Question

$A{g}_{\left(s\right)}|AgB{r}_{\left(s\right)}|B{r}_{\left(0.01M\right)}^{-}||C{l}_{\left(0.2M\right)}^{-}|AgC{l}_{\left(s\right)}|A{g}_{\left(s\right)}$
If $K_{sp}$ of $AgBr$ and $AgCl$ at ${25}^{\circ }\text{C}$ are ${10}^{-12}$ and ${10}^{-10}$ respectively, then find which of the following option(s) is/are true for above cell at ${25}^{\circ }\text{C}$.
(Given : $\log 2=0.3$)

• A. As $E_{A{g}^{+}/Ag}^{\circ }$ is not given so $E_{cell}$ can not be determined
• B. $E_{cell}$ is $0.041\text{V}$
• C. $\Delta G<0$
• D. $E_{cell}^{\circ }$ can not be determined as data is insufficient

Answer 1

Answer: (B), (C)

$\Delta G<0$

$E_{C{l}^{-}/AgCl/Ag}^0=E_{A{g}^{+}/Ag}^0+\frac{RT}{F}\ln K_{sp}.......\left(i\right)E_{B{r}^{-}/AgBr/Ag}^0=E_{A{g}^{+}/Ag}^0+\frac{RT}{F}\ln K_{sp}.......\left(ii\right)$
As $E_{A{g}^{+}/Ag}^0$ will be cancelled on substraction $\left(i\right)$ and $\left(ii\right)$, then $E_{cell}^0$ can be determined.
$E_{cell}=E_{cell}^0-\frac{0.0591}{1}\log \frac{0.2}{0.01}$
$E_{cell}^0=E_{C{l}^{-}/AgCl/Ag}^0-E_{B{r}^{-}/AgBr/Ag}^0=0.0591\times \log {10}^{-10}-0.0591\times \log {10}^{-12}=0.0591\times 2\text{V}$

$E_{cell}=0.0591\times 2-0.0591\log 20=0.041\text{V}$
$\Delta G=-nF{E}_{cell}$
$\therefore \Delta G$ is negative.
$\Delta G<0$