Chemistry Part Ii Chapter 8 Redox Reactions

(i) Extraction of metals from ores.
(ii) Electroplating of articles.

Chlorine loses as well as gain electrons in the given reaction.

Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to the more electronegative element.

.

Variable oxidation states of nitrogen:
NH₃ = (-3)
NH₂NH₂ =(-2)
NH₂OH =(-1)
N₂ = 0
N₂O =(+1)
NO =(+2)
N₂O₃ =(+3)
NO₂ =(+4)
N₂O₅ =(+5)

Oxidation number of CO =0
Therefore,
The oxidation Number of Cr in Cr(CO)_{6 .}
Cr(CO)₆
x + 6(0) = 0
or  
x = 0

Let the oxidation state of Os= x
Oxidation state of O = -2
Therefore,
The oxidation state of osmium (Os) in OsO_4.

x +(-2) x4 =0
x-8=0
x=8

Reduction (ii) is not a redox reaction. Since in the given reaction, no element undergoes a change in oxidation state.
+ 2    + 4     -2             + 2      -2        +4        -2
Ca      C        O₃        Ca       O        + C        O₂

The correct stoichiometric coefficient of  are 2, 5, 16
the balanced equation is,

                   

Let x be the oxidation number of Fe.
Oxidation state of oxygen = -2
Thus, 
3x + 4(-2) = 0
or
3x - 8 = 0
x = 8/3

(i) It acts as an inert metal connection to H₂/H⁺ system.
(ii) It allows H₂ gas to be absorbed onto its surface.

Here, Al reduces Fe₂O₃ to Fe while itself gets oxidised to Al₂O₃. Conversely, Fe₂O₃ oxidises Al to Al₂O₃ while itself gets reduced to Fe. Therefore Al acts as a reducing agent while Fe₂O₃ acts as an oxidising agent.

In reaction (i) zinc donates electrons to O to give zinc and oxide ions. Therefore, Zn acts as a reductant while oxygen acts as an oxidant.
In reaction (ii) Zn transfers its electrons to H⁺ and therefore, zinc acts as a reductant and H⁺ acts as an oxidant.

(i) H₂S is oxidised because a more electronegative element, chlorine is added to hydrogen or a more electropositive hydrogen has been removed from S and chlorine is reduced because of the addition of hydrogen to it. 
(ii) Aluminium is oxidised due to the addition of oxygen to it and ferrous ferric oxide (Fe₃O₄) is reduced due to the removal of oxygen from it.
(iii) Sodium is oxidised while hydrogen is reduced.

The redox reactions have been classified into two types:
(i) Direct redox reactions: In these reactions, oxidation and reduction take place in the same vessel. For example,
(a) Displacement of copper from CuSO₄ solution when a zinc rod is dipped in it.
(b) Reduction of HgCl₂ to Hg₂Cl₂ by SnCl₂.
(ii) Indirect redox reactions: In these reactions, oxidation and reduction take place in different vessels. These reactions form the basis of the electrochemical cells in which chemical energy is converted into electrical energy.

When a zinc rod is placed in an aqueous solution of copper sulphate, the following changes will be observed:
(i) The zinc plate loses weight gradually.
(ii) A precipitate of copper settles at the bottom of the beaker.
(iii) The blue colour of the solution gradually fades.
(iv) The solution remains electrically neutral throughout.
(v) The solution becomes hot (exothermic reaction).
The overall reaction which takes place in a beaker may be represented as:

Here, Zn is oxidised to Zn²⁺ ions by losing two electrons and Cu²⁺ ions are reduced to Cu(s).
                          

As Cu²⁺ ions from the solution are changing to Cu(s), the blue colour of the solution which is due to Cu²⁺ ions, slowly fades. Also, the number of electrons lost in the oxidation half reaction is equal to the number of electrons gained in the reduction half reaction, the solution remains electrically neutral.
Similarly, when a copper rod is placed in a silver nitrate solution, silver gets precipitated with the evolution of heat energy. The reaction taking place in the beaker is:

Cancelling the common ion, 

Hence whenever a redox reaction is carried out in a single beaker, decrease in chemical energy or free energy appears in the form of heat.

A large number of redox reactions proceed slowly in aqueous solutions. Each redox reaction can be considered as a sum of two half reactions-one involving oxidation called oxidation half reaction and the other involving reduction called reduction half reaction. For example;
(a) Let us consider the oxidation of aqueous potassium iodide by hydrogen peroxide. This reaction can be divided into two half reactions.

