Time required to deposit one millimole of aluminium metal by the passage of 9.65 amp through aqueous solution of aluminium ion is
Answer & explanation
Correct answer: option 1
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 = \(\frac{3 × 96500 × 10^{−3}}{9.65} = 30s\)