d) the pH of 0.01 M solution of acetic acid.
a) The molar conductivity of acetic acid given by;
we have given that
Ecell0 =0.36V
thus equilibrium constant is
n=2
k =1.596 x 1012
H2 ----> 2H++2e-
Ans. pH = 4.237
we have given that
electrolyitc conductivity =0.20mol/L
conductivity = 2.48 x 10-2 ohm-1cm-1
thus apply the formula
here M is molar conductivity.
Ans. 124 ohm–1 cm2 mol–1
Calculate the molar conductivity at infinite dilution of acetic acid from the following data:
Λ∞m (HCl) = 426 ohm–1 cm2 mol–1, Λ∞m CH3COONa = 91 ohm–1 cm2 mol–1 and Λ∞m(NaCl) = 126 ohm–1 cm2 mol–1.
Acetic acid is weak electrolyte than HCl and NaCl which is strong electrolyte.
Acoording to kohlransch's law
=91-126+426 =391
Ans. 391 ohm–1 cm2 mol-1
We have given
E°Ni2+/Ni = -0.25 V
E°Cu2+/Cu = + 0.34 V,
Ecell0 =
= 0.34-(-0.25)
= 0.34+0.25 =0.59
R = 8.314 J K–1 mol–1
F = 96500 C mol–1.
T=25celcius =273+25 =298 Kelvin
n= 2
= Antilog[19.956]
Ans. 9.07 x 1019
Consider the cell Zn/Zn2+ (aq) (1.0 M) || Cu2+ (aq) (0.1 M) | Cu The standard reaction potentials are + 0.35 V for 2e– + Cu2+ (aq) → Cu and – 0.763 V for 2e– + Zn2+ (aq) → Zn
(i) Write down the cell reaction.
(ii) Calculate the emf of the cell.
(iii) Is the cell reaction spontaneous or not?
(I)Cell Reaction:
Oxidation: Zn----> Zn2+ +2e-
Reduction: Cu2+ +2e------> Cu
Therefore overall reaction
Zn +Cu2+ ------> Zn2+ +Cu
(ii) =+0.35-(-0.763)
=1.113 volts
(iii) Since EMF of cell is positive, it is a spontaneous reaction.
reduction potential is given by:
From the given data PH =14 and KsP (Cu(OH2) =1.0 x 10-19
we get
H+ =10-14 M thus,
[OH]- =
ans -0.22V
The standard reduction potential of the reaction at 25°C
2H2O + 2e– H2(g) + 2OH– is – 0.8277 V
Calculate equilibrium constant for the reaction
at
Consider the given reaction as the net cell reaction
two half reaction :
Oxidation:
H2O +1/2H2 ----> H3O+ +e- Ecell0 =0
reduction:
H2O +e- -----> 1/2H2(g) + OH- Ecell0 = -0.8277V
It is evident from the cell reaction that it involves the transfer of one electron so that n= 1
We ave given that
R = 1500 Ω
K = 1.46 x 10–4 Ω–1 cm–1
Cell constant = K x R
= 1.46 x 10–4 x 1500 = 0.219
KKCl = 0.0129 s cm–1
RKCl = 58 Ω
cell constant= k x R
cell constant= 0.0129 x 58
Cell constant = 0.7482 cm–1
As AgNO3 is also in the same conductivity cell, the cell constant remains same. Therefore,
Conductivity of
Cell constant = K x R
=
M = 0.00241 M
K = 7.896 x 10-5 s cm-1
Λ° (HAC) = λ°(H+) + λ°(AC–)
= λ° (H+) + λ°(Cl–) + λ° (AC–) + λ° (Na+) – λ°(Cl–) – λ°(Na+)
= Λ° (HCl) + Λ° (NaAc) – Λ° (NaCl)
= (425.9 + 91.0 – 126.4) s cm2 mol–1 = 390.5 s cm2 mol–1
Λ°m (CH3COOH) = λ°H+ + λ°CH3coo–
= Λ°m(HCl) = λ°m (CH3COONa) – λ°NaCl
= 425.9 + 91.0 –126.4
= 390.5 s cm2 mol–1.
The molar conductivity at infinite dilution (Λ∞m) can be obtained by extrapolating the above graph of strong electrolytes to zero concentration.
Kohlrausch Law of Independent migration of ions: The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Thus, if λ°Na+ and λCl– are molar limiting conductivity of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation:
Λ°NaCl = λ°Na + λCl–
In general, if an electrolyte on dissociation gives v+ cations and v– anions then its limiting molar conductivity is given by
Λ° = v + λ°+ + V– λ°–
Here, λ°+ and λ°– are the limiting molar conductivities of the cation and anion respectively.
Kohlrausch’s law helps us to calculate
(i) Determination of molar conductivities of weak electrolytes at infinite dilution.
(ii) Determination of the degree of dissociation of electrolytes.
