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During purification of copper by electrolysis, 1.48 g of copper were deposited when current was passed through aqueous copper (II) sulphate for 2$\frac{1}{2}$ hours. Calculate the amount of current used. (Cu = 63.5; 1 F = 96500 C

During purification of copper by electrolysis, 1.48 g of copper were deposited when current was passed through aqueous copper (II) sulphate for 2$\frac{1}{2}$ hours. Calculate the amount of current used. (Cu = 63.5; 1 F = 96500 C).

Answers

John
63.5 g $\rightarrow$ (98500 x 2)C

1.48 g $\rightarrow$ $\frac{1.48\times 96500\times 2}{63.5}$ = 4498 C

4498 C = I x 2.5 x 60 x 60 $\rightarrow$ I

= $\frac{4498}{2.5\times 60\times 60}$ = 0.5 A.
johnmulu answered the question on February 16, 2017 at 06:09

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