Electrochemistry

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Question
CBSEENCH12010376

Calculate e.m.f of the following cell at 298 K:

2 Cr space left parenthesis straight s right parenthesis space plus 3 Fe to the power of 2 plus end exponent space left parenthesis 0.1 space straight M right parenthesis space rightwards arrow space 2 Cr to the power of 3 plus end exponent space left parenthesis 0.01 space straight M right parenthesis space plus 3 Fe left parenthesis straight s right parenthesis space

Given colon space straight E to the power of 0 space left parenthesis Cr to the power of 3 plus end exponent vertical line Cr right parenthesis space equals space minus 0.70 space straight V
space space space space space space space space space space space straight E to the power of 0 space left parenthesis Fe to the power of 2 plus end exponent vertical line Fe right parenthesis space equals negative 0.44 space straight V

Solution

2 Cr space left parenthesis straight s right parenthesis space space plus 3 Fe to the power of 2 plus end exponent space left parenthesis 0.1 straight M right parenthesis space rightwards arrow space 2 Cr cubed left parenthesis 0.01 straight M right parenthesis space plus 3 Fe left parenthesis straight s right parenthesis
Given colon space
straight E subscript Cr to the power of 3 plus end exponent divided by Cr end subscript superscript 0 space equals negative 0.74 straight V

straight E subscript Fe to the power of 2 plus end exponent divided by Fe end subscript superscript 0 space equals space minus 0.44 straight V

The space cell space can space be space represented space as space follows colon

Cr vertical line space Cr to the power of 3 plus end exponent space left parenthesis 0.01 straight M right parenthesis space vertical line vertical line space Fe to the power of 2 plus end exponent left parenthesis 0.1 straight M right parenthesis vertical line space Fe left parenthesis straight s right parenthesis

straight E subscript cel superscript 0 space equals space straight E subscript Fe to the power of 2 plus end exponent divided by Fe end subscript superscript 0 space minus straight E subscript Cr to the power of 3 plus end exponent divided by Cr end subscript superscript 0
straight E subscript cell superscript 0 space equals negative 0.44 minus left parenthesis negative 0.74 right parenthesis space equals space 0.30 space straight V
straight E subscript cell space equals space straight E subscript cell superscript 0 space minus fraction numerator 2.303 space RT over denominator nF end fraction space log space fraction numerator left square bracket Cr to the power of 3 plus end exponent right square bracket squared over denominator left square bracket Fe to the power of 2 plus end exponent right square bracket cubed end fraction
straight E subscript cell space equals space 0.30 space minus fraction numerator 0.0591 over denominator 6 space end fraction log space fraction numerator left parenthesis 0.01 right parenthesis squared over denominator left parenthesis 0.1 right parenthesis cubed end fraction
straight E subscript cell space equals space 0.30 space plus 0.01
straight E subscript cell space equals space 0.31 space straight V

Sponsor Area

Question
CBSEENCH12010411

straight a right parenthesis space space Calculate space straight E subscript cell superscript 0 space for space the space following space reaction space at space 298 space straight K colon
2 Al left parenthesis straight s right parenthesis space plus 3 Cu to the power of 2 plus end exponent left parenthesis 0.01 space straight M right parenthesis space rightwards arrow 2 Al to the power of 3 plus end exponent space left parenthesis 0.01 straight M right parenthesis space plus 3 Cu left parenthesis straight s right parenthesis
Given space colon space straight E subscript cell space equals space 1.98 space straight V
straight b right parenthesis space using space the space straight E to the power of 0 space values space of space straight A space and space straight B space predict space which space is space better space for space coating space the space surface space of space iron space left square bracket straight E to the power of 0 left parenthesis Fe to the power of 2 plus end exponent divided by Fe right parenthesis
equals negative 0.44 straight V right square bracket space to space prevent space corrosion space and space why ?
Given space colon space straight E to the power of 0 left parenthesis straight A to the power of 2 plus end exponent divided by straight A right parenthesis space equals space minus 2.37 space colon space straight E to the power of 0 left parenthesis straight B to the power of 2 plus end exponent divided by straight B right parenthesis space equals negative 0.14 space straight V

OR
straight a right parenthesis space The space conductivity space of space 0.001 space mol space straight L to the power of negative 1 end exponent space solution space of space CH subscript 3 COOH space space is space 3.905 space straight x space 10 to the power of negative 5 end exponent space straight S space cm to the power of negative 1 end exponent. space
Calculate space its space molar space conductivty space and space degree space of space dissociation space left parenthesis straight alpha right parenthesis

Given space straight lambda to the power of 0 space left parenthesis straight H to the power of plus right parenthesis space equals 349.6 space straight S space Cm squared space mol to the power of negative 1 end exponent space and space straight lambda to the power of 0 space left parenthesis CH subscript 3 COO to the power of minus right parenthesis space equals space 40.9 space straight S space cm squared space mol to the power of negative 1 end exponent

straight b right parenthesis space Define space electrochemical space cell space what space happen space if space external space potential space applied space
becomes space greater space than space straight E subscript cell superscript 0 space of space electrochemical space cell.

