Two electrolytic cells are connected in series containing CuSO_4 …

MCQ

Two electrolytic cells are connected in series containing $\mathrm{CuSO}_4$ solution and molten $\mathrm{AlCl}_3$. If in electrolysis 0.4 moles of 'Cu' are deposited on cathode of first cell, the number of moles of 'Al' deposited on cathode of the second cell is

A
0.6 moles
B
0.27 moles
C
0.18 moles
D
0.4 moles

✓

Answer

(b) : $\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu}$

Thus for 1 mole of $\mathrm{Cu}$, charge required $=2 \mathrm{~F}$

For 0.4 mole of $\mathrm{Cu}$, charge required $=(2 / 1) \times 0.4=0.8 \mathrm{~F}$

For $\mathrm{Al}, \mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al}$

For 1 mole of $\mathrm{Al}$, charge required $=3 \mathrm{~F}$

Thus from $0.8 \mathrm{~F}, \mathrm{Al}$ will deposit $=(1 / 3) \times 0.8=0.27$ mole

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