Solutions of $Hg_{2}(ClO_{4})_{2}$,$Hg(ClO_{4})_{2}$,$CuSO_{4}$,and $AgNO_{3}$ are connected in series. If a current of $2.58 \ A$ is passed for $1 \ hour$,calculate the moles of metal liberated from each solution at the cathode.

(N/A) Total charge passed $(Q)$ $= I \times t = 2.58 \ A \times 3600 \ s = 9288 \ C$.
Number of Faradays $(F)$ $= \frac{9288}{96500} \approx 0.0962 \ F$.
For $Hg_{2}^{2+} + 2e^{-} \rightarrow 2Hg$,$n$-factor $= 1$ per $Hg$ atom. Moles of $Hg = 0.0962 \ mol$.
For $Hg^{2+} + 2e^{-} \rightarrow Hg$,$n$-factor $= 2$. Moles of $Hg = \frac{0.0962}{2} = 0.0481 \ mol$.
For $Cu^{2+} + 2e^{-} \rightarrow Cu$,$n$-factor $= 2$. Moles of $Cu = \frac{0.0962}{2} = 0.0481 \ mol$.
For $Ag^{+} + e^{-} \rightarrow Ag$,$n$-factor $= 1$. Moles of $Ag = 0.0962 \ mol$.