Question

What is the electrode potential for `Pb^(2+)(aq) + 2e^(-) rightleftharpoons Pb(s)` when the concentration of `Pb^(2+) (aq)` is `0.05` `mol` `dm^(-3)` ?

Temperature is 298 K and the standard electrode potential for the half-cell reaction is `-0.13` V.

Answer 1

-0.16 V

Explanation:

The expression for the electrode potential of a half - cell is:

`sf(E=E^(@)-(RT)/(zF)ln([["reduced form"]]/[["oxidised form"]]))`

For this half - cell at 298K this can be simplified to:

`sf(E=E^(@)+0.0591/(2)log[Pb^(2+)])`

`:.` `sf(E=-0.13+0.051/(2)log[0.05])`

`sf(E=-0.13-0.3317=-0.16color(white)(x)V)`

We would expect the potential to become more -ve like this since Le Chatelier's Principle predicts that reducing `sf([Pb^(2+)])` will cause the equilibrium to shift to the left thus pushing out more electrons.