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Derive Nernst equation for calculating E_(cell) of Nernst equation and write the effect on E_(cell) when there is change in concentration of Zn^(2+) and Cu^(2+). |
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Answer» Solution :In DANIELL cell, the electrode POTENTIAL for any given concentration of `Cu^(2+) and Zn^(2+)` ions, its REACTION: * For Anode `Cu^(2+)|Cu`: `E_((Cu^(2+)|Cu))=E_((Cu^(2+)|Cu))^(Theta)-(RT)/(2F)ln(1)/([Cu_((aq))^(2+)])` * For Cathode `Zn^(2+)|Zn`: `E_((Zn^(2+)|Zn))=E_((Zn^(2+)|Zn))^(Theta)-(RT)/(2F)ln(1)/([Zn_((aq))^(2+)])` * Cell potential `E_(cell)=(E_("cathode")-E_("anode"))` `therefore E_(cell)=(E_((Cu^(2+)|Cu))-E_((Zn^(2+)|Zn)))` `therefore E_(cell)=(E_((Cu^(2+)|Cu))^(Theta)-(RT)/(2F)ln(1)/([Cu_((aq))^(2+)]))-(E_((Zn^(2+)|Zn))^(Theta)+(RT)/(2F)ln(1)/([Zn^(2+)]))` `therfore E_(cell)=(E_((Cu^(2+)|Cu))^(Theta)-E_((Zn^(2+)|Zn))^(Theta))-(RT)/(2F)(ln(1)/([Cu_((aq))^(2+)])-ln(1)/([Zn_((aq))^(2+)]))` `therefore E_(cell)=E_(cell)^(Theta)-(RT)/(2F)ln([Zn_((aq))^(2+)])/([Cu_((aq))^(2+)])` (Above formula is for `E_(cell)` of Daniell cell potential) * In above EQUATION if `ln=2.303log_(10)` `E_(cell)=E_(cell)^(Theta)-(2.303RT)/(2F)"log"_(10)([Zn_((aq))^(2+)])/([Cu_((aq))^(2+)])` In the above equation if `R=8.314JK^(-1)mol^(-1)`, `T=298K and F=96487" C "mol^(-1) and (RT)/(F)=0.0591` So cell potential for the Daniell cell, `E_(cell)=E_(cell)^(Theta)-(0.0591)/(2)"log"_(10)([Zn_((aq))^(2+)])/([Cu_((aq))^(2+)])` * It can be seen that `E_(cell)` depends on the concentration of both `Cu^(2+) and Zn^(2+)` ions. it incrases with the increase in the concentration of `Cu^(2+)` ions and decrease in the concentration of `Zn^(2+)` ions. |
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