The EMF of the cell $\mathrm{Zn} | {\mathrm{Zn}}^{2+} (0.01\mathrm{M}) | | {\mathrm{Fe}}^{2+} (0.001\mathrm{M}) | \mathrm{Fe}$ , at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction ${10}^x$. Find x.

For this cell, reaction is;

$\mathrm{Zn}+{\mathrm{Fe}}^{2+}\to {\mathrm{Zn}}^{2+}+\mathrm{Fe}$ 

$$\begin{gathered} \mathrm{E}=\mathrm{E}°-\frac{0.0591}{\mathrm{n}}\log \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}; \\ \mathrm{E}°=\mathrm{E}+\frac{0.0591}{\mathrm{n}}\log \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2} \end{gathered}$$

$E°=0.2905+\frac{0.0591}{2}\log \frac{{10}^{-2}}{{10}^{-3}}=0.32 \mathrm{V}$

At equilibrium,

$\mathrm{E}°=\frac{0.0591}{2}\log {\mathrm{K}}_{\mathrm{eq}}$

$$\begin{gathered} \log K_{\mathrm{eq}}=\frac{0.32\times 2}{0.0591} \\ =\frac{0.32}{0.0295} \end{gathered}$$

$\therefore {\mathrm{K}}_{\mathrm{eq}}={10}^{\frac{0.32}{0.0295}}$

Comparing the value of ${10}^x$,

$x=\frac{0.32}{0.0295}=10.847\approx 10.85$