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In the button cells widely used in watches and other devices the following reaction takes place : Zn_((S))+Ag_(2)O_((S))+H_(2)O_((l)) to Zn_((aq))^(2+)+2Ag_((S))+2OH_((aq))^(-) Determine Delta_(r)G^(@) and E^(Theta) cell for the reaction. |
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Answer» Solution :(a) CALCULATION of standard reduction potential `E_(cell)^(Theta)` : The following reaction is carried out in button cell. `Zn_((S))+Ag_(2)O_((S)) +H_(2)O_((l)) to Zn_((aq))^(2+) +2Ag_((S))+2OH_((aq))^(-)` OXIDATION : `Zn_((S)) to Zn_((aq))^(2+)+2E^(-)` `therefore E_(L)^(Theta)=E_(Zn^(2+)|Zn)^(Theta)=-0.76V` Reduction : `Ag_(2)O_((S))+2e^(-) +H_(2)O_((l)) to 2Ag_((S))+2OH_((aq))^(-)` `therefore E_(R)^(Theta)=E_(Ag_(2)O|Ag)^(Theta)=0.344V` `therefore DeltaE_(cell)^(Theta)=E_(R)^(Theta)-E_(L)^(Theta)=E_(Ag_(2)O|Ag)^(Theta)-E_(Zn^(2+)|Zn)^(Theta)` `=0.344-(-0.76)` `+1.104V` (b) Calculation of `Delta_(r)G^(@)`,`Delta_(r)G^(@)=-nFE_(cell)^(Theta)|" Where, "n=2" mol "`, `E_(cell)^(Theta)=1.104V`,`F=96500" C "mol^(-1)` `therefore Delta_(r)G^(@)=-(2mol)xx(96500" C "mol^(-1))xx(1.104V)` `=-213072CV` `=-213072J` `=-2.13xx10^(5)J` `=-213.072kJ` `=-213kJ` |
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