Degree of dissociation: It is ratio of molar conductivity at a specific countraction ‘C’ to the molar conductivity at infinite dilution. It is denoted by α.
i.e.,
|
C/mol L–1 |
Λ/s cm2 mol–1 |
|
0.000198 0.000309 0.000521 0.000989 |
148.61 148.29 147.81 147.09 |
|
C1/2 /(mol L–1)1/2 |
A/s cm2 mol–1 |
|
0.01407 0.01758 0.02283 0.03145 |
148.61 148.29 147.81 147.09 |
We have given that
m = 127 gm,
I = 50 A
Z = 0.0003294 g/c
t = ?
By Faraday's first law,
m = ZIt
or
t = 77118 sec.
We have given that
current = 0.5 ampere
weight of H2 (W) = ?
t = 1 hour = 3600 sec,
Z = 0.00001.
W = Z x c x t
= 0.00001 x 0.5 x 3600 g = 0.018 g
2g or H2 at STP occupies = 22.4 litres
0.018 g of H2 at STP occupies
Hence, Volume of H2 produced at STP = 0.2016 litres.
How much charge is required for the following reduction of
(i) 1 mol of Al3+ to Al
(ii) 1 mol of Cu2+ to Cu
(iii) 1 mol of MnO4– to Mn2+
(i) Al3+ + 3e– → Al
1 mol Al3+ for reduction requires 3 mol e– or 3F electricity
1F = 96500C
∴ 3F = 3 x 96500 = 289500 C
(ii) Reduction of 1 mol Cu2+ to Cu requires 2 mol electrons
2 mol electrons = 2F = 2 x 96500 = 193000 C
(iii) In the reduction of 1 mol MnO–4 to Mn2+, there is net gain of 5e–
MnO4 + 5e– + 8H+ → Mn2+ + 4H2O
5F = 5 x 96500 = 482500 C
The cathode reaction is Cu2+ + 2e– → Cu
I = 1.5 amperes, t = 10 x 60 = 600 s
∴ Q = It = 1.5 x 600 = 900 C
The reaction states that 2 x 96500 C are required to deposit 63.5 g Cu at cathode
900 C will deposit
t = 600 s
Charge = current x time
= 1.5 A x 600s = 900 C
According to the reaction : Cu2+(aq) + 2e– → Cu(s)
We require 2F or 2 x 96487 C to deposit 1 mol or 63 g of Cu
For 900 C, the mass of Cu deposited =
A Cell is prepared by dipping a copper rod in 0.01 M copper sulphate solution, and zinc rod in 0.02 M ZnSO4 solution. The standard reduction potentials of copper and zinc are + 0.34 V and – 0.76 V respectively.
(a) What will be the cell reaction?
(b) How will the cell be represented?
(c) What will be the emf. of the cell?
(a) As E°Zn2+/Zn < E°Cu2+/Cu . Zinc electrode acts as anode and copper electrode acts as a cathode. The cell reaction is
Zn(s) + Cu2+(aq) (0.01 M) → Zn2+(aq) (0.02 M) + Cu(s)
(b) The cell is represented as
Zn(s) | Zn2+(aq) (0.02 M) || Cu2+(aq) (0.01 M) + Cu(s)
(c) The emf of the cell can be calculated by using Nernst equation
The cell is
Mg(s) | Mg2+ (aq) (0.1 M) || Ag+ (aq) (0.01 M) | Ag(s)
and the Ecell is given by
But
Write the Nernst equation and calculate the emf of the following cell at 298 K:
Pt(s) | Br2 (l) Br– (0.01M) || H+ (0.03 M) | H2(g) (1 bar) | Pt(s)
Given E°Br2/Br– = + 1.08 V
Br2 / Br–
The cell is
and
and n = 2
∴
As K is very high, the reaction is favoured in the forward direction, so, Sn2+ can easily reduce Fe3+ ion to Fe2+ ion.
Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution:
Ag+ (aq) + e- → Ag(s) E° = +0.80 V
H+ (aq) + e- → 1/2 H2 (g) E° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
We have given:
Ag+ (aq) + e- → Ag(s) E° = +0.80 V
H+ (aq) + e-→ 1/2 H2 (g) E° = 0.00 V
The relationship between the standard free energy change and emf of a cell reaction is given by
∆ G = – nFE(cell)
Thus, the more positive the standard reduction potential of a reaction, the more negative is the standard free energy change associated with the process and, consequently, the higher is the feasibility of the reaction.
Since E 0 Ag+/Ag has a greater positive value than E0 H+ /H, the reaction which is feasible at the cathode is given by
Ag+ (aq) + e- → Ag(s)
Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with the decrease in concentration?
The limiting molar conductivity of an electrolyte is defined as its molar conductivity when the concentration of the electrolyte in the solution approaches zero.
The conductivity of an electrolyte solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted or the concentration is decreased. As a result, the conductivity of an electrolyte solution decreases with the decrease in concentration.
Calculate emf of the following cell at 25 °C:
Fe | Fe2+(0.001 M) || H+ (0.01 M) | H2 (g) (1 bar) | Pt(s)
E°(Fe2+ | Fe) = –0.44 V E°(H+ | H2 ) = 0.00 V
For the given cell representation, the cell reaction will be Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)
The standard emf of the cell will be
=0.44-0.02955(log10)
=0.44-0.02955(1)
=0.41045V 0.41V
The Nernst equation for the cell reaction at 25 º C will be
=0.44-0.02955(log10)
=0.44-0.02955(1)
=0.41045V 0.41V
Define the following terms:
(i) Limiting molar conductivity
(ii) Fuel cell
(i) When the concentration of an electrolyte approaches zero, then its molar conductivity is known as limiting molar conductivity.
