Any reaction where electrons are transferred between reactants is called an reduction-oxidation reaction or redox reaction. These are reactions where one substance wants an electron so badly that it takes it away from another substance. Whether or not it succeeds depends on who it meets. For example, consider the reaction
FeCl2(aq) + CeCl4(aq) → FeCl3(aq) + CeCl3(aq)
What happened here? If we remove the spectator ions and write the net ionic equation we find:
Fe2+(aq) + Ce4+(aq) → Fe3+(aq) + Ce3+(aq)
Ce4+ took an electron from Fe2+! This is an oxidation/reduction reaction. In this example, Fe2+ is oxidized and Ce4+ is reduced. The charge of Fe went from +2 to +3, that is, it lost an electron. This process is called oxidation.
Oxidation: The loss of an electron by a substance.
Likewise, the charge of Ce went from +4 to +3, that is, it gained an electron. This process is called reduction.
Reduction: The gain of an electron by a substance.
Half-Reactions
Oxidation/reduction reactions are important because we can exploit them as a way of generating electrical current. For example, we know that Ce4+ will pull an electron away from Fe2+ when we mix the two in solution. The trick to making a battery is to find a way to make Ce4+ pull an electron from Fe2+ when they are not mixed together in a single solution.
In order to do this we set up two 1/2 reactions in separate beakers and connect them with a salt bridge. The salt bridge electrically connects the two beakers, but prevents Fe2+ and Ce4+ from mixing.
You need two 1/2 reactions to make a reaction, so, we add them together.
Fe2+ → Fe3+ + e- |
Ce4+ + e- → Ce3+ |
__________________________________ |
Fe2+ + Ce4+ →Fe3+ + Ce3+ |
Notice that electrons on both sides of the half-reactions must cancel each other out when added together.
What about other atoms and molecules. How do you know if one chemical substance is strong enough to take an electron from another? We simply refer to a list known as the activity series.
Homework from Chemisty, The Central Science, 10th Ed.
4.19, 4.21, 4.23, 4.25, 4.27, 4.39, 4.41, 4.43, 4.45, 4.47, 4.49, 4.51, 4.53, 4.55, 4.57
To Balance a Redox Reaction Equation Carry out the following steps. This can seem a little complicated at first, but it helps to look at the example lower on the page:
- Write out the unbalanced oxidation and reduction half equations.
- Balance each half equation.
- Balance the elements being oxidized and reduced, ignoring H and O at this stage.
- Balance O atoms by adding H2O.
- Then balance H by either:
- adding H+ (for acidic solutions) or
- adding OH- (for basic solutions).
- Balance charge on the left vs the right side of each half equation by adding electrons if needed.
- Multiply the half equations by appropriate factors so that each involves the transfer of the same number of electrons.
- Combine the half reactions to eliminate the electrons from the overall reaction.
Example
Here is a redox reaction in acid solution that we're going to balance:
Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
Step A
Write the unbalanced half equations:
Fe2+(aq) → Fe3+(aq) (oxidation)
MnO4-(aq) → Mn2+(aq) (reduction)
Step B
Balance each half reaction.
Iron Half Reaction: To balance this equation, we simply need to add an electron to the right side of the equation.
Fe2+(aq) → Fe3+(aq)) + e-
Permanganate Half Reaction: We add water to the right to balance the four oxygens on the left.
MnO4-(aq) → Mn2+(aq) + 4H2O
Next, to balance the eight hydrogens in the water, we add 8H+(aq) on the left.
MnO4-(aq) + 8H+(aq) → Mn2+(aq) + 4H2O
Now, looking at the equation above, we can see the charges are unbalanced on the left vs the right side. The overall charge on the left is +7 and on the right is +2. To balance the charge, we need to add five electrons to the left side of the equation.
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O
Step C
Multiply the half equations by appropriate numbers so they each contain the
same number of free electrons on each side. In this case, we must multiply the iron reaction by 5.
Step D
Add the half reactions. The electrons on each side cancel to give the overall reaction:
5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq)) + Mn2+(aq) + 4H2O