Supplying the required number of spectator ions, the balanced redox equation is
        
Note. The ions which do not part in any reaction but are simply added to balance the charge are called spectator ions.
(b) Consider the oxidation of aqueous ferrous sulphate to ferric sulphate by aqueous acidified KMnO₄ solution.


Supplying the required spectator ions, the complete balanced redox equation is

Redox reaction (a):
(i) Oxidation half-reaction.
                 
(ii) Reduction half-reaction

For redox reaction (b):
(i) Oxidation half-reaction
               
(ii) Reduction half-reaction
          
         or 
   

 1.
(i) Na(s) is oxidised to Na⁺(g)
(ii) Cl₂(g) is the oxidising agent.
(iii) Cl₂(s) is reduced to Cl^- (aq)
(iv) Na(s) is the reducing agent.

(i) Mg(s) is oxidised to Mg²⁺ (g)
(ii) Cl₂(g) is the oxidising agent.
(iii) Cl₂(g) is reduced to Cl^-(aq)
(iv) Mg (s) is reducing agent.

(i) Zn(s) is oxidised to Zn²⁺ (aq)
(ii) H⁺ (aq) is the oxidising agent.
(iii) H⁺ (aq) is reduced to H₂(g)
(iv) Zn(i) is the reducing agent.

Oxidation number: The term oxidation number represents the positive or negative character of an atom in a compound. It may be defined as the charge which an atom has or appears to have when present in the combined state with another atom in the formula of a compound or an ion. Oxidation number can be zero, positive, negative or fraction.

Rules for assigning oxidation numbers:
(i) The oxidation number of an element in the free atomic state (Ma, HCl, Fe, Ag) or in its poly-atomic state (P₄, Sg, graphite, H₂, Cl₂, etc.) is always zero.
(ii) The oxidation number of a monatomic ion is the same as the charge on it e.g. oxidation numbers of Na⁺, Mg²⁺ and Al³⁺ are +1, +2 and +3 respectively; oxidation numbers of Cl^-, S^2- and N^3-ions are –1, –2 and –3 respectively.
(iii) In a binary compound, the more electronegative element has negative oxidation number whereas less electronegative element has positive oxidation number
 +1           -1         +3           -1
  Cl            F           Br           Cl₃

(iv) The oxidation number of hydrogen in its compounds is always +1 except in metallic hydrides (e.g. LiH, NaH, MgH₂) where it is -1.
(v) The oxidation number of oxygen in most compounds is -2. However in peroxides like H₂O₂ Na₂O₂, BaO_z etc., the oxidation number of oxygen is -1. In OF₂ the oxidation number is +2 because F is more electronegative than O.
(vi) The oxidation number of F is always -1 in all its compounds.
(vii) The oxidation number of alkali metals i.e. Li, Na, K etc. is always +1 in their compounds and that of alkaline earth metals i.e. Be, Ca, Sr and Ba are always +2 in their compounds.
(viii) The algebraic sum of the oxidation numbers of all atoms in a neutral molecule is zero.
(xi) The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge on the ion.

Oxidation: Oxidation is defined as a chemical reaction in which oxidation number of an element increases.
Reduction: Reduction is defined as a chemical reaction in which oxidation number of an element decreases.
Oxidising agent: Oxidising agent is a species which undergoes a decrease in oxidation number.
Reducing agent: Reducing agent is a species which undergoes an increase in oxidation number.
For example in the reaction:


The oxidation number of S increases from -2 to 0 while the oxidation number of Fe decreases from +3 (in FeCl₃) to +2 (in FeCl₂). Therefore H₂S is oxidised and FeCl₃ is reduced. Thus FeCl₃ acts as an oxidising agent and H₂S acts as a reducing agent.

(i) F (fluorine) exhibits only - ve oxidation state.
(ii) Cs (cesium) exhibits only +ve oxidation state.
(iii) I (iodine) exhibits both +ve and -ve oxidation states.
(iv) (neon) neither exhibits +ve nor - ve oxidation states.

a) 

(i) Oxidising agent H₂SO₄
(ii) Reducing agent HBr
(iii) Substance oxidised HBr
(iv) Substance reduced H₂SO₄

 b)

(i) Oxidising agent CuO
(ii) Reducing agent NH₃
(iii) Substance oxidised NH₃ 
(iv) Substance reduced CuO.

a)    AgBr is reduced to Ag

C₆H₆O₂ is oxidised to C₆H₄O₂

AgBr is an oxidising agent

C₆H₆O₂ is a reducing agent.

b)    [Ag(NH₃)⁺₂ reduced to Ag⁺

HCHO is oxidised to HCOO^-

[Ag(NH₃)⁺₂ is an oxidising agent.

c)   HCHO is oxidised to HCOO^-

Cu²⁺ is reduced to Cu(I)  state.

d)   N₂H₄ is reduced to H₂O

H₂O₂ is an oxidising agent

N₂H₄ is reducing agent.

e)  Pb has been oxidised to PbSO₄

PbO₂ is reduced to PbSO₄

PbO₂ is an oxidising agent

Pb is a reducing agent.

