Electrochemistry 2 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 $\Delta G$ (in $\mathrm{J} \mathrm{mol}^{-1}$ ) is
[ $\mathrm{F}$ is Faraday constant; $R$ is gas constant;
$T$ is temperature; $E^{\circ}($ cell $\left.)=1.1 \mathrm{~V}\right]$
(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) $-2.2 \mathrm{~F}$
Show Answer
Solution:
- The redox reaction is : $\mathrm{Zn}(s)+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}$
The Nernst equation is $E=E^{\circ}-\frac{2.303 R T}{2 F} \log 10$
$$ =1.1-\frac{2.303 R T}{2 F} $$
$$ \text { Also, } \begin{aligned} \Delta G & =-n E F=-2 F\left(1.1-\frac{2.303 R T}{2 F}\right) \ & =-2.2 F+2.303 R T \ & =2.303 R T-2.2 F \end{aligned} $$
