The number of electrons required to deposit 6.35 g of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate will be (Given: Atomic mass of Cu= 63.5 u, NA is Avogadro constant)

A
$(N_{A} / 20) mol$
B
$(N_{A} / 10) mol$
C
$(N_{A} / 5) mol$
D
$(N_{A} / 2) mol$

Solution:

Amount of Cu to be deposited at the cathode is $n = \frac{m}{M} = \frac{6.35 g}{63.5 g {mol}^{- 1}} = 0.10 mol$

Since 2 electrons are required to convert ${Cu}^{2 +}$ ion into Cu, the amount of electrons required to deposit 0.10mol of Cu will be 0.20 mol. The number of electrons required will be (0.20mol) $N_{A}$ i.e. $(N_{A} / 5) mol$ .

Solution:

Related content