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 $ .The equilibrium constant of the reaction would be (Given: $ F=96500\,C\,mo{{l}^{-1}}; $ $ R=8.314\,J{{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,
$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{2.303 R T}{n F} \log _{10} K$
At equilibrium $E_{\text {cell }}=0$
Given that
$\therefore R=8.315 \,JK ^{-1} mol ^{-1}$
$T=25^{\circ} C +273=298\, K$
$F=96500\, C$ 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 _{ i 0} K $
$ \because $ Given that $ E_{\text {cell }}^{\circ}=0.295 \,V $
$ \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} $