Electrochemistry - Result Question 35
####34. A cell, $Ag\left|Ag^{+} | Cu^{2+}\right| Cu$, initially contains $1 M Ag^{+}$and $1 M Cu^{2+}$ ions. Calculate the change in the cell potential after the passage of $9.65 A$ of current for $1 h$.
(1999, 6M)
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Solution:
- The number of Faraday’s passed $=\frac{9.65 \times 60 \times 60}{96500}=0.36 F$
After electrolysis : $\left[Ag^{+}\right]=1.36 M$
$$ \left[Cu^{2+}\right]=1-\frac{0.36}{2}=0.82 M $$
$E _1$ (before electrolysis) $=E^{\circ}$
$$ E _2(\text { after electrolysis })=E^{\circ}-\frac{0.0592}{2} \log \frac{\left[Ag^{+}\right]^{2}}{\left[Cu^{2+}\right]} $$
$\Rightarrow E _1-E _2=\frac{0.0592}{2} \log \frac{(1.36)^{2}}{0.82}=0.01 V$ (decreased)
