Electrochemistry - Result Question 41
####40. For the galvanic cell,
$Ag|AgCl(s), KCl(0.2 M) | KBr(0.001 M), AgBr(s)| Ag$
Calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at $25^{\circ} C$
$\left[K _{\text {sp }}(AgCl)=2.8 \times 10^{-10}, K _{\text {sp }}(AgBr)=3.3 \times 10^{-13}\right]$
(1992, 4M)
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Solution:
- $\left[Ag^{+}\right]$in left hand electrode chamber $=\frac{2.8 \times 10^{-10}}{0.2}$
$$ =1.4 \times 10^{-9} M $$
$\left[Ag^{+}\right]$in right hand electrode chamber $=\frac{3.3 \times 10^{-13}}{0.001}$
$$ \begin{aligned} & =3.3 \times 10^{-10} M \ emf & =0-0.0592 \log \frac{\left[Ag^{+}\right] _{\text {anode }}}{\left[Ag^{+}\right] _{\text {cathode }}} \ & =-0.0592 \log \frac{1.4 \times 10^{-9}}{3.3 \times 10^{-10}}=-0.037 V \end{aligned} $$
Therefore, the cell as written is non-spontaneous and its reverse will be spontaneous with emf $=0.037 V$.
