Electrochemistry

JEE Main level, numerical answer, medium

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A current of 2.0 A2.0\,\text{A} is passed for 965 seconds965\,\text{seconds} through a CuSO4CuSO_4 solution. The number of moles of copper deposited at the cathode is:

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Answer: 0.01

Q=It=2×965=1930 CQ = I t = 2 \times 965 = 1930\,\text{C}. Moles of electrons =1930/96500=0.02= 1930 / 96500 = 0.02. Since Cu2++2e−→CuCu^{2+} + 2e^- \rightarrow Cu, moles of Cu =0.02/2=0.01= 0.02 / 2 = 0.01.

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