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
Michael
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.

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