Overall reaction: Ag(s) + Fe³⁺(aq) \(\rightarrow\) Ag⁺(aq) + Fe2+(aq); E^o= -0.03 V
Since the EMF of the reaction is negative, therefore, the above reaction is not feasible.
Alternatively, the reaction between Ag(s) and Fe³⁺(aq) may occur according to the following equation
3Ag(s) + Fe³⁺(aq) \(\rightarrow\) 3Ag⁺(aq) + Fe(s) 
On similar lines, we can calculate the e.m.f. of this reaction comes to be even more negative, i.e., -0.836 V, and hence this redox reaction is also not feasible.
(e) Suppose the reaction between Br₂(aq) and Fe²⁺(aq) occurs according to the following equation:
Br₂(aq) + 2Fe²⁺(aq) \(\rightarrow\) 2Br^-(aq) + 2Fe³⁺(aq)
The above reaction can be split into the following two half reactions. Writing electrode potential for each half reaction from the Table 8.1, we have
Oxidation: Fe₂₊(aq) \(\rightarrow\) Fe3+(aq) + e^-] x 2; E^o = -0.77 V
Reduction: Br₂(aq) + 2e^- \(\rightarrow\) 2Br-(aq); E^o = +1.09 V

(a) The oxidation number of nitrogen in HNO₃ is +5 thus increase in oxidation number +5 does not occur hence HNO₃ cannot act as reducing agent but acts as an oxidising agent. In HNO₂ oxidation number of nitrogen is +3, it can decrease or increase with range of -3 to +5, hence it can act as both oxidising and reducing agent.
(b) Chlorine is in maximum oxidation state +7 in CIO₄ so it does not show the disproportionation reaction.
(c) Because it decomposes to give nascent oxygen.

\({ H }_{ 3 }\overset { +1 }{ Po_{ 2 } } \overset { heat }{ \longrightarrow } P\overset { -3 }{ { H }_{ 3 } } +{ H }_{ 3 }P\overset { +5 }{ O } _{ 4 }\)
Since P undergoes decrease as well as increase in oxidation state thus it is an example of disproportionation reaction

(a) (i) 4 x 1 - 2 x 2 + 2(x) = 0
4 - 4 + 2x = 0
2x = 0
x = 0
(ii) 2x - 12 - 2 = -2
2x = 12
x = +6

a) Ag⁺ is reduced, C₆H₆O₂ is oxidised.
Ag⁺ is oxidising agent whereas C₆H₆O₂ is reducing agent.
(b) HCHO is oxidised, Ag⁺ is reduced.
Ag+ is oxidising agent whereas HCHO is reducing agent.
(c) N₂H₄is getting oxidised it is reducing agent.
H₂O₂ is getting reduced it acts as an oxidising agent

chlorine is an oxidant and K₂MnO₄ is reductant.

a) (i) It completes the internal circuit.
 (ii) It maintains the electrical neutrality
b) Fe²⁺ + Cr₂\({ O }_{ 7 }^{ 2- }\) + H⁺ \(\rightarrow\) Fe³⁺ + Cr³⁺ + H₂O

The metals are Cu⁺, Ga⁺, In+,Mn³⁺, etc
2Cu⁺(aq) \(\rightarrow\)Cu²⁺(aq) +Cu(S)
3Ga⁺ (aq) \(\rightarrow\) Ga³⁺(aq) + 2Ga(s)
3In⁺(aq) \(\rightarrow\) In³⁺ + 2In(S)
\(2\overset { 3+ }{ M } n(aq)\) + 2H₂O(l) \(\rightarrow\) \(\overset { 4+ }{ Mn } { O }_{ 2 }(s)+\overset { 2+ }{ Mn(aq) } \) + 4H⁺(aq)

