Question

The standard e.m.f. of a galvanic cell involving cell reaction with n=2 is found to be 0.295 V at ${25}^{0}C$. The equilibrium constant of the reaction would be?

• A. $1.0\times {10}^{15}$
• B. $2.0\times {10}^{11}$
• C. $4.0\times {10}^{12}$
• D. $1.0\times {10}^{10}$

Answer 1

The correct option is D $1.0\times {10}^{10}$

given

$n=2$

$E^0=0.295V$

$T=25+273=278$

At equilibrium Gibbs energy is zero.

$\Delta G=0$

so, electrode potential also zero.

$\Delta G=E=0$

Using Nernst equation

$E=E^0-\frac{RT}{zF}InQ$ or

$E=E^0-\frac{0.0592V}{n}\log K_{eq}$ [ when temperature is $298K$ ]

$E^0=\frac{0.059}{n}\log K_{eq}$

$0.295=\frac{0.059}{2}\log K_{eq}$

$K_{eq}={10}^{10}$

Hence, option(d) is the correct answer.