What mass of copper is deposited when a current of 10.0a is passed through a solution of copper (ii) nitrate for 30.6 seconds?

Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:

According to Faraday's Law,

W = I t e / F

Putting Values,

W = (10 A * 30.6 s * 31.77 g) : 96500

W = 0.100 g

Result:

0.100 g of Cu²⁺ is deposited.