These two half reactions can be combined to get a net equation for the reaction that occurs in the cell.
But how should they be combined? You can't simply add them; that would imply that both the lead(II) and magnesium(II) ions were consuming electrons from some outside source. That isn't what is observed experimentally, either. You have to couple the two half reactions, so that one metal is donating electrons to the other metal's ions. This can be accomplished by reversing one of the reactions and THEN adding the two together.
How do you decide which of the reactions should be reversed?
The voltages are reduction potentials; the higher the number, the easier it is to reduce the metal ion. Now -.126 V is higher than -2.363 V, which means that the lead(II) ion will be easier to reduce than Mg ion. Leave the lead reaction as it is- a reduction.
The magnesium reaction should be reversed to make it an oxidation:
See where the electrons are coming from? The Mg gives them up,
and the Pb2+accepts them.
|Mg(s) = Mg2+ + 2e-
||E0 = +2.363 V
|Pb2+ + 2e- = Pb(s)
||E0 = -0.126 V
You can easily tell which half reaction is an oxidation and which is a reduction by remembering "OIL RIG": Oxidation Involves Loss of electrons, Reduction Involves Gain.
Adding the two half reactions together gives the net reaction:
Mg(s) + Pb2+ = Mg2+ + Pb(s)
Author: Fred Senese firstname.lastname@example.org