Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is : (at. mass of Cu =63.5 amu)

Engineering

Chemistry

Faraday Law

Question

Two Faraday of electricity is passed through a solution of CuSO₄. The mass of copper deposited at the cathode is : (at. mass of Cu =63.5 amu)

127 g

63.5 g

2 g

0 g

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(D) Cu²⁺ + 2e^– $\to_{}^{}$ Cu(s)

2 equivalents of Cu²⁺ = 2 × $\frac{63 .5}{2}$ = 63.5 gm