Q58 / 180 · KCET 2010 · Chemistry
$\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3}$ are the emf values of the three galvanic cells respectively.
- A$\mathrm{E}_{2}>\mathrm{E}_{3}>\mathrm{E}_{1}$
- Correct answerB$\mathrm{E}_{3}>\mathrm{E}_{2}>\mathrm{E}_{1}$
- C$\mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}$
- D$\mathrm{E}_{1}>\mathrm{E}_{3}>\mathrm{E}_{2}$
Solution
For the given cell,
$$
\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}
$$
(i)
$$
\begin{aligned}
\mathrm{E}_{1} &=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{1}{0.1} \\
&=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2}
\end{aligned}
$$
(ii) $\mathrm{E}_{2}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{1}{1}$
$$
=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \times 0
$$
$$
=\mathrm{E}_{\text {cell }}^{\circ}
$$
(iii) $\mathrm{E}_{3}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{0.1}{1}$
$$
=\mathrm{E}_{\text {cell }}^{\circ}+\frac{0.0591}{2}
$$
$\therefore \quad \mathrm{E}_{3}>\mathrm{E}_{2}>\mathrm{E}_{1}$
Hence, the correct answer is option: 'B'.