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Question

Which of the following represent the correct expressions to calculate reduction potential of the following half cells at 25 oC.
Cl−(aq)|Cl2(g,1 atm)|Pt(s)


A
ECl2/Cl=E0Cl2/Cl0.059 log 1[Cl]
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B
ECl2/Cl=E0Cl2/Cl0.059 log[Cl]
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C
ECl2/Cl=E0Cl2/Cl0.0592 log[Cl]
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D
ECl2/Cl=E0Cl2/Cl+0.059 log[Cl]
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Solution

The correct option is B ECl2/Cl=E0Cl2/Cl0.059 log[Cl]
In the given cell representation, the gas is indicated before the electrode. Hence, it is a cathode half cell where the reduction reaction occurs.
Thus, the half cell reaction is,

12Cl2(g)+eCl(aq) (reduction)
(1 atm)

Nernst equation for
aA+bBcC+dD

E=E02.303RTnFlog[C]c[D]d[A]a[B]b.....(Eqn.1)

Substituting,

R=8.314 J K1 mol1

F=96500 C

T=25+273=298 K, we get,

E=E00.059nlog[C]c[D]d[A]a[B]b....(Eqn.2)
where,
n is the number of electrons inovolved in the reaction.

12Cl2(g)+eCl(aq) (reduction)
(1 atm)

ECl2/Cl=E0Cl2/Cl0.0591log[Cl][Cl2]12


ECl2/Cl=E0Cl2/Cl0.059 log[Cl]

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