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For the cell reaction : Sn (s) + Pb^(2+) (aq) to Sn^(2+) (aq) + Pb (s) E_(Sn^(2+)|Sn)^(@) = -0.140 , E_(Pb^(2+) |Pb)^(@) = -0.126 V Calculate the ratio of concentration of Pb^(2+) to Sn^(2+) ion at which the cell reaction be reversed . |
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Answer» Solution :For the cell , `E^(@) = E_(PB^(2+)|Pb)^(@) - E_(Sn^(2+) |Sn)^(@) = -0.126 - (-0.140)` `= 0.014 V ` Applying Nernst equation `E = E^(@) - (0.059)/(2) log ([Sn^(2+)])/([Pb^(2+)])` `= 0.014 + (0.059)/(2) "log" ([Pb^(2+)])/([Sn^(2+)])` At equilibrium , E = 0 `therefore 0.014 + (0.059)/(2) log ([Pb^(2+)])/([Sn^(2+)]) = 0` or log `([Pb^(2+)])/([Sn^(2+)]) = - (0.014xx 2)/(0.059) = -0.474` `therefore ([Pb^(2+)])/([Sn^(2+)]) =` antilog `(-0.474) = 0.336` Thus ,the cell reaction will occur TILL `([Pb^(2+)])/([Sn^(2+)])` is more than `0.336` V . When `([Pb^(2+)])/([Sn^(2+)])` BECOMES less than 0.336 V , `E_(cell)` will become negative and reaction will be reversed . |
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