(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K.
Sn (s ) | Sn 2+ (0.004 M )|| H+(0.020 M ) | H 2 (g )(1bar)| Pt (s)
(Given: E0 Sn+2/Sn=-0.14V )
(b) Give reasons:
(i) On the basis of E0 values, O2 gas should be liberated at anode, but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl.
(ii) The conductivity of CH3COOH decreases on dilution.
OR
(a) For the reaction
2AgCl(s) + H2(g) (1 atm) → 2Ag(s) + 2H+ + 2H+ (0.1M) + 2Cl- (0.1M),
ΔG° = -43600J at 35° C.
Calculate the e.m.f. of the cell.
[log 10-n = -n]
(b) Define fuel cell and write its two advantages.
(a) The Half cell reaction can be written as:
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= -0.11× (log 0.004) V=0.26 V
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Ecell= 0.0+ 0.0592 × (log H+)
E cell= 0.0591 × (-1.7)
Ecell= - 0.10 V
(b) (i) The standard reduction potential of water is slightly less, and therefore it has slightly more chance of getting oxidised. However, in concentrated solution of NaCl, oxidation of chloride ions is preferred than water at the anode. The unexpected result is due higher voltage required for electrolysis due to over voltage.
The reaction at the anode:![]()
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(ii) The number of ions furnished by an electrolyte depends upon the degree of dissociation with an increase in dilution the degree of dissociation increases. As a result molar conduction of acetic and increases with dilution.
OR
(a) ΔG 0 = -43600 J at 25oC.
Calculate the e.m.f. of the cell.
[log 10–n = – n]
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According to the Nernst equation:

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Ecell =-0.45-(0.059) log(0.1× 0.1)
=-0.51× log(10-2)=-0.51× (-2)=1.02 V
Ecell=1.02V
(b) 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 fuel directly into electrical energy.
Advantage : -
(1) A fuel cell works with an efficiency of 70 %
(2) These are no objectionable by-products and therefore, its a pollution free source of energy.
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