Question

The `E_(cell)^(0)` for the reaction `Fe+Zn^(2+)hArrZn+Fe^(3+)` is `-0.32` volt at `25^(@)C`. What will be the equilibrium concentration of `Fe^(2+)`, when a piece of iron is placed in a `1MZn^(2+)` solution?

Answer 1

We have the nearest equation at equilibrium at `25^(@)C`
`E^(@)=(0.0591)/(n)logK` … (i)
Since `E_(cell^(@)` for the given reaction is negative, therefore the reverse reaction is feasible for which `E_(cell)^(@)` will be `+0.32V`, thus for `Zn+Fe^(2+)hArrFe+Zn^(2+),E_(cell)^(@)+0.32V`
Now, `E^(@)=(0.0591)/(n)log(([Zn^(2+)])/([Fe^(2+)]))` or `0.32=(0.0591)/(2)log((1)/([Fe^(2+)]))`
`log[Fe^(2+)]=-10.829` taking antilog,
`[Fe^(2+)]=1.483xx10^(-11)M`