JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Two Faraday of electricity is passed through a solution of \(CuSO_4.\) The mass of copper deposited at the cathode is ......... \(g\). (at. mass of \(Cu = 63.5\, amu\))
- A \(2\)
- B \(127\)
- C \(0 \)
- D \(63.5\)
Answer & Solution
Correct Answer
(D) \(63.5\)
Step-by-step Solution
Detailed explanation
\(C u^{2+}+2 e^{-} \longrightarrow C u\) \(2 F\) i.e. \(2 \times 96500 C\) deposit \(C u=1 \mathrm{mol}=63.5 \mathrm{g}\)
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