Electrochemistry Result Question 10
10. For the following cell,
$ \mathrm{Zn}(s)\left|\mathrm{ZnSO} _4(a q) | \mathrm{CuSO} _4(a q)\right| \mathrm{Cu}(s) $
when the concentration of $\mathrm{Zn}^{2+}$ is $10$ times the concentration of $\mathrm{Cu}^{2+}$, the expression for $\Delta G \left(\right.$ in $\mathrm{J}$ $ \mathrm{mol}^{-1}$ ) is
[F is Faraday constant; $R$ is gas constant; $T$ is temperature; $E^{\circ}$ (cell) $= 1.1 \mathrm{~V}]$
(2017 Adv.)
(a) $2.303$ $ R T + 1.1 \mathrm{~F}$
(b) $1.1 \mathrm{~F}$
(c) $2.303$ $ R T - 2.2 \mathrm{~F}$
(d) $-2.2 \mathrm{~F}$
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Answer:
Correct Answer: 10. ( c )
Solution:
- The redox reaction is : $\mathrm{Zn}(s)+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}$
The Nernst equation is $E = E^{\circ} - \frac{2.303 R T}{2 F} \log 10$
$= 1.1 - \frac{2.303 R T}{2 F}$
Also, $\Delta G=-n E F=-2 F\left(1.1-\frac{2.303 R T}{2 F}\right)$
$ \begin{aligned} & =-2.2 F+2.303 R T \\ & =2.303 R T-2.2 F \end{aligned} $
BETA
