InterviewSolution
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InterviewSolution
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`50mL` of `0.1M CuSO_(4)` solution is electrolysed with a current of `0.965A` for a period of 200sec. The reactions at electrodes are: Cathode: `Cu^(2+) +2e^(-) rarr Cu(s)` Anode: `2H_(2)O rarr O_(2) +4H^(+) +4e`. Assuming no change in volume during electrolysis, calculate the molar concentration of `Cu^(2+),H^(+)` and `SO_(4)^(2-)` at the end of electrolysis. |
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Answer» Correct Answer - `Cu^(2+) = 0.08M, H^(+) = 0.04M, SO_(4)^(2-) = 0.1M` `n_(e) = (It)/(96500) = (0.965 xx 200)/(96500) = 2xx 10^(-3)` `n_(Cu^(2+)) = 10^(-3)` `n_(Cu^(2+)) Initial = 5 xx 10^(-3)` `n_(Cu^(2+))` left in solution `=4xx 10^(-3)` `m_(Cu^(2+)) = (4 xx 10^(-3))/(50) xx 100 = 0.08M` `2H_(2)O rarr _(2) +4H^(+) +4e^(-)` `n_(E) = 2 xx 10^(-3) = n_(H^(+))` `m_(H^(+)) = (2 xx 10^(-3))/(50) xx 10^(-3) = 0.04M` Molarity of `SO_(4)^(2-)` will not be changed `m_(SO_(4)^(2-)) = 0.1` |
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