Chemistry - Electrochemistry Question with Solution | TestHub

Key-idea

Use Nernst equation, i.e.

$\mathrm{E}\left(\mathrm{emf}\right)=\mathrm{E}{°}_{\mathrm{cell}}-\frac{0.0591}{2}\log \frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}$

For reaction $\left(\mathrm{i}\right)$

${\mathrm{E}}_1=\mathrm{E}{°}_{\left(\mathrm{cell}\right)}-\frac{0.0591}{2}\log \frac{0.1}{1}$

${\mathrm{E}}_1=\mathrm{E}{°}_{\left(\mathrm{cell}\right)}+\frac{0.0591}{2}$

For reaction $\left(\mathrm{ii}\right)$

${\mathrm{E}}_2=\mathrm{E}{°}_{\left(\mathrm{cell}\right)}-\frac{0.0591}{2}\log \frac{1}{1}=0$

${\mathrm{E}}_2=\mathrm{E}{°}_{\left(\mathrm{cell}\right)}$

For reaction $\left(\mathrm{iii}\right)$

${\mathrm{E}}_3=\mathrm{E}{°}_{\left(\mathrm{cell}\right)}-\frac{0.0591}{2}\log \frac{1}{0.1}$

${\mathrm{E}}_3=\mathrm{E}{°}_{\left(\mathrm{cell}\right)}-\frac{0.0591}{2}$

Hence, ${\mathrm{E}}_1>{\mathrm{E}}_2>{\mathrm{E}}_3$.