Question

The standard emf of a galvanic cell involving cell reaction with $n=2$ is found to be $0.295V$ at $25{}^{\circ }C$ .The equilibrium constant of the reaction would be (Given: $F=96500Cmo{l}^{-1};$ $R=8.314J{K}^{-1}mo{l}^{-1}$ )

• A. $2.0\times {10}^{11}$
• B. $4.0\times {10}^{12}$
• C. $1.0\times {10}^2$
• D. $1.0\times {10}^{10}$

Answer 1

Answer: (D)

By Nernst equation,
$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{2.303RT}{nF}{\log }_{10}K$
At equilibrium $E_{\text{cell }}=0$
Given that
$\therefore R=8.315J{K}^{-1}mo{l}^{-1}$
$T=25^{\circ }C+273=298K$
$F=96500C$ and $n=2$
$\therefore E_{cell}^{\circ }=\frac{2.303\times 8.314\times 298}{2\times 96500}{\log }_{10}K$
$=\frac{0.0591}{2}{\log }_{i0}K$
$\because$ Given that $E_{\text{cell }}^{\circ }=0.295V$
$\therefore 0.295=\frac{0.0591}{2}{\log }_{10}K$
${\log }_{10}K=\frac{0.295\times 2}{0.0591}=10$
or antilog of ${\log }_{10}K=$ antilog $10$
$K=1\times 10^{10}$