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  • Four Faraday of electricity is passed through a solution of   .The mass of co

Four Faraday of electricity is passed through a solution of  \mathrm{CuSO_4} .The mass of copper deposited at the cathode is

Option: 1

127 g


Option: 2

63.5 g


Option: 3

37 g


Option: 4

100 g


Answers (1)

best_answer

\mathrm{Cu}^{+2}+2 e^{-} \rightarrow \mathrm{Cu} \\

2 \text { Faraday }=2 \text { equivalents of } \mathrm{Cu} \\
4 \text { Faraday } =\frac{2}{2} \times 4=4 \text { equivalents of Cu } \\

\text { Mass of Cu deposits } =4 \times \frac{63.5}{2} \\
\mathrm{127 g}.

Posted by

Ritika Jonwal

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