InterviewSolution
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InterviewSolution
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Copper reduced `NO_(3)^(-)` into `NO` and `NO_(2)` depending upon conc.of `HNO_(3)` in solution. Assuming `[Cu^(2+)] = 0.1M`, and `P_(NO) = P_(NO_(2)) = 10^(-3)` atm and using data answer the following questions: `E_(Cu^(2+)//Cu)^(@) =+ 0.34` volt, at `298K (RT)/(F) (2.303) = 0.06` volt `E_(cell)` for reduction of `NO_(3)^(-) rarr NO` by `Cu(s)` when `[HNO_(3)] = 1M` is [At `t = 298]`A. `10^(1.23)M`B. `10^(0.56)M`C. `10^(0.66)M`D. `10^(0.12)M` |
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Answer» Correct Answer - C `(e^(-) +2H^(+) +NO_(3)^(-) rarr NO_(2) +H_(2)O) xx 2` `Cu rarr Cu^(+2) +2e^(-)` `4H^(+) +2NO_(3)^(-) +Cu rarr Cu^(+2) +2NO_(2) +2H_(2)O` `E = 0.79-0.34 -(0.0591)/(2) log.([NO_(2)]^(2)[Cu^(+2)])/([NO_(3)^(-)]^(2)[H^(+)]^(4))`..(ii) Let `[HNO_(3)] = xM` so `[H^(+)] = [NO_(3)^(-)] = x` equation (i) & (ii) `0.62 -(0.0591)/(2) log.(10^(-9))/(x^(10)) = 0.45 -(0.0591)/(2)log.(10^(-7))/(x^(6))` `0.17 =(0.591)/(6) [-9-10 logx] -(0.0591)/(2)[-7-6logx]` `0.0518 = 0.0788 logx` `x = 10^(0.657)M` |
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