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 )

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

Tardigrade Admin · Accepted Answer

Given i=2.5 A t=6 min 26 s=6 × 60+26=386 s No. of coulomb passed =i × t =2.5 × 386 =965 C C u2++2 e arrow C u ∴ 2 × 96500 C charge deposits C u=63.5 g ∴ 965 C charge deposits C u=(63.5/2 × 96500) × 965 =0.3175 g

Q. 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 $C u = 63.5, 1 F = 96500 C$ )

0.3175 g

3.175 g

0.635 g

6.35 g

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

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