Question

CuSO4(aq) electrolysed using platinum electrodes
A current is passed until 1.6 g of O2 liberated at
anode. The amount of Cu deposited at the cathode
during same time period

(1) 6.35 g
(2) 63.5 g
(3) 12.7g
(4) 3.2g​

Answer 1

Answer: Option 1

Explanation:

Answer 2

Answer : The correct option is, (1) 6.35 g

Explanation:

According to the Faraday's second law:

$\frac{\text{Weight of }Cu}{\text{Weight of }O_2}=\frac{\text{Equivalent weight of }Cu}{\text{Equivalent weight of }O_2}$

Equivalent weight : It is the molar mass of a substance divided by the number of equivalents in the substance.

Given :

Weight of $O_2$ = 1.6 g

Molar mass of solute $O_2$ = 32 g/mole

$\text{Equivalent weight of }O_2=\frac{\text{Molar mass of }O_2}{\text{Number of valency}}=\frac{32}{4}=8g.eq$

Molar mass of solute Cu = 63.55 g/mole

$\text{Equivalent weight of Cu}=\frac{\text{Molar mass of Cu}}{\text{Number of valency}}=\frac{63.55}{2}=31.78g.eq$

Now put all the given values in the above law, we get:

$\frac{\text{Weight of }Cu}{\text{Weight of }O_2}=\frac{\text{Equivalent weight of }Cu}{\text{Equivalent weight of }O_2}$

$\frac{\text{Weight of }Cu}{1.6}=\frac{31.78}{8}$

$\text{Weight of }Cu=6.35g$

Therefore, the amount of Cu deposited at the cathode  during same time period is 6.35 grams.