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Balancing a reduction-oxidation
reaction is a bit trickier than balancing a normal chemical equation. Follow
these steps:
- Break the reaction into an oxidation 1/2 reaction and a reduction 1/2
reaction. Do this by comparing the oxidation
numbers on the products and reactants.
- Balance
each 1/2 reaction
- Recombine the balanced 1/2 reactions so that the electrons cancel out.
The last step is necessary since there can be no net charge change
during a redox reaction.
An example is the clearest way to do this: let's balance the reaction
Fe+3(aq) + Cu(s) -> Fe(s) + Cu+2
To begin, we need to seperate the reaction into a reduction and oxidation.
The iron is being reduced, the copper oxidized. The two unbalanced half
reactions are
- Reduction: Fe+3(aq) -> Fe(s)
- Oxidation: Cu(s) -> Cu+2
Once these
half
reactions are balanced, we have the reactions
- Fe+3(aq) +3e- -> Fe(s)
- Cu(s) -> Cu+2 +2e-
We need to recombine these reactions to cancel the electrons. The iron
reduction has three electrons, the copper has two. If we multiply the top
equation by 2 and the bottom by three, each has six electrons. We can now
combine them and cancel the electrons
- 2Fe+3(aq) +6e- -> 2Fe(s)
- + 3Cu(s) -> 3Cu+2 +6e-
-
- = 2Fe+3(aq) + 3Cu(s) -> 2Fe(s) +
3Cu+2(aq)
The reaction is now balanced.
Example: Balance the following redox reaction in acid
- Co(s) + NO3-(aq) -> Co+2(aq) +
NO(g)
Solution: Start by breaking the reaction up into two half reactions
- Oxidation: Co(s) -> Co+2(aq)
- Reduction: NO3-(aq) ->
NO(g)
Balance
each half reaction in acid. When done, you have
- Co(s) -> Co+2(aq) + 2e-
- NO3-(g) + 3e- + 4H+(aq) ->
NO(g) + 2H2O(l)
The cobalt oxidation has two electrons, the
nitrate reduction has three. Multiply the top reaction by 3 and the bottom by 2
to get 6 electrons on both sides, then add
- 3Co(s) -> 3Co+2(aq) + 6e-
- + 2NO3-(g) + 6e- +
8H+(aq) -> 2NO(g) + 4H2O(l)
-
- = 3Co(s) + 2NO3- + 8H+(aq)
-> 3Co+2(aq) + 2NO(g) + 4H2O(l)
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