The passage of 0.95 A for 40 minutes, deposited 0.7493g of Cu from CuSO₄solution. Calculate the Molar Mass of Cu.

Dear User,
Current strength (I) = 0.95 A
Time(t) = 40 min=40 x 60 =2400 s
Quantity of electricity passed,Q = I x t = (0.95) x (2400) = 2280 C
Now, Q = nF
where, n=number of electrons involved in the reaction
F=Faraday's constant =95600 C
Therefore,
$$\begin{gathered} \mathrm{n}=\frac{\mathrm{Q}}{\mathrm{F}}=\frac{2280}{96500}=0.0236 \\ \mathrm{The} \mathrm{half} \mathrm{reaction} \mathrm{at} \mathrm{cathode} \mathrm{is} \\ {\mathrm{Cu}}^{2+}\left(\mathrm{l}\right) + 2{\mathrm{e}}^{-} \to \mathrm{Cu}\left(\mathrm{s}\right) \\ \mathrm{Therefore}, \mathrm{mole} \mathrm{ratio} =\frac{\mathrm{Moles} \mathrm{of} \mathrm{Cu} \mathrm{produced}}{\mathrm{moles} \mathrm{of} \mathrm{electrons} \mathrm{required}}=\frac{1}{2}=0.5 \mathrm{mol} \mathrm{Cu}/\mathrm{mol} {\mathrm{e}}^{-} \\ \mathrm{Moles} \mathrm{of} \mathrm{Cu} \mathrm{produced} = \mathrm{moles} \mathrm{of} \mathrm{electrons} \mathrm{actually} \mathrm{passed}\times \mathrm{mole} \mathrm{ratio} \\ = 0.0236\times 0.5 = 0.0118 \mathrm{mol} \mathrm{Cu} \\ \mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{Cu} =\frac{\mathrm{Mass} \mathrm{of} \mathrm{Cu} \mathrm{formed}}{\mathrm{moles} \mathrm{of} \mathrm{Cu} \mathrm{formed}}=\frac{0.7493}{0.0118}=63.5 \\ \mathrm{Therefore}, \mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{Cu} \mathrm{is} 63.5 \mathrm{g} {\mathrm{mol}}^{-1} \end{gathered}$$