Question

The solubility product of $AgBr\left(s\right)$ is $5\times {10}^{-13}$ at $298K$. If the standard reduction potential of the half cell $E_{B{r}^{-}|AgBr|Ag}^0$ is $0.07V$.
Find the value of the $E_{A{g}^{+}/Ag}^0$

Answer 1

$AgBr\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)......\left(1\right)$
$\Delta G_1=\Delta G_1^0+RTlnQ$
At equilibrium,
$\Delta G_1=0$
$Q=K_{sp}$
$\Delta G_1^0=-RTln{K}_{sp}$

$A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)........\left(2\right)$
$\Delta G_2^0=-nF{E}_{A{g}^{+}/Ag}^0$

Adding equation (1) and (2)
$AgBr\left(s\right)+e^{-}\to Ag\left(s\right)+B{r}^{-}.......\left(3\right)$
$\Delta G_3^0=-nF{E}_{B{r}^{-}/AgBr/Ag}^0$

Adding equation (1) and (2) gives (3)
$\therefore$
$\Delta G_3^0=\Delta G_1^0+\Delta G_2^0$
$-nF{E}_{B{r}^{-}/AgBr/Ag}^0=-RTln{K}_{sp}-nF{E}_{A{g}^{+}/Ag}^0$

$nF{E}_{B{r}^{-}/AgBr/Ag}^0=RTln{K}_{sp}+nF{E}_{A{g}^{+}/Ag}^0$

$E_{B{r}^{-}/AgBr/Ag}^0=\frac{RT}{nF}ln{K}_{sp}+E_{A{g}^{+}/Ag}^0$

$n=1$ for given electrode reaction,

$E_{B{r}^{-}/AgBr/Ag}^0=\frac{RT}{F}ln{K}_{sp}+E_{A{g}^{+}/Ag}^0$

$E_{B{r}^{-}/AgBr/Ag}^0=\frac{2.303RT}{F}log{K}_{sp}+E_{A{g}^{+}/Ag}^0$

$E_{B{r}^{-}/AgBr/Ag}^0=0.0591log{K}_{sp}+E_{A{g}^{+}/Ag}^0$

$E_{B{r}^{-}/AgBr/Ag}^0=0.0591log5-0.0591\times 13log10+E_{A{g}^{+}/Ag}^0$

$0.07=0.041-0.77+E_{A{g}^{+}/Ag}^0$

$E_{A{g}^{+}/Ag}^0=0.799V$

$E_{A{g}^{+}/Ag}^0\approx 0.8V$