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)

$1.0\times {{10}^{10}}$

Answer 2

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 $10$ $K =1 \times 10^{10}$