Balancing Redox Reaction

Balance the following reaction in an acidic medium.
Identify the oxidizing agent and the reducing agent.

MnO₄^- + Fe²⁺ -> Mn²⁺ + Fe³⁺

Click for the balanced reaction.

Step 1
Determine the oxidation number of each species in the unbalanced reaction.

Left-hand-side Right-hand-side

Mn +7 +2

Fe +2 +3

Fe²⁺ is oxidized. Mn in MnO₄ is reduced.
Step 2
Write two separate equations, one using only the substances that are involved in the reduction, and another using only the substances involved in oxidation. (When balanced, these are called half-reaction equations.)

Oxidation: Fe²⁺ -> Fe³⁺

Reduction: MnO₄^- -> Mn²⁺
Step 3
Balance each kind of atom other than H and O by using coefficients.
- They are balanced in these equations.
Step 4
Balance oxygen atoms by using H₂O.

Oxidation: Fe²⁺ -> Fe³⁺

Reduction: MnO₄^- -> Mn²⁺ + 4 H₂O
Step 5
Balance H atoms by using H⁺ ions.

Oxidation: Fe²⁺ -> Fe³⁺

Reduction: MnO₄^- + 8 H⁺ -> Mn²⁺ + 4 H₂O
Step 6
Balance the charge on each side of the reactions by adding electrons as necessary.
Electrons appear on the right-hand-side for the oxidation half-reaction.
Electrons appear on the left-hand-side for the reduction half-reaction.

Oxidation half-reaction: Fe²⁺ -> Fe³⁺ + e^-

Reduction half-reaction: MnO₄^- + 8 H⁺ + 5 e^- -> Mn²⁺ + 4 H₂O
Step 7
Multiply the half-reactions by the simplest set of whole numbers so that
electrons gained = electrons lost

Oxidation half-reaction: 5[Fe²⁺ -> Fe³⁺ + e^-]

Reduction half-reaction: 1[MnO₄^- + 8 H⁺+ 5 e^- -> Mn²⁺ + 4 H₂O]
Step 8
Combine the oxidation half-reaction with the reduction half-reaction. Cancel electrons and equal amounts of any substance that appears on both sides of the equation.
The balanced redox reaction is:

MnO₄^- (aq) + 5 Fe²⁺ (aq) + 8 H⁺ (aq) -> Mn²⁺ (aq) + 5 Fe³⁺ (aq) + 4 H₂O (l)

MnO₄^- is the oxidizing agent.

Fe²⁺ is the reducing agent.
Step 9
Check to make sure all atoms and charges are balanced, and to see that all substances have whole number coefficients that are in their simplest ratios.

Left-hand-side Right-hand-side

Mn 1 1

Fe 5 5

H 8 8

O 4 4

charge + 17 + 17

Left-hand-side = Right-hand-side

Oxidation:	Fe²⁺ -> Fe³⁺
Reduction:	MnO₄^- -> Mn²⁺ + 4 H₂O

Oxidation half-reaction:	Fe²⁺ -> Fe³⁺ + e^-
Reduction half-reaction:	MnO₄^- + 8 H⁺ + 5 e^- -> Mn²⁺ + 4 H₂O

Oxidation half-reaction:	5[Fe²⁺ -> Fe³⁺ + e^-]
Reduction half-reaction:	1[MnO₄^- + 8 H⁺+ 5 e^- -> Mn²⁺ + 4 H₂O]

	Left-hand-side	Right-hand-side
Mn	+7	+2
Fe	+2	+3

	Left-hand-side	Right-hand-side
Mn	1	1
Fe	5	5
H	8	8
O	4	4
charge	+ 17	+ 17