Balancing Redox Reactions
To fully balance a redox reaction, you must balance:
Atoms (as you would in a regular equation)
Electrons lost/gained
Total charge
Often, redox reactions are shown and balanced as net ionic equations. In this case, balancing them is often a simple matter of making sure that the same number of electrons are produced by the oxidation half-reaction and consumed by the reduction of half-reaction.
For example, consider the unbalanced net ionic equation:
Al (s) + Zn 2+ (aq) —> Al 3+ (aq) + Zn (s)
In this reaction, Al is oxidized from Al 0 to Al +3, and Zn is reduced from Zn 2+ to Zn 0. The atoms appear balanced, but Zn 2+ needs only 2 electrons to form Zn 0, but Al 0 produces 3 electrons when oxidized to Al 3+.
Al 0 —> Al 3+ + 3 e-
Zn 2+ + 2 e- —> Zn 0
To balance the electrons, we need to multiply the first half-reaction by 2, and the second one by 3, giving:
2 Al 0 —> 2 Al 3+ + 6 e-
3 Zn 2+ + 6 e- —> 3 Zn 0
If we combine these and cancel the electrons (because we have the same number on both sides), we get the balanced net ionic equation:
2 Al 0 (s) + 3 Zn 2+ (aq) —> 2 Al 3+ (aq) + 3 Zn 0 (s)
Balancing Redox Reactions in Acidic or Basic Solutions
In some redox reactions that take place in acid or base, H+ or OH- ions form the solution and participate in the redox reaction. This makes balancing them a little more complicated.
In these reactions, in addition to balancing atoms and electrons, we also need to balance the total positive and negative charges on both sides, using H+ or OH- ions from the solution and forming H2O
In order to keep track of everything, I use the mnemonic AEIOU
A = Atoms. Balance the atoms (except for H and O) in each half-reaction.
E = Electrons. Add the electrons gained or lost to each half-reaction.
I = Ions. Add H+ (if the reaction is in acid) or OH- (if the reaction is in base) to balance the charges on each side of each half-reaction
O = Oxygen and Hydrogen. Balance oxygen and hydrogen in each half-reaction by adding H2O molecules as needed
U = Unite. Unite the two half-reactions by multiplying all coefficients in each one by the factor needed to make the electrons in one cancel the electrons in the other
Then cancel anything that appears on both sides to get the balanced net ionic equation.
For example:
Balance the following reaction in basic solution:
Ca (s) + Cr2O7 2- (aq) —> CaO (s) + Cr 3+ (aq)
The half-reactions are:
Ca 0 (s) —> CaO (s)
Cr2O7 2- (aq0 —> Cr 3+ (aq)
Atoms in each half-reaction:
The first half reaction is already balanced for calcium atoms.
Ca 0 —> CaO
We need 2 Cr 3+ ions to balance the 2 Cr atoms in Cr2O7 2-
Cr2O7 2- —> 2 Cr 3+
Electrons in each half-reaction:
Ca is being oxidized from Ca 0 to Ca +2. This produces 2 electrons:
3 Ca 0 —> Ca+2O-2 + 6 e-
Cr is being reduced from Cr +6 in Cr2O7 2- to Cr 3+. This consumes 3 electrons per Cr atom, or 6 electrons in total
Cr2+6O72- + 6 e- —> 2 Cr 3+
Ions of H+ and OH- in each half-reaction to balance charges.
The total charge (ions and electrons) in the first half-reaction is 0 on the left and -2 on the right.
6 OH- (aq) + 3 Ca 0 —> 3 CaO + 6 e-
In the second-half reaction, the total charge is -8 on the left and +6 on the right. So we add 14 OH- ions to the right to bring the total charge on both sides to -8.
Cr2O7 2- + 6 e- —> 2 Cr 3+ + 14OH- (aq)
Oxygen and Hydrogen atoms in each half-reaction:
In the first half-reaction, there are 2 unbalanced H atoms and one unbalanced O atom on the left. So to balance, we add 3 H2O molecules on the right.
3 Ca 0 + 6 OH- —> 3 CaO + 6 e- + H2O
The second half-reaction has 14 unbalanced H atoms and 7 unbalanced O atoms on the right. So we add 7 H2O molecules on the left.
7 H2O + Cr2O7 2- + 6 e- —> 2 Cr 3 + + 14 OH-
Unite the half-reactions with equal numbers of electrons produced/consumed: (we already did this part, so now all we need to do is unite the two half-reactions and cancel anything on both sides)
To unite together:
3 Ca 0 + 6 OH- + 7 H2O + Cr2O7 2- + 6 e- —> 3 CaO + 6 e- + 3 H2O + 2 Cr 3+ + 14 OH-
Cancel everything that occurs on both sides to get net ionic form:
3 Ca 0 (s) + 4 H2O (l) + Cr2O7 2- (aq) —> 3 CaO (s) + 2 Cr 3+ (aq) + 8 OH- (aq)