InterviewSolution
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InterviewSolution
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`Cu^(2+)+2e^(-)toCu,E^(@)=+0.34V,Ag^(+)+1e^(-)toAg,E^(@)=+0.80V` (i) Construct a galvanic cell using the above data. (ii) For what concentration of `Ag^(+)` ions will the emf of the cell be zero at `25^(@)C`, if the concentration of `Cu^(2+)` is 0.01 M? (log 3.919=0.593). |
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Answer» Correct Answer - `5xx10^(-9)M` (i) For `E_(cell)^(@)` to be +ve, oxidation will occur at copper electrode and reduction at silver electrode. Hence, the cell will be represented as : `Cu|Cu^(2+)||Ag^(+)|Ag` (ii) `Cu+2Ag^(+)toCu^(2+)+2Ag,n=2` `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))`. hence, `0.46V-(0.0591)/(2)"log"(0.1)/([Ag^(+)]^(2))`. calculate `[Ag^(+)]` |
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