Chemistry - Electrochemistry Question with Solution | TestHub

$\text{ Charge }\left(\mathrm{q}\right)=\frac{\text{ it }}{96500}\mathrm{F}=\frac{1\times 15\times 60}{96500}=\frac{900}{96500}=\frac{9}{965}\mathrm{F}=0.0093 \mathrm{F}$

No. of moles of ${\mathrm{Au}}^{+}=0.025$ & No. of moles of ${\mathrm{Ag}}^{+}=0.025$

Species with higher value of $\mathrm{SRP}$ will get deposited first at cathode.

$\left(\mathrm{i}\right) {\mathrm{Au}}^{+}\left(\mathrm{aqs}\right) + {\mathrm{e}}^{-}⟶ \mathrm{Au}\left(\mathrm{s}\right)$

$0.025 0.0093$ mole 

so only $\mathrm{Au}$ will get deposited.