(ii) Fuel cells are the galvanic cells in which the energy of combustion of the fuels likes hydrogen, methanol. etc is directly converted into electrical energy.
The resistance of a conductivity cell filled with 0.1 mol L-1 KCl solution is 100 . If the resistance of the same cell when filled with 0.02 mol L-1 KCl solution is 520 
, calculate the conductivity and molar conductivity of 0.02 mol L-1 KCl solution. The conductivity of 0.1 mol L-1 KCl solution is 1.29x 10-2 
-1cm-1
Given that:
Concentration of the KCl solution = 0.1 mol L-1
Resistance of cell filled with 0.1 mol L-1 KCl solution = 100 ohm
Cell constant = G* = conductivity x resistance
1.29x10-2 ohm-1 cm-1 x 100 ohm = 1.29 cm-1 = 129 m-1
Cell constant for a particular conductivity cell is a constant.
Conductivity of 0.02 mol L-1 KCl solution = =0.248 Sm-1
Concentration = 0.02 mol-1
= 1000x 0.02 mol m-3 = 20 mol m-3
Now,
Molar conductivity =
State Faraday's first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+ to Cu.
Faraday's first law of electrolysis states that 'the amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolytic solution or melt'.
The reduction of one mol of Cu2+ to Cu can be represented as:
Cu2++ 2e-----> Cu
Since, in this reaction, there are two moles of electrons involved, so the amount of charge required is 2F.
Calculate emf of the following cell at 298 K: Mg(s) | Mg2+(0.1 M) || Cu2+ (0.01) | Cu(s)
[Given E0 cell = +2.71 V, 1 F = 96500 C mol-1]
The cell reaction can be represented as:
Mg(s) + Cu2+(aq.) ---> Mg+(aq.) + Cu(s)
Given:
=+2.71 V
T = 298 K
According to the Nernst equation:
=2.71-0.0295 log 10 = 2.71-0.0295
=2.6805 V
a) What type of a battery is the lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it.
(b) In the button cell, widely used in watches, the following reaction takes place
Determine E° and G° for the reaction 
A lead storage battery has a secondary cell. Thus, it can be recharged by passing direct current through it. Therefore, it can be reused. It is used in automobiles.
In a lead storage cell, the anode is made of spongy lead and the cathode is a grid of lead packed with lead dioxide. The electrolyte used is H2SO4.
(a) Define molar conductivity of a solution and explain how molar conductivity changes with a change in concentration of solution for a weak and a strong electrolyte.
(b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 
1500 . What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146x10-3 S cm-1?
Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.
Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.
Variation of molar conductivities with dilution:
For strong electrolytes, molar conductivity slowly increases with dilution.
For weak electrolytes, molar conductivity increases steeply on dilution, especially near lower concentrations. The variation of with for strong and weak electrolytes is shown in the following plot:
b) Given,
Conductivity, k = 0.146 x 10-3 S cm-1
Resistance, R = 1500 ohm
The molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length, the conductivity of the solution is given by the following relation.
The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere.
The process of corrosion is a redox reaction and involves simultaneous oxidation and reduction reactions. It can, therefore, be referred to as an electrochemical reaction.
In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is can be written as follows.
Anode reaction: 
Electrons released at the anodic spot move through the metallic object and go to another spot of the object. There, in the presence of H+ ions, the electrons reduce molecular oxygen. This spot behaves as the cathode. These H+ ions come either from H2CO3, which are formed due to the dissolution of carbon dioxide from the air into water or from the dissolution of other acidic oxides from the atmosphere in water.
The reaction corresponding at the cathode is written as follows.
Cathode reaction:
Also, ferrous ions are further oxidised by atmospheric oxygen to ferric ions.
These ferric ions combine with moisture, present in the surroundings, to form a hydrated ferric oxide (Fe2O3, x H2O) i.e., rust.
How much charge is required for the reduction of 1 mol of Zn2+ to Zn?
Zn2+ + 2e- ---> Zn
Number of electrons involved = 2
Charge required for the reduction of Zn2+ = 2F
We know
1F = 96,487 C
Thus,
2F = 2 x 96487 = 1,92,974 C
Define rate of reaction? Write two factors that affect the rate of reaction.
Rate of Reaction: rate of reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed:
Factors Affecting Rate of Reaction
1. Concentration of reactants
2. Temperature
3. Catalyst
The conductivity of 0.20 mol L-1 solution of KCl is 2.48 x 10-2 S cm-1. Calculate its molar conductivity and degree of dissociation (K+) = 73.56 S cm2 mol-1 and (Cl-)= 76.5 S
(b) What type of battery is mercury cell? Why is it more advantageous than dry cell?
