A current of 0.25 A is passed through CuSO 4 solution placed in voltameter for 45 minutes. The amount of Cu deposited on cathode is (at. weight of Cu =63.6)

Question

A current of 0.25 A is passed through CuSO 4 solution placed in voltameter for 45 minutes. The amount of Cu deposited on cathode is (at. weight of Cu =63.6) (A) 0.20 g (B) 0.22 g (C) 0.25 g (D) 0.30 g

Tardigrade Admin · Accepted Answer

I =0.25 A , t=45 × 60 s Z =(63.5/96500 × 2) w =(63.5/2 × 96500) × 0.25 × 45 × 60=0.22 g

Q. A current of $0.25 A$ is passed through $C u S O_{4}$ solution placed in voltameter for $45$ minutes. The amount of $C u$ deposited on cathode is (at. weight of $C u = 63.6)$

$0.20 g$

$0.22 g$

$0.25 g$

$0.30 g$

$I = 0.25 A, t = 45 \times 60 s$

$Z = \frac{63.5}{96500 \times 2}$

$w = \frac{63.5}{2 \times 96500} \times 0.25 \times 45 \times 60 = 0.22 g$

Q. A current of 0.25A is passed through CuSO4​ solution placed in voltameter for 45 minutes. The amount of Cu deposited on cathode is (at. weight of Cu=63.6)

0.20g

0.22g

0.25g

0.30g