InterviewSolution
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InterviewSolution
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`Zn+Cu^(2+)(aq)hArrCu+Zn^(2+)(aq).` Reaction quotient is `Q=([Zn^(2+)])/([Cu^(2+)])` . Variation of `E_(cell)` with log `Q` is of the type with `OA=1.10` `V.E_(cell) ` will be `1.1591V` when A. `[Cu^(2+)]//[Zn^(2+)]=0.01`B. `[Zn^(2+)]//[Cu^(2+)]=0.01`C. `[Zn^(2+)]//[Cu^(2+)]=0.1`D. `[Zn^(2+)]//[Cu^(2+)]=1` |
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Answer» Correct Answer - b `E_(cell)=E^(c-)._(cell)-(0.0591)/(2)log.([Zn^(2+)])/([Cu^(2+)])` From line `OA=E^(c-)._(cell)=1.10V` If `([Zn^(2+)])/([Cu^(2+)])=10^(-2)M,` then `E_(cell)=1.1591V` |
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