Conductivity of KCl solution = 2.48 x 10-2 S cm-1
Concentration of KCl solution = 0.20 mol L-1
= 0.20 x 1000 mol cm -3
= 200 mol cm-3
Molar conductivity
b) Mercury cell is a type of primary battery. In primary batteries , the charging reaction occurs only once and after it has been used over a period of time , the battery becomes dead and cannot be refused mercury cell is more advantage than dry cell has a very short life span due to the conversion of zinc to zinc chloride that makes the zinc casing porous . Due to this porous casing, the substance inside the cell leaks out and corrodes the metal, reducing the lifetime of the cell. While, in the case of mercury cell, the overall reaction does not involve the formation of any ion in the solution whose concentration can change during its life time.
The standard electrode potential (E°) for Daniell cell is +1·1 V. Calculate the G° for the reaction
Zn(s) + Cu2+ (aq) ----> Zn+ (aq) + Cu (s)
(1 F = 96500 C mol-1).
E° for Daniel cell = 1.1 V
Zn(s) + Cu2+ (aq)---> Zn2+ (aq) + Cu(s)
1F = 96500 C mol-1
G° =?
n = 2 (no. of e-s exchanged)
Since, G° = -nFE°
Therefore, G° = -2 x 96500 x 1.1
G° = -212300 J mol-1
G° = -212.3 kJ mol-1
Express the relation among cell constant, the resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity?
The conductivity (k) of the solution in a cell is the reciprocal of its resistivity.
The quantity 1/2 is cell constant.
L --> Distance between 2 electrodes
a --> Area of cross section
R --> Resistance
The electrical resistance of a column of 0.05 M NaOH solution of diameter 1cm and length 50
cm is 5.55 x 103 ohm. Calculate its resistivity, conductivity and molar conductivity.
A = πr2 = 3.14 x (0.5)2 cm2 = 0.785 cm2
l = 50 cm
From the given cells:
Lead storage cell, Mercury cell, Fuel cell and Dry cell Ans the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii)Which cell is used in automobiles and inverters?
(iv)Which cell does not have long life?
(i) Mercury cell is used in hearing aids.
(ii)Fuel cell was used in the Apollo space programme.
(iii)Lead storage cell is used in automobiles and inverters.
(iv) Dry cell does not have a long life.
a)
b) A is better for coating the surface of the iron because its E0 value is more negative.
Or
a) 
b) A device used for the production of electricity from energy released during the spontaneous chemical reaction and the use of electrical energy to bring about a chemical change. The reaction gets reversed / It starts acting as an electrolytic cell & vice – verse.
(a) What type of a battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery.
(b) Calculate the potential for half-cell containing 0.10 M K2Cr2O7 (aq), 0.20 M Cr3+(aq) and 1.0 x 10-4 M H+ (aq)
The half-cell reaction is 
And the standard electrode potential is given as E0 = 1.33 V.
OR
(a) How many moles of mercury will be produced by electrolysing 1.0 M?
Hg (NO3)2 solution with a current of 2.00 A for 3 hours?
[Hg (NO3)2 = 200.6 g mol-1]
(b) A voltaic cell is set up at 25°C with the following half-cells Al3+ (0.001 M) and Ni2+ (0.50 M). Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(a) A lead storage battery is a secondary battery.
The following chemical equations take place in a lead storage battery.
: 
When a battery is charged, the reverse of all these reactions takes place.
Hence, on charging, PbSO4(s) present at the anode and cathode is converted into Pb(s) and PbO2(s) respectively.
b) 
Or
(a) Quantity of electricity passed = (2A) x (3 x 60 x 60s) = 21600 C
Thus, 2F i.e. 2 x 96500 C deposit Hg = 1 mole 21600 C will deposit Hg
= 0.11 mole
Or
At anode: Al (s) --> Al3+ (aq) + 3e-] x2
At cathode: Ni2+ + 2e- --> Ni(s) ] x3
Cell reaction: 2Al(s) + 3Ni2+(aq) ---> 2Al3+(aq) + 3Ni (s)
Applying nernst equation to the above cell reaction
State Kohlrausch's law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?
Kohlrausch's law of independent migration of ions: It states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of its anion and cation.
A conductivity of a solution decreases with dilution because it leads to decrease in a number of ions per unit volume.
(a) Calculate 
for the reaction
Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s)
Given : E°cell = + 2.71 V, 1 F = 96500 C mol−1
(b) Name the type of cell which was used in Apollo space programme for providing electrical power.
(a) For the cell reaction,
Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s) E°cell = + 2·71 V
The change in the standard Gibbs free energy is given as:
= −nFE°cel. = −2 x 96500 x 2.71 = -523030 Jmol−1
b) The fuel cell which uses the reaction of hydrogen with oxygen to form water was used in Apollo space programme for providing electrical power.
Calculate the degree of dissociation (a) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2mol–1. Given λo(H+) = 349.6 S cm2 mol–1 and λo(CH3COO–) = 40.9 S cm2 mol–1
The degree of dissociation is given by,
α = Λm / λo
here, α is the degree of dissociation,
Λm is the molar conductivity
λo is the molar conductivity at infinite dilution.
