Electrochemistry 1 Question 35
34. A cell, $\mathrm{Ag}\left|\mathrm{Ag}^{+} | \mathrm{Cu}^{2+}\right| \mathrm{Cu}$, initially contains $1 \mathrm{M} \mathrm{Ag}^{+}$and $1 \mathrm{M} \mathrm{Cu}^{2+}$ ions. Calculate the change in the cell potential after the passage of $9.65 \mathrm{A}$ of current for $1 \mathrm{h}$.
(1999, 6M)
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Answer:
Correct Answer: 34. $0.01 \mathrm{V}$
Solution:
- The number of Faraday’s passed $=\frac{9.65 \times 60 \times 60}{96500}=0.36 \mathrm{F}$
After electrolysis : $\left[\mathrm{Ag}^{+}\right]=1.36 \mathrm{M}$
$ \left[\mathrm{Cu}^{2+}\right]=1-\frac{0.36}{2}=0.82 \mathrm{M} $
$E _{1}$ (before electrolysis) $=E^{\circ}$
$ E _{2}(\text { after electrolysis })=E^{\circ}-\frac{0.0592}{2} \log \frac{\left[\mathrm{Ag}^{+}\right]^{2}}{\left[\mathrm{Cu}^{2+}\right]} $
$\Rightarrow E _{1}-E _{2}=\frac{0.0592}{2} \log \frac{(1.36)^{2}}{0.82}=0.01 \mathrm{V}$ (decreased)
