For the galvanic cell at ${25}^{o}C$,
$Ag|AgCl\left(s\right),KCl\left(0.2M\right)||KBr\left(0.001M\right),AgBr\left(s\right)|Ag$

Which of the following(s) is/are true for the given cell?
Given:
$K_{sp}\left(AgCl\right)=2.8\times {10}^{-10}$
$K_{sp}\left(AgBr\right)=3.3\times {10}^{-13}$

The correct option is C At ${25}^{o}C,E_{cell}=-0.037V$
From cell representation,
At anode,
$A{g}_{L.H.S}\to A{g}_{L.H.S}^{+}$

At cathode,
$A{g}_{R.H.S}^{+}+e^{-}\to A{g}_{R.H.S}$

The cell reaction is
$A{g}_{L.H.S}+A{g}_{R.H.S}^{+}\to A{g}_{L.H.S}^{+}+A{g}_{R.H.S}$

Nernst equation for the given concentration cell is
$E_{cell}=E_{cell}^0-0.0591log\frac{A{g}_{L.H.S}^{+}}{A{g}_{R.H.S}^{+}}$
For concentration cell,
$E_{cell}^0=0$

$E_{cell}=0.0591log\frac{[A{g}^{+}{]}_{RHS}}{[A{g}^{+}{]}_{LHS}}.....eqn\left(1\right)$

For
$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$
$K_{sp}\left(AgCl\right)=2.8\times {10}^{-10}$
$K_{sp}\left(AgCl\right)=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$\left[A{g}^{+}\right]=\frac{K_{sp}\left(AgCl\right)}{\left[C{l}^{-}\right]}$
Concentration of $KCl$ is $0.2M$
$\therefore$
$\left[C{l}^{-}\right]=0.2M$
$\left[A{g}^{+}\right]=\frac{2.8\times {10}^{-10}}{0.2}$

$[A{g}^{+}{]}_{LHS}=14\times {10}^{-10}M$

For
$AgBr\rightleftharpoons A{g}^{+}+B{r}^{-}$
$K_{sp}\left(AgBr\right)=3.3\times {10}^{-13}$
$K_{sp}\left(AgBr\right)=\left[A{g}^{+}\right]\left[B{r}^{-}\right]$
$\left[A{g}^{+}\right]=\frac{K_{sp}\left(AgBr\right)}{\left[B{r}^{-}\right]}$
Concentration of $KBr$ is $0.001M$
$\therefore$
$\left[B{r}^{-}\right]=0.001M$
$\left[A{g}^{+}\right]=\frac{3.3\times {10}^{-13}}{0.001}$

$[A{g}^{+}{]}_{RHS}=3.3\times {10}^{-10}M$

Substituting $A{g}^{+}$ concentrations in equation (1) gives,

$E_{cell}=0.0591log\frac{3.3\times {10}^{-10}}{14\times {10}^{-10}}$

$E_{cell}=0.0591log0.235$

$E_{cell}\approx -0.037V$

The emf of cell is negative, hence, the cell reaction is non-spontaneous.