(a) Ag ions are reduced to Ag which is precipitated.
(b) Cu²⁺ (aq) are reduced to Cu which is precipitated.
(c) Ag⁺ (aq) present in the complex are reduced to Ag which gets precipitated as shining mirror.
(d) Cu²⁺ (aq) ions are not reduced by C₆H₅CHO (benzaldehyde) which is a very weak reducing agent.
Therefore from the above reactions, we conclude that Ag⁺ ion is a stronger oxidising agent than Cu²⁺ ions.

(i) Let the oxidation number of Si = x
Writing the oxidation number of each atom at the op of its symbol,
  x               -1
Si                H₄
The algebraic sum of oxidation number of various atoms = 0
x + 4 (-1) = 0
     x  = 4
(ii) Let the oxidation number of B =x
Writing the oxidation number of each atom at the top of its symbol
x               -1
B               H₃
The algebraic sum of the oxidation number of various atoms = 0
        x  + 3(-1) = 0
                    x = 3
(iii) Let the oxidation number of B = x
Writing the oxidation number of each atom at the top of its symbol,
 x      -1 
B        F₃
The algebraic sum of the oxidation number of various atoms = 0
                    x +  3 (-1) = 0
                                x = 3

Redox reactions have been divided into the following classes.
(a) Combination reactions. In such reactions, two or more elements combine chemically to form compounds. The chemical reactions involving the participation of oxygen (combustion reactions) and also few others which involve a change in the oxidation number of some atoms are the examples of such redox reactions.


(b) Decomposition reactions: In such reactions, a compound either of its own or upon heating decomposes to produce two or more components. Out of such components, at least one component must be in the elemental state. For example,

There are also some decomposition reactions which are not redox reactions in nature because in such reactions there is no change in Oxidation number of the reactant species. 

(c) Displacement reactions: In such reactions, a weak atom/ion present in a compound gets replaced by strong atom/ion of another element. such as,

In this reaction, A has replaced B from the compound BC. These reactions are of two types.
(i) Metal displacement reactions. In such reactions, a metal present in a compound gets displaced by a metal in the uncombined state provided it occupies a position higher in the activity series. For example,

(ii) Non-metal displacement reactions Such reactions include hydrogen displacement and very few occurring reactions involving oxygen displacement, All alkali metals and few alkaline earth Sr metals(Ca, Sr and Ba) known to be very good reducing agents will displace hydrogen from cold water.


(d) Disproportionation reaction: Such reactions, the same substance acts as oxidant as well as reductant simultaneously i.e. a Chemical reaction in which same species undergoes Oxidation (oxidation number increases) as well as reduction (oxidation number decreases);
For example:
(i) Hydrogen peroxide (H₂O₂) is quite unstable and undergoes? disproportionation.

Oxygen of H₂O=₂ in -1 oxidation state is converted to , (0) oxidation state as well as (-2) oxidation state.
(ii) In alkaline solution, phosphorus (P₄) undergoes disproportionation.

     In the above reaction oxidation state of P₄(0) is converted to (-3) in PH₃ and to (+1) in 

(a) It is a combination redox reaction:
          0                 0                  +2 - 2
      N₂(g)              O₂(g)         2NO(g)
because, the compound nitric oxide is formed by the combination of the elemental substances, like nitrogen and oxygen. Since the oxidation number of nitrogen increases from 0 (in N₂) to +2(in NO) and that of oxygen decreases from zero (in 0₂) to -2 (in NO),
hence, it is a combination redox reaction.
(b) It is a decomposition redox reaction,

because on heating lead nitrate decomposes to form a lead oxide, nitrogen dioxide and oxygen. Since the oxidation number of nitrogen decreases from +5 (in lead nitrate) to +4 (in NO₂) and that of oxygen increases from -2 (in lead nitrate) to zero (in O₂), hence, it is a decomposition redox reaction.

because hydrogen of water has been displaced by hydride ion to produce dihydrogen gas. Also, the oxidation number of hydrogen increases from -1 in NaH) to zero (in H₂) while that of hydrogen decreases from +1 (in water) to zero (in H₂), hence it is a displacement redox reaction.

