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Two electrochemical cells, `Zn^(2+)|Zn^(2+)||Cu^(2+)|Cu` and `Fe|Fe^(2+)||Cu^(2+)|Cu` are connected in series. What will be the net e.m.f. of the cell at `25^(@)C` ? Given : `E^(@)` of `Zn^(2+)|Zn=-0.76" V"`, `Cu^(2+)|Cu=+0.34" V",Fe^(2+)|Fe=-0.41" V"`A. `+1.85" V"`B. `-1.85" V"`C. `+0.83" V"`D. `-0.83" V"` |
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Answer» Correct Answer - A (a) emf of the cell,`Zn(s)|Zn^(2+)(aq)||Cu^(2+)(aq)|Cu(s)` is : `E_(Cu^(2+)//Cu)^(@)-E_(Zn^(2+)//Zn)^(@)=+0.34-(-0.76)=1.10" V"` emf of the cell, `Fe(s)|Fe^(2+)(aq)||Cu^(2+)(aq)|Cu(s)` is : `E_(Cu^(2+)//Cu)^(@)-E_(Fe^(2+)//Fe)^(@)=+0.34-(0.41)=0.75" V"` Since the two electrodes are connected in series, the net emf of cell is : `=1.10+0.75=+1.85" V"` |
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