Question

A current of strength $2.5\, A$ was passed through CuSO $_{4}$ solution for $6 $ min $26 s$. The amount of copper deposited is :
(At. wt. of $Cu =63.5,1 \,F =96500\, C$ )

• A. 0.3175 g
• B. 3.175 g
• C. 0.635 g
• D. 6.35 g

Answer 1

Answer: (A)

Given $i=2.5 \,A$
$t=6$ min $26 s=6 \times 60+26=386\, s$
No. of coulomb passed $=i \times t$
$=2.5 \times 386$
$=965\, C$
$C u^{2+}+2 e \rightarrow C u$
$\therefore 2 \times 96500 C$ charge deposits $C u=63.5\, g$
$\therefore 965\, C$ charge deposits
$C u=\frac{63.5}{2 \times 96500} \times 965$
$=0.3175\, g$