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Derive Nernst equation for the following galvanic cell. Ni_((S))|Ni_((aq))^(2+)||Ag_((aq))^(+)|Ag |
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Answer» SOLUTION :* The cell reaction is as follows: Anode: `Ni_((S))toNi_((AQ))^(2+)+2e^(-)`. ..(Reduction) Cathode: `2Ag_((aq))^(+)+2e^(-) to 2Ag_((S))`. . . (Oxidation) Redox reaction: `Ni_((S))+2Ag_((aq))^(+) to Ni_((aq))^(2+)+2Ag_((S))` If `E_(cell)`=Non standard electrode potential. `E_(cell)^(Theta)`=Standard electrode potential. n=Number of electrons TAKING part in the reaction at equilibrium. * `E_(cell)=E_(cell)^(Theta)-(RT)/(2F)LN([Ni_((aq))^(2+)])/([Ag_((aq))^(+)]^(2))` `therefore E_(cell)=E_(cell)^(Theta)-(0.0591)/(2)"log"_(10)([Ni_((aq))^(2+)])/([Ag_((aq))^(+)]^(2))`. |
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