Electrochemistry 1 Question 41
40. For the galvanic cell,
$\mathrm{Ag}|\mathrm{AgCl}(s), \mathrm{KCl}(0.2 \mathrm{M}) | \mathrm{KBr}(0.001 \mathrm{M}), \mathrm{AgBr}(s)| \mathrm{Ag}$
Calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at $25^{\circ} \mathrm{C}$
$\left[K_{\text {sp }}(\mathrm{AgCl})=2.8 \times 10^{-10}, K_{\text {sp }}(\mathrm{AgBr})=3.3 \times 10^{-13}\right]$
(1992, 4M)
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Solution:
- $\left[\mathrm{Ag}^{+}\right]$in left hand electrode chamber $=\frac{2.8 \times 10^{-10}}{0.2}$
$$ =1.4 \times 10^{-9} \mathrm{M} $$
$\left[\mathrm{Ag}^{+}\right]$in right hand electrode chamber $=\frac{3.3 \times 10^{-13}}{0.001}$
$$ \begin{aligned} & =3.3 \times 10^{-10} \mathrm{M} \ \mathrm{emf} & =0-0.0592 \log \frac{\left[\mathrm{Ag}^{+}\right]{\text {anode }}}{\left[\mathrm{Ag}^{+}\right]{\text {cathode }}} \ & =-0.0592 \log \frac{1.4 \times 10^{-9}}{3.3 \times 10^{-10}}=-0.037 \mathrm{~V} \end{aligned} $$
Therefore, the cell as written is non-spontaneous and its reverse will be spontaneous with emf $=0.037 \mathrm{~V}$.