We have given
λo(H+) = 349.6 S cm2mol-1
and
λ0(CH3COO-) = 40.9 S cm2mol-1
then,
λoCH3COOH = λo CH3COO-+ λoH+
λoCH3COOH = 349.6 + 40.9 = 390.5
Now, degree of dissociation (α),
α = Λm /Λo
= 39.05 / 390.5 = 0.1
Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3 for 15 minutes.
(Given : Molar mass of Ag = 108 g mol–1 1F = 96500 C mol–1)
Given,
Molar mass of Ag = 108 g mol–1
1F = 96500 C mol–1
Q=It
Q= 2×15×60=1800 C
W(weight of substance deposited)= ZQ
Z= M/nF
n-factor here is 1
So Z = 108/1×96500
=2.01 g
The mass of Ag deposited at the cathode is2.01 g
Define fuel cell.
A fuel cell is a device that produces electrical energy through a chemical reaction between a source fuel and an oxidant.
Define fuel cell and write its two advantages.
These are voltaic cells in which the reactants are continuously supplied to the electrodes. There are designed to convert the energy from the combustion of a fuel like a hydrogen, methane, methanol, etc directly into electrical energy.
Advantage : -
(1) A fuel cell works with an efficiency of 60 to 70 %
(2) They are pollution free
Galvanization is applying a coating of:
Cr
Cu
Zn
Pb
C.
Zn
Zinc metal is the most stable metal to cover iron surfaces. The process of coating the iron surface by zinc is called galvanization.
Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is: (at. mass of Cu = 63.5 amu)
0 g
63.5 g
2 g
127 g
B.
63.5 g
Atomic mass of Cu = 63.5 u
Valency of the metal Z= 2
We have,
CuSO4 → Cu2+ + SO42-
Cu2+ + 2e- → Cu
Given below are the half-cell reactions
Mn2+ + 2e- → Mn; Eo = - 1.18 eV
2(Mn3+ + e- →Mn2+); Eo = +1.51 eV
The Eo for 3Mn2+ → Mn + 2Mn3+ will be
-2.69 V; the reaction will not occur
-2.69 V; the reaction will occure
-0.33 V; the reaction will not occur
-0.33 V; the reaction will occur
A.
-2.69 V; the reaction will not occur
Standard element potential of reaction [ Eo] can be calculated as
Eocell = ER-EP
where ER = SRP of reactant
EP = SRP of product
If Eocell = +ve, then the reaction is spontaneous otherwise non-spontaneous.
therefore, For Mn2+ disproportionation
Eo = - 1.51 V - 1.18 V = - 2.69 V < 0
Given,
Based on the data given above, strongest oxidising agent will be
Cl
Cr3+
Mn2+
MnO4-
D.
MnO4-
Higher the SRP, better is an oxidising agent, among the given 
is highest, hence, MnO4- is a strongest oxidising agent.
How many litres of water must be added to 1 L to an aqueous solution of HCl with a pH of 1 create an aqueous solution with PH of 2?
0.1 L
0.9 L
2.0 L
9.0 L
D.
9.0 L
Initial pH = 1, i.e. [H+] = 0.1 mole/litre
New pH = 2, i.e. [H+] = 0.01 mole/litre
In case of dilution: M1V1 = M2V2
0.1 ×1 =0.01 × V2
V2 = 10 litre.
A volume of water added = 9 litres.
The reduction potential of hydrogen half-cell will be negative if
p(H2) = 1 atm and [H+] = 2.0 M
p(H2) = 1 atm and [H+] = 1.0 M
p(H2) = 2 atm and [H+] = 1.0 M
p(H2) = 2 atm and [H+] = 2.0 M
C.
p(H2) = 2 atm and [H+] = 1.0 M
Given,
Among the following, the strongest reducing agent is
Cr
Mn2+
Cr3+
Cl-
A.
Cr
Cr+3 is having least reducing potential, therefore Cr is the best Reducing agent.
On the basis of the following thermochemical data: (∆f G° H(aq)+=0)
H2O(l) → H+(aq) + OH–(aq); ∆H =57.32kJ
H2(g) + 1/2O2(g) → H2O(l); ∆H = –286.20 kJ
The value of enthalpy of formation of OH–ion at 25°C is
–22.88 kJ
–228.88 kJ
+228.88 kJ
–343.52 kJ
B.
–228.88 kJ
By adding the two given equations, we have
Aluminium oxide may be electrolysed at 1000o C to furnish aluminium metal (Atomic mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is
Al3+ +3e- → Alo
To prepare 5.12 kg of aluminium metal by this method would require
5.49 x 107 C of electricity
1.83 x 107 C of electricity
1.49x 107 C of electricity
5.49 x 101 C of electricity
A.
5.49 x 107 C of electricity
Calomel (Hg2Cl2) on reaction with ammonium hydroxide gives
HgNH2Cl
NH2 – Hg – Hg – Cl
Hg2O
HgO
A.
HgNH2Cl
Hg2Cl2 + 2NH4OH → Hg + Hg(NH2)Cl + NH4Cl + 2H2O
Which among the following factors is the most important in making fluorine the strongest oxidizing halogen?
Electron affinity
Bond dissociation energy
Hydration enthalpy
Ionization enthalpy
B.
Bond dissociation energy
Oxidising power depends on the (+ve) reduction potential value, and (-ve) value of Gibbs free energy.
