# (a) Write the cel

(a) The Half cell reaction can be written as:

= -0.11× (log 0.004) V=0.26 V

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:

(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]

According to the Nernst equation:

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.

(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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