InterviewSolution
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InterviewSolution
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The `E^(@)` values for the changes given below are measured againest `NHE` at `27^(@)C`. `Cu^(2+) + e rarr Cu^(+), E^(@) = + 0.15 V`, `Cu^(+) + e rarr Cu, E^(@) = + 0.50 V`, `Zn^(2+) + 2e rarr Zn, E^(@) = - 0.76 V` The temperature coefficient of emf a cell designed as `Zn"|"underset(1 M)(Zn^(2+))"||"underset(0.1 M)(Cu^(2+))"|"Cu` is `-1.4 xx 10^(-4)V` per degree. For a cell reaction in equilibrium `DeltaG = 0` and `DeltaG^(@) = -2.303 RTlog_(10)K_(c)`. The heat of reaction and entropy change during the reaction are are related by `DeltaG = DeltaH - TDeltaS`. If `9.65` ampere current is passed through making `Cu` anode `Zn` cathode for `1000` sec in the cell `Zn"|"underset(1 M)(Zn^(2+))"||"underset(0.1 M)(Cu^(2+))"|"Cu`, the e.m.f. of cell after after passage of current would be :A. `1.066`B. `1.076`C. `1.086`D. `1.056` |
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Answer» Correct Answer - A `(w)/(E) = (i .t)/(96500) = (9.65 xx 1000)/(96500) = 0.1` `2e + Zn^(2+) rarr Zn` (Cathode) `Cu rarr Cu^(2+)+2e` (Anode) `:. [Zn^(2+)] = 1 - 0.10 = 0.9` `[Cu^(2+)] = 0.1 + 0.1 = 0.2` `:. E_(cell) = E_(cell)^(@) + (0.059)/(2)"log"([Cu^(2+)])/([Zn^(2+)])` `= 1.085 + (0.059)/(2)"log"(0.2)/(0.9) = 1.066 V` |
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