Question

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 cach electrode for a spontaneous process after taking into account the cell reaction at $25^{\circ} C$.
$\left[K_{ sp }( AgCl )=2.8 \times 10^{-10}, K_{ sp }( AgBr )=3.3 \times 10^{-13}\right] $

Answer 1

$\left[ Ag ^{+}\right]$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} $
$=3.3 \times 10^{-10} M$
emf $=0-0.0592 \log \frac{\left[ Ag ^{+}\right]_{\text {anode }}}{\left[ Ag ^{+}\right]_{\text {cathode }}}$
$=-0.0592 \log \frac{1.4 \times 10^{-9}}{3.3 \times 10^{-10}}=-0.037\, V$
Therefore, the cell as written is non-spontaneous and its reverse will be spontaneous with emf $=0.037 \,V$.