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.

1 Answer
Dec 5, 2017

-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.