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  • In an electrolysis a constant current was passed for 4 hours through two cells connected in series. The first cell contains a solution of gold salt and the second cell contains copper sulphate solu

In an electrolysis a constant current was passed for 4 hours through two cells connected in series. The first cell contains a solution of gold salt and the second cell contains copper sulphate solution 8.85g of gold was deposited in the first cell. If the oxidation number of gold is +3 , find the amount of copper deposit e on the cathode in the second cell.

Option: 1

5.64 g


Option: 2

4.36 g


Option: 3

4.27 g


Option: 4

2.64 g


Answers (1)

best_answer

\mathrm{\frac{\text { Mass Au deposited }}{\text { Mass of } \mathrm{Cu} \text { deposited }}=\frac{Eq \text { mass of } \text { Au }}{\text { Eq Mass of } \mathrm{Cu}}}

\mathrm{ \text { Eq mass of } \mathrm{Au}=\frac{197}{3} \\ }

\mathrm{ \text { Eq mass of } \mathrm{Cu}=\frac{63.5}{2} \\ }

\mathrm{ \frac{8.85}{\text { Mass of Cu deposited }}=\frac{\frac{197}{3}}{\frac{63.5}{2}} }

\mathrm{ \text { Mass of } \mathrm{Cu} \text { deposited }=8.85 \times \frac{63.5}{2} \times \frac{3}{197} }

\mathrm{ =4.27 \mathrm{~g} }

Posted by

Irshad Anwar

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