Question

A current being passed for two hour through a solution of an acid liberating $11.2$ litre of oxygen at NTP at anode. What will be the amount of copper deposited at the cathode by the same current when passed through a solution of copper sulphate for the same time?

• A. $16g$
• B. $63g$
• C. $31.5g$
• D. $8g$

Answer 1

The correct option is B $63g$

Solution:- (B) $63g$

At STP,

$1$ mole of oxygen gas $=32g=22.4L$

Therefore,

$11.2L$ of $O_2=\frac{32}{22.4}\times 11.2=16g$

Therefore,

$\frac{M_{Cu}}{M_{O_2}}=\frac{E_{Cu}}{E_O}$

$\Rightarrow \frac{M_{Cu}}{16}=\frac{\left(\frac{63}{2}\right)}{\left(\frac{16}{2}\right)}$

$\Rightarrow M_{Cu}=63g$

Hence the amount of copper deposited at the cathode is $63g$.