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Question

Three faradays of electricity is passed through molten Al2O3, aqueous solutions of CuSO4 and molten NaCl. The amounts of Al, Cu and Na deposited at the cathodes will be in the molar ratio of

Options

A.1:2:3
B.3:2:1
C.1:1.5:3
D.6:3:2

Solution

Number of moles, n = neq/n-factor

$\therefore n_{Al}:n_{Cu}:n_{Na} = \frac{3}{3}:\frac{3}{2}:\frac{3}{1} = 2:3:6$

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