Calculate the emf of the following cell at ${25}^{\circ }C$.
$Ag\left(s\right)|AgN{O}_3\left(0.01{\text{mol kg}}^{-1}\right)||AgN{O}_3\left(0.05{\text{mol kg}}^{-1}\right)|Ag\left(s\right)$

The correct option is D $0.0412\text{V}$
It is a concentration cell. The standard emf of a concentration cell is zero.
$\therefore$
$Ag\left(s\right)+A{g}^{+}\text{(cathode)}\to A{g}^{+}\text{(anode)}+Ag$
$\text{E}=-\frac{0.059}{\text{n}}\log \frac{A{g}^{+}\text{(anode)}}{A{g}^{+}\text{(cathode)}}\text{at}{25}^{\circ }C$
$\Rightarrow \text{E}=\frac{0.059}{\text{n}}\log \frac{A{g}^{+}\text{(cathode)}}{A{g}^{+}\text{(anode)}}$
$n=1$
$\text{E}=0.059\log \frac{0.05}{0.01}=0.0412\text{V}$