Time required to deposit one millimole of aluminium metal by the passage of 9.65 amp through aqueous solution of aluminium ion is

30 s

Quantity of electricity in coulombs = Current in amperes × Time (s)

For aluminium metal,

\(Al^{3+} + 3e^- \rightarrow Al\)

Quantity of charge required for 1 mol of \(Al^{3+}\) = \(3 × 96500C\)

Quantity of charge required for \(10^{−3}\) mol of \(Al^{3+}\) = \(3 × 96500 × 10^{−3}\)

Time = \(\frac{3 × 96500 × 10^{−3}}{9.65} = 30s\)