Question

what is current efficiency during electrodeposition of copper metal in which 8.8834g copper is deposited by passage of 3A current for 10000s​

Answer 1

Explanation:

n=2 [Cu

2+

+2e

−

→Cu]

Atomic mass=63.5 g

t=10000sec

i=3 A

Mass deposited=8.8834 g

m=

n×F

Atomic Mass

×i×t

m=

2×96500

63.5

×3×10000

m=9.87 g

Current efficency is the ratio of the actual mass deposited to the theoretical mass liberated according to Faraday's law.

Cuurent Efficency=

Theoretical Mass

Actual Mass

×100

=

9.87

8.8834

×100

=90%

The current efficency is 90%