A current of strength 2.5 A was passed through CuSO4 solution for 6 minute 265 seconds. 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 CuSO4 solution for 6 minute 265 seconds. The amount of copper deposited is (At. wt. of Cu = 63.5; 1 F = 96500 C) (A) 0.3175 g (B) 1.028 g (C) 0.514 g (D) 6.35 g

Tardigrade Admin · Accepted Answer

W=ZIt =(E/96500)× I× t =t=6minute+265 sec =360 +265 =625 sec I=2.5A E=(63.5/2) W=(63.5/2×96500)×2.5×625=0.514 g

Q. A current of strength $2.5 A$ was passed through $C u S O_{4}$ solution for $6$ minute $265$ seconds. The amount of copper deposited is (At. wt. of $C u = 63.5; 1 F = 96500 C$ )

$0.3175$ g

$1.028$ g

$0.514$ g

$6.35$ g

$W = Z I t$
$= \frac{E}{96500} \times I \times t$
$= t = 6 min u t e + 265$ sec
$= 360 + 265 = 625$ sec
$I = 2.5 A$
$E = \frac{63.5}{2}$
$W = \frac{63.5}{2 \times 96500} \times 2.5 \times 625 = 0.514 g$

Q. A current of strength 2.5A was passed through CuSO4​ solution for 6 minute 265 seconds. The amount of copper deposited is (At. wt. of Cu=63.5;1F=96500C)

0.3175 g

1.028 g

0.514 g

6.35 g