F2 has the most negative ΔG° value which dependent on hydration enthalpy.
In hydrogen-oxygen fuel cell, combustion of hydrogen occurs to
generate heat
remove adsorbed oxygen from electrode surfaces
produce high purity water
create potential difference between the two electrodes
D.
create potential difference between the two electrodes
Any cell (such as fuel cell), works when a potential difference is developed.
The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at 25°C. The equilibrium constant of the reaction is (F = 96,500 C mol-1: R = 8.314 JK-1 mol-1)
1.0×101
1.0×1030
1.0×1010
1.0×1010
D.
1.0×1010
Relation between Keq and Ecell is
The limiting molar conductivities Λ° for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol-1 respectively. The Λ° for NaBr is
128 S cm2 mol-1
302 S cm2 mol-1
278 S cm2 mol-1
176 S cm2 mol-1
A.
128 S cm2 mol-1
By Kohlrausch's law
In a cell that utilises the reaction Zn(s) + 2H+ (aq) → Zn2+(aq) + H2(g) addition of H2SO4 to cathode compartment, will
lower the E and shift equilibrium to the left
increases the E and shift equilibrium to the left
increase the E and shift equilibrium to the right
Lower the E and shift equilibrium to the right
C.
increase the E and shift equilibrium to the right
The 
values for Cr, Mn, Fe and Co are – 0.41, +1.57, + 0.77 and +1.97 V respectively. For which one of these metals the change in oxidation state form +2 to +3 is easiest?
Cr
Co
Fe
Mn
A.
Cr
indicates better reducing agent thus easily oxidised. Thus, oxidation of Cr2+ to Cr3+ is the easiest
How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u)
1.6 hours
6.4 hours
0.8 hours
3.2 hours
D.
3.2 hours
According to the balanced equation:
27.66 g B2H6 i.e. 1 mole B2H6 requires 3 moles of O2. Now, this oxygen is produced by electrolysis of water.
1 mole O2 is produced by 4 F charge
therefore, 3 mole O2 will be produced by 12 F charge
hence, Now applying
Q = It
12 x 96500 C = 100 x t (s)
t = 3.2
The pressure of H2 required to make the potential of H2-electrode zero in pure water at 298 K is
10-12 atm
10-10 atm
10-4 atm
10-14 atm
D.
10-14 atm
From the question, we have an equation
2H+ +2e- --> H2(g)
According to Nernst equation
A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as
fuel cell
electrolytic cell
dynamo
Ni-Cd cell
A.
fuel cell
A fuel cell is a device that converts the energy of combustion of fuels like hydrogen and methane, directly into electrical energy.
Electrolytic cell converts electrical energy into chemical energy. Dynamo is an electrical generator that produces direct current with the use of a commutator.
Ni-Cd cell is the type of rechargeable battery which consists of a cadmium anode and a metal grid containing NiO2 acting as a cathode.
When 0.1 mol MnO42- is oxidised, the quantity of electricity required to completely oxidise MnO42- to MnO42- is
95600C
2 x 96500C
9650 C
96.50 C
C.
9650 C
Using the Gibbs energy change, ΔG0 = +63.3 KJ for the following reaction,
Ag2CO3 (s) r2Ag+ (aq) + CO32- (aq)
the Ksp of Ag2CO3 (s) in water at 250 C is (R= 8.314 JK-1 mol-1)
3.2 x 10-26
8.0 x 10-12
2.9 x 10-3
7.9 x 10-2
B.
8.0 x 10-12
ΔG0 is related to Ksp by the equation,At 250 molar conductance of 0.1 molar aqueous solutions of ammonium hydroxide is 9.54 ohm-1 cm2 mol-1 and at infinite dilution, its molar conductance is 238 ohm-1 cm2 mol-1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is
2.080%
20.800%
4.008%
40.800%
C.
4.008%
Given molar conductance at 0.1 M concentration,
λc = 9.54 Ω-1 cm2 mol-1
Molar conductance at infinite dilution,
A button cell used in watches functions as following
The cell potential will be
1.10 V
0.42 V
0.84 V
1.34 V
A.
1.10 V
Anode is always the site of oxidation thus anode half cell is
Zn2+ (aq) +2e- --> Zn (s); E0 =-0.76 V
Cathode half cell is
Ag2O (s) +H2O (l) +2e- ---> 2Ag(s) +2OH- (aq); E0 =0.34 V
E0cell = E0cathode -E0anode
= 0.34 -(-0.76) = +1.10 V
Limiting molar conductivity of NH4OH (i.e., 
is equal to
D.
According to Kohlrausch's law, limiting molar conductivity of NH4OH
Standard reduction potentials of the half-reactions are given below.
F2 (g) +2e- → 2F- (aq) ; Eo = +2.85 V
Cl2 (g) +2e- →2Cl- (aq) ; Eo = +1.36V
Br2 (l) +2e- → 2Br- (aq) ; Eo = +1.06 V
I2 (s) +2e- →2I- (aq); Eo = +0.53 V
The strongest oxidising and reducing agents respectively are
F2 and I-
Br2 and Cl-
Cl2 and Br-
Cl2 and I2
A.
