Question

For the galvanic cell : $Ag\left|AgCl\right(s\left)\left|KCl\right(0.2M\left)\right|\right|KBr\left(0.001M\right)\left|AgBr\right(s\left)\right|Ag$,Calculate the EMF in V, generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at ${25}^{\circ }C$.

$[K_{sp}=2.8\times {10}^{-10};K_{sp\left(AgBr\right)}=3.3\times {10}^{-13}]$ The answer is -m.n $\times {10}^{-2}$ Then n-m =

Answer 1

At anode :
$Ag+C{l}^{-}\to AgCl+e^{-}$
At cathode:
$AgBr+e^{-}\to Ag+B{r}^{-}$
$E=E_{Ag/AgCl/C{l}^{-}}^{\circ }+E_{B{r}^{-}/AgBr/Ag}^{\circ }-0.0591\log \frac{\left[B{r}^{-}\right]}{\left[C{l}^{-}\right]}$
....(i)
For reaction :
$Ag+C{l}^{-}\to AgCl-e^{-}$
$A{g}^{+}+e^{-}\to Ag$
$E=O=E_{Ag/AgCr/C{l}^{-}}^{\circ }+E_{A{g}^{+}/Ag}^{\circ }-0.0591$
$\log \frac{1}{\left[A{g}^{+}\right]\left[C{l}^{-}\right]}$
$E_{Ag/AgCr/C{l}^{-}}^{\circ }=E_{Ag/A{g}^{+}}^{\circ }-0.0591\log K_{sp}\left(AgCl\right)$ ....(ii)
for reaction
$AgBr+e^{-}\to Ag+B{r}^{-}$
$Ag\to A{g}^{+}+e^{-}$
$E=0=E_{B{r}^{-}/AgBr/Ag}^{\circ }+E_{Ag/A{g}^{+}}^{\circ }-0.0591\log \left[A{g}^{+}\right]\left[B{r}^{-}\right]$
$E_{B{r}^{-}/AgBr/Ag}^{\circ }=E_{A{g}^{+}/Ag}^{\circ }+0.0591\log K_{sp}\left(AgBr\right)$ ...(iii)
from equation (1), (2) & (3)
$E=0.0591\log \frac{K_{sp}\left(AgBr\right)}{K_{sp}\left(AgCl\right)}-0.0591\log \frac{\left[B{r}^{-}\right]}{\left[C{l}^{-}\right]}$
$E=0.0591\log \frac{3.3\times {10}^{-13}\times 0.2}{2.8\times {10}^{10}\times 0.001}=-0.037V$