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Balancing Redox Reactions
Redox reactions are chemical equations where oxidation and reduction are happening simultaneously, meaning at the same time.
There are six steps to balancing a redox reaction. To fully understand this, you first must read about oxidation and reduction.
Background:
What is oxidation and reduction?
Oxidation - When a substance loses electrons (oxidizes). Oxidation is signified by an increase in oxidation number. In other context oxidation means the removal of hydrogen and addition of oxygen.
Reduction - When a substance gains electrons (reduces). Reduction is signified by a decrease in oxidation number. In other context reduction means the addition of hydrogen and loss of oxygen.
We will be using:
Fe + MnO → Fe + Mn
2+ - 3+ 2+
How to Find Oxidation Number
4
It is necessary to know the oxidation number of both the reactants and products.
Take Mn as "X". The charge of oxygen (2-) is given. There are four oxygen atoms in MnO . Four oxygen atoms will have a charge of 4 x -2 = -8.
The negative sign above MnO tell us that it has a charge of -1, so we will set our equation equal to -1. Solve X-8=-1. X=7
The charge of MnO in the reactant side is 7+ while the charge of Mn on the product side is 2+. This tells us that MnO is getting reduced while Fe is getting oxidized.
Fe has an oxidation number of +2 on the reactant side and +3 on the product side. The increase in oxidation number means Fe is oxidized.
Read: Oxidation and Reduction
-
4
4
4
4
-
-
-
Step 1: Split in to half reactions + balance:
This means split the equation in to two half reaction. One half reaction will be oxidation, and the other will be reduction.
Oxidation: Fe → Fe
Reduction: MnO → Mn
2+
- 2+
3+
4
In this case, the equations are already balanced. If not, balance as you would in an ionic equation. How to balance ionic equations.
This part takes place in the half reaction in which oxygen is already present.
2
Step 2: Balance oxygen by adding H O:
There are four oxygen atoms on the reactant side so we must have four on the product side as well. This will be done by adding four water molecules (4H O).
4
Reduction: MnO → Mn
- 2+
Reduction: MnO → Mn + 4H O
- 2+
4
2
2
Step 3: Balance hydrogen:
Now that we have added 4H O, we must balance the hydrogen on both sides. Add 8 hydrogen ions to the reactant side
2
Reduction: 8H + MnO → Mn + 4H O
+ - 2+
4
2
Step 4: Balance the charges:
4
Now we will address the charges. First count the charges of both sides of the reduction equation. 8H has a charge of +8 and MnO has a charge of -1. Together they make a charge of +7.
Mn has a charge of +2 and 4H O is neutral meaning it has a charge of 0. Together they have a charge of +2.
To have an equal charge on both sides, add five electrons to the reactant side.
Reduction: 8H + MnO + 5e → Mn + 4H O
+ - - 2+
4
2
Step 5: Multiply oxidation half reaction:
Take the oxidation half reaction and multiply everything by five to balance the electrons. The half reaction Fe → Fe can be written as Fe → Fe + e .
However, we need the electrons to cancel out so we multiply the whole half reaction by five to get five electrons.
2+
3+
2+ 3+ -
Oxidation: 5 Fe → 5 Fe + 5 e
2+ 3+ -
Step 6: Combine half reactions:
Lastly, we will combine the two half reactions again. Keep in mind that the electrons will cancel out.
Recount charges and atoms to double check.
8H + 5 Fe + MnO → 5 Fe + Mn +4H O
+ 2+ - 3+ 2+
4
2
Remember To:
Balancing Redox Reactions
- Find necessary oxidation numbers
- Carefully count charges and atoms
- Cancel out atoms when finished
- Check in between steps