How many minutes would be required to deposit copper in 500 mL of 0.25 N CuSO4 by a current of 75 milliampere?

Question

How many minutes would be required to deposit copper in 500 mL of 0.25 N CuSO4 by a current of 75 milliampere? (A) 1340 min (B) 670 min (C) 2680 min (D) 5360 min

Tardigrade Admin · Accepted Answer

V=500 mL =0.5 L Equivalents = 0.25 Ã 0.5 = 0.125 Current =75×10-3 ampere; Q = It 0.125=(75×10-3× t/96500) ⇒ t=2680 min

Q. How many minutes would be required to deposit copper in 500 mL of 0.25 N $C u S O_{4}$ by a current of 75 milliampere?

1340 min

670 min

2680 min

5360 min

$V = 500 m L = 0.5 L$
Equivalents = 0.25 × 0.5 = 0.125
Current = $75 \times 1 0^{- 3}$ ampere; Q = It
$0.125 = \frac{75 \times 1 0 ^{- 3} \times t}{96500} \Rightarrow t = 2680 min$

Q. How many minutes would be required to deposit copper in 500 mL of 0.25 N CuSO4​ by a current of 75 milliampere?