Redox (reduction-oxidation) reactions include all chemical reactions in which atoms have their oxidation state changed. Oxidation is the loss of electrons—or the increase in oxidation state—by a molecule, atom, or ion. Reduction is the gain of electrons—or the decrease in oxidation state—by a molecule, atom, or ion. Figure 1 Describing the overall electrochemical reaction for a redox process requires balancing the component half-reactions for oxidation and reduction. In general, for reactions in aqueous solution, this involves adding H+, OH−, H2O, and electrons to compensate for oxidation changes.
There are several general rules to follow when balancing redox equations:
- Write the oxidation and reduction half reactions for the whole compound, not just the element that is reduced/oxidized.
- Balance both reactions for all elements except oxygen and hydrogen.
- If the oxygen atoms are not balanced in either reaction, add water molecules to the side missing the oxygen. If the hydrogen atoms are not balanced, add hydrogen ions.
- Multiply the half reactions by the appropriate number so that they have equal numbers of electrons.
- Add the two equations to cancel out the electrons, as in the previous method. The equation should be balanced.
If the reaction occurs in base, proceed as if it is in an acid environment, but, after step 4, add a hydroxide ion to both sides of the equation for each hydrogen ion added. Then, combine the hydroxide and hydrogen ions to form water. Next, cancel all of the water molecules that appear on both sides.
Both acidic and basic media conditions are explored more in depth below.
In acidic media, H+ ions and water are added to half reactions to balance the overall reaction. For example, when manganese(II) reacts with sodium bismuthate:
- Unbalanced reaction: Mn2+(aq) + NaBiO3(s) → Bi3+(aq) + MnO4− (aq)
This reaction can be split into its two half reactions, one representing the oxidation and one representing the reduction.
- Oxidation: 4 H2O(l) + Mn2+(aq) → MnO4-(aq) + 8 H+(aq) + 5 e−
- Reduction: 2 e− + 6 H+ + BiO3-(s) → Bi3+(aq) + 3 H2O(l)
The reaction is balanced by scaling the two half-cell reactions to involve the same number of electrons. Multiply the oxidation reaction by the number of electrons in the reduction step and vice versa:
- 8 H2O(l) + 2 Mn2+(aq) → 2 MnO4-(aq) + 16 H+(aq) + 10 e−
- 10 e− + 30 H+ + 5 BiO3-(s) → 5 Bi3+(aq) + 15 H2O(l)
Adding these two reactions eliminates the electrons terms and yields the balanced reaction:
- 14 H+(aq) + 2 Mn2+(aq) + 5 NaBiO3(s) → 7 H2O(l) + 2 MnO4-(aq) + 5 Bi3+(aq) + 5 Na+(aq)
In basic media, OH− ions and water are added to half reactions to balance the overall reaction. For example, in the reaction between potassium permanganate and sodium sulfite:
Unbalanced reaction: KMnO4 + Na2SO3 + H2O → MnO2 + Na2SO4 + KOH
As in acidic media, the unbalanced reaction can be separated into its two half reactions, each representing either the reduction or oxidation.
- Reduction: 3 e− + 2 H2O + MnO4− → MnO2 + 4 OH−
- Oxidation: 2 OH− + SO32− → SO42− + H2O + 2 e−
Balancing the number of electrons in the two half-cell reactions gives:
- 6 e− + 4 H2O + 2 MnO4− → 2 MnO2 + 8 OH−
- 6 OH− + 3 SO32− → 3 SO42− + 3 H2O + 6 e−
Adding these two half-cell reactions together gives the balanced equation:
- 2 KMnO4 + 3 Na2SO3 + H2O → 2 MnO2 + 3 Na2SO4 + 2 KOH