F2 and I-
Higher the value of standard reduction potential, stronger will be the oxidising agent. Therefore, F2 will act as stronger oxidising agent.
Similarly, lower the value of standard reduction potential stronger will be the reducing agent. Therefore, I- will act as strongest reducing agent.
Molar conductivities (Λom) at infinite dilution of NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 S Cm2 mol-1 respectively. Λom for CH3COOH will be
425.5 S cm2 mol-1
180.5 S cm2 mol-1
290.8 S cm2 mol-1
390.5 S cm2 mol-1
D.
390.5 S cm2 mol-1
CH3COONa +HCl → NaCl + CH3COOH
91 + 425.9 = 126.4 + x
x= 516.9 -126.4
= 390.5 S cm2 mol-1
The Gibb's energy for the decomposition of Al2O3 at 500o C is as follow
2/3 Al2O3 → 4/3 Al + O2;
ΔrG = +960 kJ mol-1
The potential difference needed for the electrolytic reduction aluminium oxide (Al2O3) at 5000 C is at least
4.5 V
3.0 V
2.5 V
5.0 V
C.
2.5 V
at anode
2O2- +4e- →O2] x 3
at cathode
Al3+ → Al +3e-] x4
Net reaction
4Al +6O2- → 3O2 +4Al
Or
4/3Al +2O2- → O2 +4/3Al
n= 12/3 = 4
ΔGo = -nFEo
ΔGo = +960 kJ mol-1
= 960 x 1000 J mol-1
n=4
F =96500 Coulomb
Standard electrode potential of three metal X, Y and Z are -1.2 V, +0.5 V and -3.0 V respectively. The reducing power of these metals will be
Y > X > Z
Z> X> Y
X > Y > Z
Y > Z > X
B.
Z> X> Y
Exo = -1.2 V;
If the Eocell for a given reaction has a negative value then which of the following gives the correct relationships for the values of ΔGo and Keq ?
ΔGo < 0; Keq > 1
ΔGo < 0; Keq < 1
ΔGo > 0; Keq < 1
ΔGo > 0; Keq > 1
C.
ΔGo > 0; Keq < 1
ΔGo = - nFEo
When Eo is negative, then ΔGo > 0
ΔGo = - RT ln Keq
When ΔGo >0, Keq = 10-x which is less than ones ,ie, Keq < 1.
A solution contains Fe2+ Fe3+, and I- ions. This solution was treated with iodine at 35o C.Eo for Fe3+/ Fe2+ is +0.77 V and Eo for I2/ 2 I- = 0.536 V. The favourable redox reaction is
I2 will be reduced to I-
There will be No redox reaction
I- will be oxidised to I2
Fe2+ will be oxidised to Fe3+
C.
I- will be oxidised to I2
For the reaction of silver ions with copper metal, the standard cell potential was found to be +0.46 V at 25o C. The value of standard Gibbs energy, ΔGo will be (F = 96500 C mol-1 )
-89.0 kJ
-89.0 J
-44.5 kJ
-98.0 kJ
A.
-89.0 kJ
We know that,
standard Gibbs energy, ΔGo = -nFEocell
For the cell reaction,
An increase in equivalent conductance of strong electrolyte with dilution is mainly due to
the increase in ionic mobility of ions
100% ionisation of electrolyte at normal dilution
the increase in both, ie, the number of ions and ionic mobility of ions
the increase in the number of ions
A.
the increase in ionic mobility of ions
Which of the following expression correctly represents the equivalent conductance at infinite dilution of Al2(SO4)3. Given that 
are the equivalent conductance at infinite dilution of the respective ions.
B.
Al2(SO4)3 ⇌ 2 Al3+ + 3SO42-
Since equivalent conductances are given only for ions, the equivalent conductance at infinite dilution,
For vaporisation of water at 1 atm pressure, the values of ΔH and ΔS are 40.63 kJ mol- and 108.8 JK-1 mol-1, respectively. The temperature when Gibbs energy change (ΔG) for this transformation will be zero, is
273.14 K
393.4 K
373.4 K
293.4 K
C.
373.4 K
ΔG = ΔH - TΔS
ΔG = 0 (given)
ΔH = TΔS,
T = 40.63 x 103 / 108.8 = 373.4 K
Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.5 x 104 An of current is passed through molten Al2O3 for 6h, what mass of aluminium is produced? (Assume 100% current efficiency, at.mass of Al = 27 g mol-1)
9.0 x 103 g
8.1 x 104 g
2.4 x 105 g
1.3 x 104 g
B.
8.1 x 104 g
Al2O3 ionises as,
The equivalent conductance of M/32 solution of weak monobasic acid is 8.0 ohm-cm2 and at infinite dilution is 400 ohm-cm2. The dissociation constant of this acid is
1.25 x 10-5
1.25 x 10-6
6.25 x 10-4
1.25 x 10-4
A.
1.25 x 10-5
Degree of dissociation,
Where, 
and 
are equivalent conductances at a given concentration and at infinite dilution respectively.
