InterviewSolution
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InterviewSolution
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Zinc granules are added in excess to 500 mL of 1M `Ni(NO_(3))_(2)` solution of `25^(@)C` untill the equilibrium is reached. If `E_(Zn^(2+)//Zn)^(@)` and `E_(Ni^(2+)//Ni)^(@)` are `-0.75V` and `-0.24V` respectively, find out the `[Ni^(2+)]` at equilibrium. |
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Answer» `{:("The redox change is "Zn+ni^(2+)hArrZn^(2+)+Ni),("mM before equilibrium 500 0"),("mM at equilibrium a (500-a)"):}` `E_(cell)^(0)=E_(OP_(ZN//Zn^(2+)))+E_(RP_(Ni^(2+)//Ni))` `E_(cell)^(0)=E_(Nn//Zn^(2+))^(0)+E_(RP_(Ni^(2+)//Ni))^(0)+(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)])` At equilibrium `E_(cell)=0` `thereforeE_(OP_(Zn//Zn^(2+)))^(0)+E_(RP_(Ni^(2+)//Ni))^(0)=-(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)])` `or 0.75+(-0.24)=-(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)])` `([Ni^(2+)])/([Zn^(2+)])="anti log"((-0.51xx2)/(0.059))=5.15xx10^(-18)` `therefore(a)/(500-a)=5.15xx10^(-18)` `thereforea=500xx5.15xx10^(-18)` `therefore[Ni^(2+)]=(mV)/(V)=(500xx5.15xx10^(-18))/(500)` `=5.15xx10^(-18)M` |
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