InterviewSolution
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InterviewSolution
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`100mL CuSO_(4)(aq)` was electrolyzed using inert electrodes by passing `0.965A `till the `pH` of the resulting solution was 1. The solution after electrollysis was neutralized, treated with excess `KI` and titrated with `0.04M Na_(2)S_(2)O_(3)`. Volume of `Na_(2)S_(2)O_(3)` required was `35mL`. Assuming no volume change during electrolysis, calculate: (a) duration of electrolysis if current efficiency is `80%` (b) initial concentration `(M)` of `CuSO_(4)`. |
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Answer» Correct Answer - `1250 S, 0.064M` `2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)` `n_(H^(+)) =10^(-2) n_(E) = 10^(-2)` `n_(E)` actual `= (0.10 xx 100)/(80) rArr 0.0125` `I = (q)/(t)` `0.965 = (0.0125 xx 96500)/(t)` `t =(0.0125 xx 96500)/(0.965) rArr 1250 S` `2CuSO_(4) +4Ki rarr 2CuI +I_(2) +2K_(2)SO_(4)` `I_(2) +2Na_(2)S_(2)O_(3) rarr 2NaI +Na_(2)S_(4)O_(6)` Mili moles of `Na_(2)S_(2)O_(3) = 0.04 xx 35 rArr 1.4` Mili moles of `I_(2) = (1.4)/(2)` Mili moles of `CuSO_(4) = 2xx (1.4)/(2) rarr 1.4` `Cu^(2+) +2e^(-) rarr Cu` `n_(CU^(2+)) = (1)/(2) xx 10^(-2)|n_(E) = 10^(-2)` Total moles of `Cu^(2+) = 5 xx 10^(-3) +14.2 xx 10^(-3)` = total moles of `Cu^(2+) = 5 xx 10^(-3) +1.2 xx 10^(-3)` `= 6.42 xx 10^(-3)` `=(6.42 xx 10^(-3))/(100) = 0.064M` |
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