D39.3 Balancing Redox Reactions

Redox reactions frequently occur in aqueous solutions, which can be acidic, basic, or neutral. Moreover, H2O molecules may actively participate in the reaction, and depending on the conditions, H3O+ (present under acidic conditions) or OH (present under basic conditions) may also be a reactant or a product.

Half-reactions make it easier to balance redox reactions because you can balance the oxidation half-reaction separately from the reduction half-reaction, and make certain that the number of electrons lost in the oxidation process equals the number of electrons gained in the reduction process.

We will balance an example redox reaction in acidic solution and one in basic solution to illustrate the process of balancing redox reactions, and highlight how the nature of the solution can play a role. (A neutral solution may be treated as acidic or basic, though treating it as acidic is usually easier.)

Acidic Solution

Consider the unbalanced reaction:

MnO4‾(aq) + Fe2+(aq) ⟶ Mn2+(aq) + Fe3+(aq)

where iron underwent oxidation because Fe2+ has lost an electron to become Fe3+, and manganese underwent reduction because it gained five electrons to change from an oxidation state of +7 to an oxidation state of +2.

Oxidation (unbalanced): Fe2+(aq) Fe3+(aq)
Reduction (unbalanced): MnO4‾(aq) Mn2+(aq)

For redox reactions, it is a useful simplification to represent H3O+(aq) as H+(aq): where there is a need for balancing O atoms, we can involve H2O(ℓ), and then use H+(aq) to balance the H atoms.

For instance, in the above reduction half-reaction, there are four O atoms on the reactant side and none on the product side. To balance the elements, we can add 4 H2O(ℓ) to the product side, and then to balance H atoms, add 8 H+(aq) to the reactant side:

Reduction (charge not balanced):          MnO4(aq) + 8H+(aq) ⟶ Mn2+(aq) + 4H2O(ℓ)

Once the atoms have been balanced, we need to balance the electric charge for each half-reaction. For the oxidation half-reaction, the total charge on the reactant side is +2 and the total charge on the product side is +3, so charge is unbalanced. We use electrons to balance the charge. Adding a single electron on the product side gives a balanced oxidation half-reaction:

Oxidation (balanced):          Fe2+(aq) ⟶ Fe3+(aq) + e‾

In oxidation half-reactions, electrons appear on the product side of the equation. Because iron is oxidized, iron is the reducing agent in this redox reaction.

You should always check that the half-reaction is balanced for the number of atoms of each element and the total charge:

Fe: (1 atom in Fe2+)·(1 Fe2+) = 1 (1 atom in Fe3+)·(1 Fe3+) = 1 1 = 1
Charge: 1·(+2) = +2 1·(+3) + 1·(-1) = +2 +2 = +2

If the atoms and charges are balanced, then the half-reaction itself is balanced.

For the reduction half-reaction, we have balanced the atoms but not the charge. The total charge on the reactant side is +7, the total charge on the product side is +2. Therefore, it is necessary to add five electrons to the reactant side to achieve charge balance:

Reduction (balanced):          MnO4(aq) + 8H+(aq) + 5e ⟶ Mn2+(aq) + 4H2O(ℓ)

In reduction half-reactions, electrons appear on the reactant side. The species that was reduced, MnO4, is the oxidizing agent in this redox reaction.

Again, check that the half-reaction is balanced for the number of atoms of each element and the total charge:

Mn: (1 atom in MnO4)·(1 MnO4) = 1 (1 atom in Mn2+)·(1 Mn2+) = 1 1 = 1
O: (4 atoms in MnO4)·(1 MnO4) = 4 (1 atom in H2O)·(4 H2O) = 4 4 = 4
H: (1 atom in H+)·(8 H+) = 8 (2 atoms in H2O)·(4 H2O) = 8 8 = 8
Charge: 1·(-1) + 8·(+1) + 5·(-1) = +2 1·(+2) + 4·(0) = +2 +2 = +2

We now have two balanced half-reactions:

Oxidation: Fe2+(aq) Fe3+(aq) + e‾
Reduction: MnO4(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(ℓ)

The key to combining the half-reactions is the electrons: the number of electrons generated by the oxidation half-reaction must equal the number of electrons consumed by the reduction half-reaction. Here, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five. Therefore, it is necessary to first multiply the oxidation half-reaction by five, then sum the resulting half-reactions:

Oxidation: 5 × [Fe2+(aq) Fe3+(aq) + e‾]
Reduction: MnO4(aq) + 8 H+(aq) + 5e Mn2+(aq) + 4 H2O(ℓ)
overall: 5 Fe2+(aq) + MnO4(aq) + 8 H+(aq) 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(ℓ)

This is the overall balanced equation in an acidic solution. Electrons do not appear in the overall reaction equation because all electrons lost in the oxidation half-reaction are gained in the reduction half-reaction. The number of atoms of each element and the total electric charge should be the same on both side of the overall equation. Be sure to carefully check each side to verify everything has been balanced correctly.

