Question

For the galvanic cell,
$Ag|AgCl(s),KCl(0.2M)\Vert KBr(0.001M),AgBr(s)|Ag$
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

$\left[A{g}^{+}\right]$in left hand electrode chamber $=\frac{2.8\times 10^{-10}}{0.2}$
$=1.4\times 10^{-9}M$
$\left[A{g}^{+}\right]$ in right hand electrode chamber $=\frac{3.3\times 10^{-13}}{0.001}$
$=3.3\times 10^{-10}M$
emf $=0-0.0592\log \frac{{\left[A{g}^{+}\right]}_{\text{anode }}}{{\left[A{g}^{+}\right]}_{\text{cathode }}}$
$=-0.0592\log \frac{1.4\times 10^{-9}}{3.3\times 10^{-10}}=-0.037V$
Therefore, the cell as written is non-spontaneous and its reverse will be spontaneous with emf $=0.037V$.