WBJEE · Chemistry · Electrochemistry
On passing 'C ampere of current for time 't' sec through 1 L of 2 (M) \(\mathrm{CuSO}_{4}\) solution (atomic weight of \(\mathrm{Cu}=63.5\) ), the amount \({ }^{\prime} \mathrm{m}^{\prime}\) of Cu (in gram) deposited on cathode will be
- A \(m=C t /(63.5 \times 96500)\)
- B \(m=C t /(31.25 \times 96500)\)
- C \(m=(C \times 96500) /(31.25 \times t)\)
- D \(m=(31.25 \times C \times t) / 96500\)
Answer & Solution
Correct Answer
(D) \(m=(31.25 \times C \times t) / 96500\)
Step-by-step Solution
Detailed explanation
\(m=\frac{E G}{F}\) \(=\frac{\frac{63.5}{2} \times C \times t}{96500}=\frac{31.75 \times C \times t}{96500}\)
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