Q79 / 200 · NEET 2009 (Screening) · Chemistry
Given,
(i) $\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}$,
$\mathrm{E}^{\circ}=0.337 \mathrm{~V}$
(ii) $\mathrm{Cu}^{2+}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}^{+}$,
$\mathrm{E}^{\circ}=0.153 \mathrm{~V}$
Electrode potential, $\mathrm{E}^{\circ}$ for the reaction, $\mathrm{Cu}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}$, will be
- Correct answerA$0.52 \mathrm{~V}$
- B$0.90 \mathrm{~V}$
- C$0.30 \mathrm{~V}$
- D$0.38 \mathrm{~V}$
Solution
Key Idea Gibb's free energy is an additive property.
$$
\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ}
$$
For reaction, $\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}$;
$$
\Delta G^{\circ}=-2 \times \mathrm{F} \times 0.337
$$
For reaction, $\mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{e}^{-}$;
$$
\Delta \mathrm{G}^{\circ}=+1 \times \mathrm{F} \times 0.153
$$
Adding Eqs. (i) and (ii), we get
$$
\begin{gathered}
\mathrm{Cu}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu} ; \quad \Delta \mathrm{G}^{\circ}=-0.521 \\
\quad \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\
-0.521 \mathrm{~F}=-\mathrm{nFE}^{\circ} \\
\therefore \quad \mathrm{E}^{\circ}=0.52 \mathrm{~V}
\end{gathered}
$$
Hence, the correct answer is option: 'A'.