Chemistry - Electrochemistry Question with Solution | TestHub

$1$ Faraday electricity was passed through ${\mathrm{Cu}}^{2+}(1.5$ $\mathrm{M},1\mathrm{L})/\mathrm{Cu}$ and $0.1$ Faraday was passed through ${\mathrm{Ag}}^{+}(0.2\mathrm{M},1\mathrm{L})/\mathrm{Ag}$ electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is-

Given:${\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\mathrm{o}}=0.34\mathrm{V}$
${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^0=0.8\mathrm{V}$
$\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.06\mathrm{V}$