Solution

a)
straight E subscript cell space equals space straight E subscript cell superscript 0 space minus space fraction numerator 0.0591 over denominator straight n end fraction space log space fraction numerator left square bracket Al to the power of 3 plus end exponent right square bracket squared over denominator left square bracket Cu to the power of 2 plus end exponent right square bracket cubed end fraction

straight E subscript cell superscript 0 space equals 1.98 plus fraction numerator 0.0591 over denominator 6 end fraction log space fraction numerator left parenthesis 0.01 right parenthesis squared over denominator left parenthesis 0.01 right parenthesis cubed end fraction

straight E subscript cell superscript 0 space equals space 1.98 space plus fraction numerator 0.0591 over denominator 6 end fraction log space 10 squared

straight E subscript cell superscript 0 space equals space 1.98 space plus fraction numerator 0.0591 over denominator 6 end fraction space straight x space 2 space straight x space log space 10
straight E subscript cell superscript 0 space equals space 1.98 space plus fraction numerator 0.0591 over denominator 6 end fraction space straight x space 2
straight E subscript cell superscript 0 space equals space 1.98 space straight V space plus 0.0197 straight V
straight E subscript cell superscript 0 space equals 1.9997 space straight V
b) A is better for coating the surface of the iron because its E0  value is more negative.

Or
a)  straight capital lambda subscript straight m space equals fraction numerator space straight k space straight x space 1000 over denominator straight C end fraction
space equals fraction numerator 3.905 space straight x space 10 to the power of negative 5 end exponent space straight x space 1000 over denominator 0.001 end fraction
equals 39.05 space straight S space cm squared divided by mol

CH subscript 3 COOH space rightwards arrow CH subscript 3 COO to the power of minus space plus straight H to the power of plus
straight capital lambda to the power of 0 space CH subscript 3 COOH space equals space straight lambda to the power of 0 CH subscript 3 COO to the power of minus space plus space straight lambda to the power of 0 straight H to the power of plus

equals space 409 plus 349.6
straight capital lambda to the power of 0 CH subscript 3 COOH space equals 390.5 space straight S space cm squared divided by mol
straight alpha space equals space fraction numerator straight capital lambda subscript straight m over denominator straight capital lambda subscript straight m superscript 0 end fraction
equals 39.05 divided by 390.5 equals 0.1
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.

Question
CBSEENCH12010440

(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  Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript left parenthesis aq right parenthesis straight space plus straight space 14 straight space straight H to the power of plus subscript left parenthesis aq right parenthesis end subscript straight space plus straight space 6 straight e to the power of minus straight space rightwards arrow straight space 2 Cr subscript left parenthesis aq right parenthesis end subscript superscript 3 plus end superscript straight space plus straight space 7 straight H subscript 2 straight O subscript left parenthesis straight l right parenthesis end subscript

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.

  Error converting from MathML to accessible text.

Solution

(a) A lead storage battery is a secondary battery.

The following chemical equations take place in a lead storage battery.

At space anode colon space Pb subscript left parenthesis straight s right parenthesis end subscript space plus space SO subscript 4 superscript 2 minus end superscript subscript left parenthesis aq right parenthesis end subscript space rightwards arrow space PbSO subscript 4 left parenthesis straight s right parenthesis space plus 2 straight e to the power of minus

At space cathode colon space PbO subscript 2 space left parenthesis straight s right parenthesis end subscript space plus space SO subscript 4 superscript 2 minus end superscript subscript left parenthesis aq right parenthesis end subscript space plus space 4 straight H to the power of plus space plus 2 straight e to the power of minus space rightwards arrow space PbSO subscript 4 subscript space left parenthesis straight s right parenthesis end subscript space plus space 2 straight H subscript 2 straight O subscript left parenthesis straight l right parenthesis end subscript
The space overall space cell space reaction space is space given space by comma

Pb subscript left parenthesis straight s right parenthesis end subscript space plus space PbO subscript 2 space left parenthesis straight s right parenthesis end subscript space plus space 2 straight H subscript 2 SO subscript 4 space rightwards arrow space 2 PbSO subscript 4 subscript space left parenthesis straight s right parenthesis end subscript space plus space space 2 straight H subscript 2 straight O subscript left parenthesis straight l right parenthesis end subscript
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)  Error converting from MathML to accessible text.
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

  fraction numerator 1 over denominator 2 cross times 96500 end fraction cross times 21600

= 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 
Error converting from MathML to accessible text.

Question
CBSEENCH12010452

State Kohlrausch's law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?

Solution

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.
straight lambda subscript infinity space equals space straight lambda subscript straight a space plus straight lambda subscript straight c
where space straight lambda subscript straight a space and space straight lambda subscript straight c space are space anion space and space cation
A conductivity of a solution decreases with dilution because it leads to decrease in a number of ions per unit volume.