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 .

Explanation

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+)]) = - (0.014 xx 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 .