From Ostwald's dilution law (for weak monobasic acid)
Given,
(i) Cu2+ + 2e- → Cu, Eo = 0.337 V
(ii) Cu2+ +e- → Cu+, Eo = 0.153
Electrode potential, Eo for the reaction,
Cu +e- →Cu, will be
0.52 V
0.90 V
0.30 V
0.38 V
A.
0.52 V
Gibb's free energy is an additive property.
ΔGo = -nFEo
For reaction, Cu2+ +2e- → Cu;
ΔGo = 2 x F x 0.337 ... (i)
For reaction, Cu+ → Cu2+ +e-;
ΔGo = +1 xF x 0.153
Adding Eqs. (i) and (ii), we get
Cu+ + e- → Cu; ΔGo = - -0.521 F
ΔGo = - nFEo
-0.521 F = -nFEo
Eo = 0.52 V
The values of ΔH and ΔSfor the reaction, C(graphite) + CO2 → 2 CO (g) are 170 kJ and 170 JK-1 respectively. This reaction will be spontaneous at
710 K
910 K
1110 K
510 K
C.
1110 K
For spontaneous process, ΔG < 0
ΔG = ΔH - TΔS
Given , ΔH = 170 KJ = 170 x 103 J
ΔS = 170 JK-1
T = ?
ΔG = ΔH - TΔS
Kohlrausch's law states that at
finite dilution, each ion makes a definite contribution to the equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.
infinite dilution, each ion makes a definite contribution to the equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte.
infinite dilution, each ion makes a definite contribution to the conductance of an electrolyte whatever be the nature of the other ions of the electrolyte.
infinite dilution, each ion makes a definite contribution to the equivalent conductance of an electrolyte. whatever be the nature of the other of the electrolyte.
D.
infinite dilution, each ion makes a definite contribution to the equivalent conductance of an electrolyte. whatever be the nature of the other of the electrolyte.
According to Kohlrausch's law 'at infinite dilution when the dissociation is complete, each ion makes a definite contribution towards equivalent condctivity of the electrolyte irrespective of the nature of the order ion with which it is associated.
The equilibrium constant of the reaction:
Cu (s) + 2 Ag+ (aq) → Cu2+ (aq) + 2 Ag (s);
Eo = 0.46 V at 298 K
2.4 x 1010
2.0 x 1010
4.0 x 1010
4.0 x 1015
D.
4.0 x 1015
Cu (s) + 2 Ag+ (aq) → Cu2+ (aq) + 2 Ag (s)
Eo = 0.46 V at 298 K
If 
the standard emf of the reaction:
Fe + 2 Fe3+ →3Fe2+
will be:
0.330 V
1.653 V
0.111 V
C.
Given that,
So, on the basis of cell reaction following half-cell reactions are written
At anode:
Fe → Fe2+ + 2e- (oxidation)
At cathode:
2Fe3+ + 2e- →2Fe2+ (reduction)
So,
A hypothetical electrochemical cell is shown below
A- | A+ (xM)|| B+ (yM)|B+
The emf measured is +0.20 V. The cell reaction is:
A+ + B → A + B+
A+ + e- → A ; B+ + e- → B-
the cell reaction cannot be predicted
A+ + B → A+ + B
D.
A+ + B → A+ + B
Electrochemical cell
A0 | A+ (xM) ||B+ yM) B+
The emf of cell is +.20 V. so cell reaction is possible. The half cell reaction are given as follows:
In the electrochemical cell :
Zn|ZnSO4(0.01M)||CuSO4(1.0 M)|Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E2. From the following, which one is the relationship between E1 and E2?
(Given, RT/F= 0.059)
E1= E2
E1< E2
E1> E2
E2= 0 ≠ E1
C.
E1> E2
Zn|ZnSO4(0.01 M)||CuSO4(1.0 M)|Cu
When a lead storage battery is discharged;
SO2 is evolved
lead sulphate is consumed
lead is formed
sulphuric acid is consumed
D.
sulphuric acid is consumed
H2SO3 is consumed during discharging of the lead storage battery as Pb changes PbSO4 and PbO2 changes to PbSO4.
The value of reaction quotient [Q], for the following cell
Zn(s)|Zn2+ (0.01 M)|| Ag+ (1.25 M | Ag (s) is
156
125
1.25 x 10-2
6.4 x 10-3
D.
6.4 x 10-3
The cell reaction is
Zn(s) → Zn2+ (0.01 M) + 2e-
[Ag+ (1.25 M) + e- → Ag (s)] x 2
________________________________________
Zn(s) + 2Ag+ (1.25 M) → Zn2+ (0.01 M) + 2Ag(s)
value of Q = 6.4 x 10-3
The standard reduction potential for Zn2+/Zn, Ni2+/Ni and Fe2+/Fe are -0.76, -0.23 and -0.44 V, respectively.
The reaction X + Y2+ → X2+ + Y will be spontaneous when
X = Ni, Y = Fe
X = Ni, Y = Zn
X =Fe, Y = Zn
X = Zn, Y = Ni
D.
X = Zn, Y = Ni
The reaction is spontaneous when ΔG < 0 and E°cell > 0.
Sponsor Area
Sponsor Area
G° for the following reaction: 
value?
The value of standard electrode potential for the change 
will be
will be 
