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lana66690 [7]
3 years ago
12

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?

Chemistry
1 answer:
asambeis [7]3 years ago
3 0
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 Cu
²⁺ is deposited.
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