InterviewSolution
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InterviewSolution
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An acidic solution of `Cu^(2+)` salt contaning `0.4` of `Cu^(2+)` is electrolyzed untill all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100mL and the current at `1.2` amp. Calculate the volume of gases evolved at NTP during the entire electrolysis. |
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Answer» Correct Answer - `V(O_(2)) = 99.68mL V(H_(2)) = 58.46 mL`, Total vol `=158.1 mL` In beginning At anode: `2e^(-) +Cu^(+2) rarr Cu` At anode: `2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)` Eq. of `O_(2) = Eq. of Cu^(+2)` `=(0.4 xx 2)/(63.5) = 0.012598` vol of `O_(2) = (0.012598 xx 22400)/(4) = 70.55mL` Later on: At anode: `2H_(2)O +2e^(-) rarr H_(2) +2OH^(-)` At anode: `2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)` Faraday passed `= (1.2 xx 7 xx 60)/(96500) = 0.00522` moles of `O_(2) = (0.00522)/(4)` moles of `H_(2) = (0.00522)/(2)` Volume of `O_(2) = (0.00522)/(4) xx 22400 = 29.24mL` Volume of `H_(2) = (0.00522)/(2) xx 22400 = 58.464 mL` Total volume of `O_(2)` `= 29.24 + 70.55 = 99.79 mL` Total volume of `H_(2)` `= 58.464 mL` |
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