InterviewSolution
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InterviewSolution
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Zinc granules are added in excess to `500mL` OF `1.0m` nickel nitrate solution at `25^(@)C` until the equilibrium is reached. If the standard reduction potential of `Zn^(2+)|Zn` and `Ni^(2+)|Ni` are `-0.75V` and `-0.24V`, respectively, find out the concentration of `Ni^(2+)` in solution at equilibrium. |
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Answer» The redox change is `{:(,Zn+,Ni^(2+),hArr,Zn^(2+)+,Ni),("mM before equilibrium",,500,,0,),("mM at equilibrium",,a,,(500-a),):}` `:. E_(cell) = E_(OP_(Zn//Zn^(2+)))+E_(RP_(Ni^(2+)//Ni))` `E_(cell) = E_(Zn//Zn^(2+))^(@)+E_(RP_(Ni^(2+)//Ni))^(@)+(0.059)/(2)"log"_(10)([Ni^(2+)])/([Zn^(2+)])` At equilibrium `E_(cell) = 0` `:. E_(OP_(Zn//Zn^(2+)))^(@)+E_(RP_(Ni^(2+)//Ni))^(@) =-(0.059)/(1)log_(10).([Ni^(2+)])/([Zn^(2+)])` or `0.75+(-0.24)=-(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)]) ` `([Ni^(2+)])/([Zn^(2+)]) = "antilog" (-(0.51 xx 2)/(0.059))=5.15 xx 10^(-18)` `:. (a)/(500-a) = 5.15 xx 10^(-18)` `:. a = 500 xx 5.15 xx 10^(-18)` `:. [Ni^(2+)] = (mM)/(V) = (500 xx 5.15 xx 10^(-18))/(500)` `= 5.15 xx 10^(-18)M` |
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