KCET · Chemistry · Electrochemistry
\(\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3}\) are the emf values of the three galvanic cells respectively.
- A \(\mathrm{E}_{2}>\mathrm{E}_{3}>\mathrm{E}_{1}\)
- B \(\mathrm{E}_{3}>\mathrm{E}_{2}>\mathrm{E}_{1}\)
- C \(\mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}\)
- D \(\mathrm{E}_{1}>\mathrm{E}_{3}>\mathrm{E}_{2}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{E}_{3}>\mathrm{E}_{2}>\mathrm{E}_{1}\)
Step-by-step Solution
Detailed explanation
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) \(\mathrm{E}_{3}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{0.1}{1}\)
\[
=\mathrm{E}_{\text {cell }}^{\circ}+\frac{0.0591}{2}
\]
\(\therefore \quad \mathrm{E}_{3}>\mathrm{E}_{2}>\mathrm{E}_{1}\)
\[
\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}
\]
(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) \(\mathrm{E}_{3}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{0.1}{1}\)
\[
=\mathrm{E}_{\text {cell }}^{\circ}+\frac{0.0591}{2}
\]
\(\therefore \quad \mathrm{E}_{3}>\mathrm{E}_{2}>\mathrm{E}_{1}\)
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