Question

The standard $EMF$ of a galvanic cell involving cell reaction with $n = 2$ is found to be $0.295\, V$ at $25^\circ C. $ th $n = 2$ is found to be $0.295\, V$ at $25^{\circ}C$. The equilibrium constant of the reaction would be
(Given $F = 96500\, C\, mol^{-1}, R=8.314 \, JK^{-1}mol^{-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^{\circ}_{\text{cell}}- \frac {2.303RT}{nF}\log _{10} \, K $
At equilibrium, $E_{\text{cell}}=0 $
Given that,
$R=8.314 \, JK^{-1}mol^{-1} $
$T=25^{\circ}C+273=298 \, K $
$F=96500 \, C $ and $ n=2 $
$\therefore E^{\circ}_{\text{cell}}= \frac {2.303\times 8.314 \times 298}{2 \times 96500}\log_{10} \, K $
$= \frac {0.0591}{2}\log_{10} \, K $
Given that $E^{\circ}_{\text{cell}}=0.295 \, V $
$\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} $