Zinc rod is dipped in 0.1 M solution of ZnSO₄. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential. [Given that: E°_Zn²⁺/Zn = – 0.76 V.]

Concentration of ${\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)$
             = 0.1 + $\frac{95}{100}=0.095$
           ${\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \to \mathrm{Zn}$
According to Nernst equation,
            $\mathrm{E} = \mathrm{E}°+\frac{0.0591}{\mathrm{n}}\log \frac{\left[{\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)\right]}{\left[\mathrm{Zn}\right]}$
             $$\begin{gathered} = -0.76 + \frac{0.0591}{\mathrm{n}}\log \frac{0.095}{1} \\ = 0.76+0.02953 \times (-1.0223) \\ = -0.76 - 0.03021 = -0.79 \mathrm{V}. \end{gathered}$$