(d) It is a disproportionation reaction

Because oxidation state of nitrogen decreases from +4 (In NO₂) to +3 (In NO^-₂ ion), as well as increases from +4 (in NO₂) to +5 (In NO₃^- ion), hence it is disproportionation reaction.

In a disproportionation reaction, one of the reacting substances in a reaction always contains an element that can exist in at least three oxidation states.

ClO^- (hydrochlorite ion):

ClO₄^-1 (perchlorate ion) cannot show any disproportionation reaction. The oxidation state of Cl is ClO₄^- ion is +7. It is the maximum oxidation state which it can have. It is the maximum oxidation state which it can have. It can decrease the same by undergoing reduction and not increase it anymore hence ClO₄^- ion does not undergo disproportionation reactions.

The oxidation number of carbon in (CN)², CN^- and CNO^- are +3, +2 and +4 respectively. These are obtained as shown below:

(CN)₂: Let the oxidation number of C =x

2(x-3) =0

X= 3

CN^-

x-3= -1

x=2

CNO^-

X-3-2 =-1

X=4

(i) The reaction involves the decomposition of cyanogen (CN)₂ in alkaline medium.
(ii) Both (CN)₂ (i.e. cyanogen) and CN^- (i.e., cyanide ions) are pseudo halogens in nature i.e. both behave like halogens in characteristics.
(iii) Cyanogen undergoes disproportionation in the reaction.
(iv) It is an example of a redox reaction.

In reaction the compound nitric oxide is formed by the combination of the elemental substances, nitrogen and oxygen, therefore, this is an example of combination redox reactions. 

The oxidation number of carbon in (CN)², CN^- and CNO^- are +3, +2 and +4 respectively. These are obtained as shown below:

(CN)₂: Let the oxidation number of C =x

2(x-3) =0

X= 3

CN^-

x-3= -1

x=2

CNO^-

X-3-2 =-1

X= 4

(i) The reaction involves the decomposition of cyanogen (CN)₂ in alkaline medium.
(ii) Both (CN)₂ (i.e. cyanogen) and CN^- (i.e., cyanide ions) are pseudo halogens in nature i.e. both behave like halogens in characteristics.
(iii) Cyanogen undergoes disproportionation in the reaction.
(iv) It is an example of a redox reaction.

In hydrogen peroxide H₂O₂, the oxidation number of O is -1 and the range of the Oxidation number that O can have are from O to -2 can sometimes also attain the oxidation numbers +1 and +2. Hence, H₂O₂ can act  as an oxidising as well as reducing agent.

Oxidation number of S in SO₄^2-=+6

Since Br₂ is a stronger oxidant than I₂, it oxidises S of S₂O₃^2- to a higher oxidation state of +6 and hence forms SO₄^2- ions. As a result, thiosulphate ions behave differently with I₂ and Br_2.

Writing the oxidation number of all the atoms above their respective symbols, 
      
Since the O.N.of F increases from -1 in F^- to zero in F₂ and therefore it is oxidised and hence acts as a reductant. The O.N. of Xe decreases from +8 in XeO₆^4- to 6+ in XeO₃ and therefore it is reduced and acts as an oxidant. This reaction occurs since Na4XeO₆^4- (or XeO₆^4- is a stronger oxidant than F₂).

a) Toluene, being a non-polar liquid will not dissolve in an aqueous alkaline or acidic KMnO₄. Therefore alcoholic KMnO₄ is preferred in the manufacture of benzoic acid from toluene.

A bromide (NaBr) will also form a vapour of hydrogen bromide (HBr) but be a strong reducing agent, HBr will be oxidised by sulphuric acid to evolve red vapours of bromine. 
   

(i) H₂SO₄:
Let the oxidation number of S = x
Writing the oxidation number of each atom at the top of its symbol
+1             x             -2
H₂             S              O₄
The algebraic sum of the oxidation number of various atoms = 0
2(+1) + x + 4(-2) = 0
              x - 6  = 0
                   x = +6
(ii) Na₂S₂O₂:
Let the oxidation number of S = x
Writing the oxidation number of each atom at the top of its symbol.
               +1             x            -2
              Na₂            S₂            O₃
The algebraic sum of the oxidation number of various atoms  = 0
                       2(+1) + 2(x) + 3(-2) = 0
or                           2 + 2x - 6 = 0
                                 x = +2

C₂H₆: Let the oxidation number of C= x
Oxidation number of H = 1
The algebraic sum of the oxidation number of various atoms = 0
       

