The standard e.m.f of a galvanic cell involving cell reaction with $\mathrm{n}=2$ is found to be $0.295…
The standard e.m.f of a galvanic cell involving cell reaction with $\mathrm{n}=2$ is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction would be:
(Given $\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}, \mathrm{R}=8.314 \mathrm{~J}$ $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$
$2.0 \times 10^{11}$
$4.0 \times 10^{12}$
$1.0 \times 10^2$
$1.0 \times 10^{10}$
Solution
$\mathrm{E}=\mathrm{E}^{\circ}-\frac{0.0591}{n} \log _{10} \mathrm{Q}$ at $25^{\circ} \mathrm{C}$
At equilibrium, $\mathrm{E}=0, \mathrm{Q}=\mathrm{K}$
$0=\mathrm{E}^{\circ}-\frac{0.0591}{n} \log _{10} \mathrm{~K}$
or, $\mathrm{K}=$ Antilog $\left[\frac{n \mathrm{E}^{\circ}}{0.0591}\right]$
or, $\mathrm{K}=$ Antilog $\left[\frac{2 \times 0.295}{0.0591}\right]$
$\begin{aligned}
& =\text { Antilog }\left[\frac{0.590}{0.0591}\right] \\
& =\text { Antilog } 10=1 \times 10^{10} .
\end{aligned}$
Related Theory
The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the standard electromotive force (emf) of the cell when no current is drawn through the cells.