Question

A solution of $CuS{O}_4$ is electrolysed for $10\text{minutes}$ with a current of $1.5$ amperes. What is the mass of copper deposited at the cathode?

Answer 1

Given:
Time $\left(t\right)=10\text{min}=10\times 60=600s$

Current $\left(I\right)=1.5A$
According to the question reaction at cathode:

$C{u}^{2+}\left(aq\right)+2e^{-}=Cu\left(s\right)$

Here,$2$ electrons are exchanged so,valency
factor $(v.f.)=2$

Applying faraday's first law:

$w=ZQ$

We know, $z=\frac{Eq.wt.}{F},Eq.wt.=\frac{\text{Mol}.wt.\left(M\right)}{v.f}$
and
$Q=I\times t$
So,mass of copper $\left(w\right)=\frac{M\times I\times t}{v.f.\times F}$
By putting the values,

$w=\frac{63.5gmo{l}^{-1}\times 1.5A\times 600s}{2\times 96487C}$

$w=0.296g$

Final answer:
The mass of copper deposited at the cathode is $0.296g.$