(ii) C₄H₁₀:
 Let the oxidation number of C = x
Writing the oxidation number of each atom at the top of its symbol,
                           x                +1
                         C₄                 H₁₀
The algebraic sum of the oxidation number of various atoms = 0
                 4x  + 10(+1) = 0
or                        4x = -10
or                         
(iii) CO:
Let the oxidation number of C = x
Writing the oxidation number of each atom at the top of its symbol,
                   x               -2
                   C                O
The algebraic sum of the oxidation number of various atoms = 0
                                   x - 2  = 0
                        or             x = 2
(iv) CO₂:
Let the oxidation number of C = x
Writing the oxidation number of each atom at the top of its symbol,
               x              -2
               C              O₂
The algebraic sum of the oxidation number of various atoms = 0
             x + 2(-2) = 0
or                  x = 4

                   
 

 (i) KMnO₄.
Let the oxidation number of Mn=x
Writing the oxidation number of each atom at the top of its symbol,
          +1             x           -2
          K             Mn           O₄
The algebraic sum of the oxidation number of various atoms = 0


(ii) 
Let the oxidation number of Cr = x
Writing the oxidation number of each atom at the top of tis symbol,
                         +1             x            -2
                         K₂             Cr₂           O₇
The algebraic sum of the oxidation number of various atoms = 0


Let the oxidation number of Cl = x
Writing the oxidation number of each atom at the top of its symbol,
                     +1         x            -2
                      K          Cl            O₄
The algebraic sum of the oxidation number ofvarious atoms = 0

 

Let the oxidation state of O =x
we know that,
the oxidation state of Ca=+2
Thus, oxidation state of O is,
2+2(x) =0
2x=-2
x=-1
The oxidation state of O is -1.

Let the oxidation number of B =x
Oxidation number of Na =+1
Oxidation number of H =-1
Therefore the oxidation number of B is,
1+ x+4(-1) =0
x-3=0
x= 3

Let the oxidation number of S =x
Oxidation number of H =1
Oxidation number of O= (-2)
Therefore the oxidation number of S in H₂S₂O₇ is,
1(2) +2x+7(-2)=0
2x-12=0
x=6

Let the oxidation number of S =x
Oxidation number of K=+1
Oxidation number of Al =+3
Oxidation number of O= (-2)
In this case, we ignore water molecule since it is a neutral molecule.
1+3+2x+[2(-2)4] =0
4+2x+(-16) =0
2x-12=0
x=6

C₂H₆O: The oxidation number of carbon in the ethanol can be given as,

C₂H₄O₂: The oxidation number of C in the acetic acid is given as,
Let the oxidation number of carbon= x
oxidation number of H = +1
Oxidation number of O= (-2)
Thus,
The algebraic sum of all oxidation number gives,
2x+ 4(1) +2(-2)=0
2x+ 4-4=0
2x=0
x=0
However, 0 is average oxidation number of carbon. The two carbon atoms present in this molecules are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and -2 in CH₃COOH.

(i) Oxidation number of S in H₂SO₅
By conventional method
2 x 1 + x + 5 x (-2) = 0
or 2 + x - 10 = 0 or x = + 8 (wrong)
But this cannot be true as maximum oxidation number for sulphur cannot exceed +6. The exceptional value is due to the fact that two oxygen atoms in H₂SO₅ show peroxide linkage.

This is wrong because maximum oxidation of Cr cannot be more than +6 since it has only six electrons (3d⁵ 4s¹) to take part in bond formation. Actually, four out of five oxygen atoms in CrO₅ are present in peroxide linkage.
By chemical bonding method.

(i) Let the oxidation number of Cr in dichromate ion 
Writing the oxidation number of each atom at the top of its symbol
                          
The algebraic sum of oxidation number of various atoms = 2
                    2(x) + 7(-2) = 0
or                     2x - 14 = -2
                              2x =  12
                          x = +6
(ii) Let the oxidation number of Mn in manganate ion 
         Writing the oxidation number of each atom at the top of its symbol
                               x             -2
                              [Mn          O₄]^2-
The algebraic sum of the oxidation number of various atoms = -2
                 x + 4(-2) = -2
                     x - 8 = -2
                      x = +6
(iii) Let the oxidation number of S in tetrathionate ion 
           Writing the oxidation number of each atom at the top of its symbol.
                             x              -2
                            [S₄             O₆]^2-
          The algebraic sum of oxidation number of various atoms = -2
                        4(x) + 6(-2) = -2
                   or     4x - 12 = -2
                   or            4x = 10
Therefore ,   x = 5/2.