(i) P₄ is a reducing agent and  Cl₂ is an oxidising agent.
\((a) \ \overset { 0 }{ { P }_{ 4 } } \ (s)+6{ Cl }_{ 2 }(g)\rightarrow 4\overset { +3 }{ P } \ { Cl }_{ 3 }\)
Excess                              Lower oxidation state of P
\((b) \ \overset { 0 }{ { P }_{ 4 } } (2)+10{ Cl }_{ 2 }\rightarrow 4\overset { +5 }{ P } \ { Cl }_{ 5 }\)
Excess        Higher oxidation state of P
Therefore when P₄ (reducing agent) is in excess, PCl₃ is formed in which oxidation state of P is +3 and if Cl₂ (oxidising agent) is an excess, PCl₅ is formed in which oxidation state of P is +5. other two examples are
(II) C is a reducing agent while O₂ is an oxidising agent.
\((a) \ 2\overset { 0 }{ C } (s) \ +{ O }_{ 2 }(g)\rightarrow \overset { +2 }{ { 2CO }(g) } \)
Excess        
\((b) \ 2\overset { 0 }{ C } (s) \ +{ O }_{ 2 }\rightarrow \overset { 0 }{ { CO }_{ 2 }(g) } \)   
Excess
When reducing agent C is in excess, a compound CO of lower oxidation state is formed if oxidising agent O₂ is in excess, a compound CO₂ of higher oxidation state is formed
(iii) Na is a reducing agent while O₂ is an oxidising agent.
\((a) \ 4Na(s)+\overset { 0 }{ { O }_{ 2 }(g) } \rightarrow { Na }_{ 2 } \overset { -2 }{ 0 } (s)\)
Excess                        Lower oxidation state
\((b) \ 2Na(s)+2{ O }_{ 2 }(g)\rightarrow { Na }_{ 2 } \ \overset { -1 }{ { O }_{ 2 } } (s)\)
 Excess      Lower oxidation state

Mercury (II) chloride: Hg (II) Cl₂
Nickel (II) sulphate : NiSO₄
Tin(IV) oxide : SnO₂
Thallium(I) sulphate : Tl₂SO₄
Iron(III) sulphate : Fe₂(SO₄)₃
Chromium(III) oxide Answer : Cr₂O₃

Overall reaction: Cu(s) + 2Fe³⁺(aq) \(\rightarrow\) Cu²⁺(aq) + 2Fe²⁺(aq); E^o = +0.43 V
Since the EMF of the reaction is positive, therefore, the above reaction is feasible. Alternatively, if the reaction between Fe³⁺(aq) and Cu(s) occurs according to the following equation.
3Cu(s) + 2Fe³⁺(aq) \(\rightarrow\) 3Cu²⁺(aq) + 2Fe(s)
The EMF of the reaction comes out to be - Ie, i.e., -0.376 V (-0.34 V - 0.036 V) and hence this reaction is not feasible.
(d) Suppose the reaction between Ag(s) and Fe3+ (aq) occurs according to the following equation:
Ag(s) + Fe³ + (aq) \(\rightarrow\) Ag + (aq) + Fe²⁺(aq)
The above reaction can be split into the following two half reactions. Writing electrode potential for each half reaction from Table 8.1, we have,
Oxidation: Ag(s) \(\rightarrow\) Ag⁺(aq) + e^-; E^o = -0.80 V
Reduction: Fe + (aq) + e^- \(\rightarrow\) Fe²⁺(aq); E^o = +0.77 V

Overall reaction: Cu(s) + 2Ag⁺(aq) \(\rightarrow\) Cu²⁺(aq) + 2Ag(s); po = +0.46 V
Since the EMF of the above reaction comes out to be positive, therefore, the above reaction is feasible.
(c) Suppose the reaction between Fe³⁺(aq) and Cu(s) occurs according to the following equation.
Cu(s) + 2Fe³⁺(aq) \(\rightarrow\) 3Cu²⁺(aq) + 2Fe²⁺(aq)
The above reaction can be split into the following two half reactions. Writing electrode potential for each half reaction from Table 8.1, we have,
Oxidation: Cu(s) \(\rightarrow\) Cu²⁺(aq) + 2e-; E^o = -0.34 V
Reduction: Fe³⁺(aq) + e^- \(\rightarrow\) Fe²⁺(aq)] x ²; E^o= +0.77 V