Electrolytic cells containing \(\mathrm{Zn}\) and \(\mathrm{Al}\) salt solutions are connected in series. If \(6.5 \mathrm{~g}\) of \(\mathrm{Zn}\) is deposited in one cell calculate mass of Al deposited in second cell (molar mass : \(\mathrm{Zn}=65\), \(\mathrm{Al}=27\) ) by passing definite quantity of electricity?

Cell 1:
\(
\begin{aligned}
& \mathrm{Zn}_{(\mathrm{m})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}_{(\mathrm{s})} \\
& (\text { mole ratio) })_1=\frac{1 \mathrm{~mol}}{2 \mathrm{~mole}^{-}}
\end{aligned}
\)
Cell 2:
\(
\begin{aligned}
& \mathrm{Al}_{(\mathrm{sq})}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}_{(\mathrm{s})} \\
& (\text { mole ratio })_2=\frac{1 \mathrm{~mol}}{3 \mathrm{~mole}^{-}}
\end{aligned}
\)
\(\begin{aligned} & \frac{\mathrm{W}_1}{(\text { mole ratio })_1 \times \mathrm{M}_1}=\frac{\mathrm{W}_2}{(\text { mole ratio })_2 \times \mathrm{M}_2} \\ & \frac{6.5 \mathrm{~g}}{1 \mathrm{~mol} / 2 \mathrm{~mol} \mathrm{e}^{-} \times 65 \mathrm{~g} \mathrm{~mol}^{-1}} \\ & =\frac{\mathrm{W}_2}{1 \mathrm{~mol} / 3 \mathrm{~mol} \mathrm{e}^{-} \times 27 \mathrm{~g} \mathrm{~mol}^{-1}} \\ & \frac{6.5 \mathrm{~g} \times 2}{65}=\frac{\mathrm{W}_2 \times 3}{27} \\ & \mathrm{~W}_2=\frac{6.5 \times 2 \times 27}{65 \times 3}=1.8 \mathrm{~g}\end{aligned}\)