In electrolytic refining of Cu, when 9.65 A of current is passed through an electrolytic cell for 30000 seconds to purify 100 g of an impure Cu metal at the cathode. What is % of impurity in the metal? (Atomic mass of Cu = 63.5 amu)

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Solution

Charge passed = 9.65 x 30000 C
= $\frac{(9.65 \times 30000)}{96500 F}$ = 3 F
So, 3 equivalents = 3/2=1.5 mol of Cu is pure
That is 63.5 x 1.5 = 95.25 gm of pure Cu
% of impure Cu = 100 - 95.25= 4.75%

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In electrolytic refining of Cu, when 9.65 A of current is passed through an electrolytic cell for 30000 seconds to purify 100 g of an impure Cu metal at the cathode. What is % of impurity in the metal? (Atomic mass of Cu = 63.5 amu)

Charge passed = 9.65 x 30000 C = (9.65×30000)96500F = 3 F So, 3 equivalents = 3/2=1.5 mol of Cu is pure That is 63.5 x 1.5 = 95.25 gm of pure Cu % of impure Cu = 100 - 95.25= 4.75%

Charge passed = 9.65 x 30000 C
= $\frac{(9.65 \times 30000)}{96500 F}$ = 3 F
So, 3 equivalents = 3/2=1.5 mol of Cu is pure
That is 63.5 x 1.5 = 95.25 gm of pure Cu
% of impure Cu = 100 - 95.25= 4.75%