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Zinc granules are added in excess to 500mL of 1M nickel nitrate solution at 25^@C until the equilibrium is reached. If the standard reduction potentials 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» SOLUTION :The cell REACTION for the cell `Zn,Zn^(2+)||Ni^(2+), Ni` is `Zn+Ni^(2+) leftrightarrow Zn^(2+)+Ni` for which `E_(cell)=E_(cell)^@-0.0591/2 log""([Zn^(2+)])/([Ni^(2+)])` At equilibrium `E_(cell)=0` `therefore E_(cell)^@=0.0591/2 log""([Zn^(2+)])/([Ni^(2+)])` `E_(Ni^(2+))^@,Ni-E_(Zn^(2+))^@, Zn=0.0591/2 log""([Zn^(2+)])/([Ni^(2+)])` `=-0.24-(-0.75)= 0.0591/2 log""([Zn^(2+)])/([Ni^(2+)])` `log""([Zn^(2+)])/([Ni^(2+)])=17.259` TAKING antilog `([Zn^(2+)])/([Ni^(2+)])=1.816 times 10^17` This concentration ratio shows that almost whole of the `Ni^(2+)` ions are reduced to Ni and therefore the concentration of `Zn^(2+)` PRODUCED from Zn would be nearly 1M `[because Ni(NO_3)_2=1M]` Thus, `1/([Ni^(2+)])=1.816 times 10^17` or `[Ni^(2+)]=5.5 times 10^-18 M` |
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