Basic Solution

The simplest way to balance a redox reaction in a basic solution is to start with the balanced equation in acidic solution, then “convert” H+(aq) to OH(aq) (there is an excess of OH ions instead of H3O+ ions in basic solutions). For example, when balancing the following reaction in basic solution:

MnO4(aq) + Cr(OH)3(s) ⟶ MnO2(s) + CrO42-(aq)

start by collecting the species given into unbalanced oxidation and reduction half-reactions:

Oxidation (unbalanced): Cr(OH)3(s) CrO42-(aq)
Reduction (unbalanced): MnO4‾(aq) MnO2(s)

For the oxidation half-reaction, we can add one H2O molecule to the reactant side to balance oxygen atoms, and then balance hydrogen atoms with five H+(aq) on the product side (again, we do the initial balancing by assuming acidic solution):

Oxidation (charge not balanced):          Cr(OH)3(s) + H2O(ℓ) ⟶ CrO42-(aq) + 5H+(aq)

The reactant side of the equation has a total charge of 0, and the product side a total charge of +3. Adding three electrons to the product side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):

Oxidation (balanced):          Cr(OH)3(s) + H2O(ℓ) ⟶ CrO42-(aq) + 5H+(aq) + 3e‾

Checking the half-reaction:

Cr: (1 atom in Cr(OH)3)·(1 Cr(OH)3) = 1 (1 atom in CrO42-)·(1 CrO42-) = 1 1 = 1
O: (3 atoms in Cr(OH)3)·(1 Cr(OH)3)
+ (1 atom in H2O)·(1 H2O) = 4
(4 atoms in CrO42-)·(1 CrO42-) = 4 4 = 4
H: (3 atoms in Cr(OH)3)·(1 Cr(OH)3)
+ (2 atoms in H2O)·(1 H2O) = 5
(1 atom in H+)·(5 H+) = 5 5 = 5
Charge: 1·(0) + 1·(0) = 0 1·(-2) + 5·(+1) + 3·(-1) = 0 0 = 0

For the reduction half-reaction, we need two H2O molecules on the product side to balance oxygen atoms. Then to balance hydrogen atoms, we need to add four H+(aq) on the reactant side:

Reduction (charge not balanced):          MnO4(aq) + 4H+(aq) ⟶ MnO2(s) + 2H2O(ℓ)

Then add three electrons to the reactant side to balance the charge:

Reduction (balanced):          MnO4(aq) + 4H+(aq) + 3e‾ ⟶ MnO2(s) + 2H2O(ℓ)

Make sure to check the half-reaction:

Mn: (1 atom in MnO4)·(1 MnO4) = 1 (1 atom in MnO2)·(1 MnO2) = 1 1 = 1
O: (4 atoms in MnO4)·(1 MnO4) = 4 (2 atoms in MnO2)·(1 MnO2)
+ (1 atom in H2O)·(2 H2O) = 4
4 = 4
H: (1 atom in H+)·(4 H+) = 4 (2 atoms in H2O)·(2 H2O) = 4 4 = 4
Charge: 1·(-1) + 4·(+1) + 3·(-1) = 0 1·(0) + 2·(0) = 0 0 = 0

Collecting what we have so far:

Oxidation: Cr(OH)3(s) + H2O(ℓ) CrO42-(aq) + 5H+(aq) + 3e‾
Reduction: MnO4(aq) + 4H+(aq) + 3e‾ MnO2(s) + 2H2O(ℓ)

In this case, both half-reactions involve the same number of electrons, and therefore we can simply add the two half-reactions together and simplify:

Cr(OH)3(s) + H2O(ℓ) + MnO4(aq) + 4 H+(aq) + 3 e‾ CrO42-(aq) + 5 H+(aq) + 3 e‾ + MnO2(s) + 2 H2O(ℓ)
Cr(OH)3(s) + MnO4(aq) CrO42-(aq) + H+(aq) + MnO2(s) + H2O(ℓ)

This is the balanced redox equation in an acidic solution. To “convert” to basic solution, add OH(aq) to both sides of the equation to “react” with all the H+(aq). This converts H+(aq) to H2O(ℓ) on one side and adds OH(aq) on the other side:

Cr(OH)3(s) + MnO4(aq) + OH(aq) CrO42-(aq) + H+(aq) + OH(aq) + MnO2(s) + H2O(ℓ)
Cr(OH)3(s) + MnO4(aq) + OH(aq) CrO42-(aq) + MnO2(s) + 2H2O(ℓ)

This is the balanced equation in basic solution. Checking each side of the equation:

Cr: (1 atom in Cr(OH)3)·(1 Cr(OH)3) = 1 (1 atom in CrO42-)·(1 CrO42-) = 1 1 = 1
Mn: (1 atom in MnO4)·(1 MnO4) = 1 (1 atom in MnO2)·(1 MnO2) = 1 1 = 1
O: (3 in Cr(OH)3)·(1) + (4 in MnO4)·(1)
+ (1 in OH)·(1) = 8
(4in CrO42-)·(1) + (2 in MnO2)·(1)
+ (1 in H2O)·(2) = 8
8 = 8
H: (3 in Cr(OH)3)·(1) + (1 in OH)·(1) = 4 (2 atoms in H2O)·(2 H2O) = 4 4 = 4
Charge: 1·(0) + 1·(-1) + 1·(-1) = -2 1·(-2) + 1·(0) + 2·(0) = -2 -2 = -2

Exercise: Balancing Oxidation-Reduction Reactions

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Chemistry 109 Fall 2021 by John Moore, Jia Zhou, and Etienne Garand is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.