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)

$1.0 \times 10^{10}$

Answer 2

By Nernst equation, $E_{cell}=E^0_{cell}- \frac {2.303RT}{nF}log _{10} \, K$ At equilibrium, $E_{cell}=0$ Given that, \hspace5mm R=8.314 \, JK^{-1}mol^{-1} \hspace5mm T=25^0C+273=298 \, K \hspace5mm F=96500 \, C \, and \, n=2 $\therefore \, \, \, \, \, E^0_{cell}= \frac {2.303\times 8.314 \times 298}{2 \times 96500}log_{10} \, K$ \hspace5mm = \frac {0.0591}{2}log_{10} \, K Given that $E^0_{cell}=0.295 \, V$ $\therefore \, \, \, \, \, \, \, \, \, 0.295= \frac {0.0591}{2}log_{10} \, K$ $\, \, \, \, \, \, log_{10}K= \frac {0.295 \times 2}{0.0591}=10$ $antilog \, log_{10}K= antilog \, 10$ \hspace5mm K=1 \times 10^{10}