Question

A certain quantity of electricity is passed through aqueous Al2( SO4)3 and CuSO4 solutions connected in series. 0.09 g of Al is deposited on cathode during electrolysis. The amount of copper deposited on cathode in grams is:

(Atomic mass of Al = 27 u, Cu = 63.5 u)

• 0.318
• 31.8
• 0.636
• 3.18

Answer 1

0.318

Explanation:

Aluminium sulphate: Al2( SO4)3

[\ce{Al^3+ + 3e^- -> Al}]

(mole ratio)1 = $(" of product")/(" of e"^-) = 1/3$

Copper sulphate: CuSO4

[\ce{Cu^2+ + 2e^- -> Cu}]

(mole ratio)2 = $(" of product")/(" of e"^-) = 1/2$

$"W"_1/(("mole ratio")_1 xx "M"_1) = "W"_2/(("mole ratio")_2 xx "M"_2)$

$0.09/(1/3 xx 27) = "W"_2/(1/2 xx 63.5)$

W2 = $0.09/2 xx 3/27 xx 63.5$ = 0.3175 g 0.318 g