Question

The standard $EMF$ of a galvanic cell involving cell reaction with $n=2$ is found to be $0.295V$ at $25^{\circ }C.$ th $n=2$ 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,
$R=8.314J{K}^{-1}mo{l}^{-1}$
$T=25^{\circ }C+273=298K$
$F=96500C$ and $n=2$
$\therefore E_{\text{cell}}^{\circ }=\frac{2.303\times 8.314\times 298}{2\times 96500}{\log }_{10}K$
$=\frac{0.0591}{2}{\log }_{10}K$
Given that $E_{\text{cell}}^{\circ }=0.295V$
$\therefore 0.295=\frac{0.0591}{2}{\log }_{10}K$
${\log }_{10}K=\frac{0.295\times 2}{0.0591}=10$
anti log ${\log }_{10}K=$ anti log $10$
$K=1\times 10^{10}$