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The standard electrode potential for the half cell reactions are `Zn^(2+)+2e^(-)rarrZn ,E^(@)=-0.76V` `Fe^(2+)+2e^(-)rarrFe,E^(@)=-0.44V` The EMF of the cell reaction `Fe^(2+)+ZnrarrZn^(++)+Fe` isA. `-0.32V`B. `-1.20V`C. `+1.20V`D. `+0.32V` |
Answer» Correct Answer - D `Fe^(2+)+ZnrarrZn^(2+)+Fe` `ZnrarrZn^(2+)+2e^(-)` `ZnrarrZn^(2+)+2e^(-)` `Fe^(2+)+2e^(-)rarrFe` `Zn|Zn^(2+)||Fe^(2+)|Fe` `EMF^(@)=E_(Fe^(2+)//Fe)^(@)-E_(Zn^(2+)//Zn)^(@)` `=-0.44-(-0.76)` `=0.76-0.44=+0.32V` |
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