Chemistry - Electrochemistry Question with Solution | TestHub

$\mathrm{E}={\mathrm{E}}^{\circ }-\frac{0.0591}{n}{\log }_{10}\mathrm{Q}$ 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 $\left[\frac{2\times 0.295}{0.0591}\right]$ $$\begin{array}{rl}& =\text{ Antilog }\left[\frac{0.590}{0.0591}\right]\\ & =\text{ Antilog }10=1\times 10^{10}.\end{array}$$ 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.