- Steps
- - General
- Check oxidation numbers of reactants and products
- Identify the oxidizing and reducing species
- Write the half-reactions and balance atoms and charges
- Substitute coefficients of the ionic equation into the general equation
- - Balance in acid medium
- - Balance in basic medium
- Examples
- Exercises
- Exercise 1
- General steps
- Balance in basic medium
- Exercise 2
- General steps
- Balance in acid medium
- References
The redox balancing method is one that allows balancing the chemical equations of redox reactions, which would otherwise be a headache. Here one or more species exchange electrons; the one that donates or loses them is called the oxidizing species, while the one that accepts or gains them, the reducing species.
In this method it is essential to know the oxidation numbers of these species, since they reveal how many electrons they have gained or lost per mole. Thanks to this, it is possible to balance the electric charges by writing the electrons in the equations as if they were reactants or products.
General semi-reactions of a redox reaction together with the three protagonists during their balancing: H +, H2O and OH-. Source: Gabriel Bolívar.
The upper image shows how effectively electrons, e - are placed as reactants when the oxidizing species gains them; and as products when the reducing species loses them. Note that to balance this type of equations it is necessary to master the concepts of oxidation and oxidation-reduction numbers.
The H +, H 2 O and OH - species, depending on the pH of the reaction medium, allow redox balancing, which is why it is very common to find them in exercises. If the medium is acidic, we resort to the H +; but if on the contrary the medium is basic, then we use the OH - for balancing.
The nature of the reaction itself dictates what the pH of the medium should be. That is why, although it can be balanced assuming an acidic or basic medium, the final balanced equation will indicate whether the H + and OH - ions are really dispensable or not.
Steps
- General
Check oxidation numbers of reactants and products
Assume the following chemical equation:
Cu (s) + AgNO 3 (aq) → Cu (NO 3) 2 + Ag (s)
This corresponds to a redox reaction, in which a change occurs in the oxidation numbers of the reactants:
Cu 0 (s) + Ag + NO 3 (aq) → Cu 2+ (NO 3) 2 + Ag (s) 0
Identify the oxidizing and reducing species
The oxidizing species gains electrons by oxidizing the reducing species. Therefore, its oxidation number decreases: it becomes less positive. Meanwhile, the oxidation number of the reducing species increases, since it loses electrons: it becomes more positive.
Thus, in the previous reaction, copper is oxidized, since it passes from Cu 0 to Cu 2+; and silver is reduced, as it goes from Ag + to Ag 0. Copper is the reducing species, and silver the oxidizing species.
Write the half-reactions and balance atoms and charges
Identifying which species gain or lose electrons, the redox half-reactions for both the reduction and oxidation reactions are written:
Cu 0 → Cu 2+
Ag + → Ag 0
Copper loses two electrons, while silver gains one. We place the electrons in both half-reactions:
Cu 0 → Cu 2+ + 2e -
Ag + + e - → Ag 0
Note that the loads remain balanced in both half-reactions; but if they were added together, the law of conservation of matter would be violated: the number of electrons must be equal in the two half-reactions. Therefore, the second equation is multiplied by 2 and the two equations are added:
(Cu 0 → Cu 2+ + 2e -) x 1
(Ag + + e - → Ag 0) x 2
Cu 0 + 2Ag + + 2e - → Cu 2+ + 2Ag 0 + 2e -
The electrons cancel out because they are on the sides of the reactants and products:
Cu 0 + 2Ag + → Cu 2+ + 2Ag 0
This is the global ionic equation.
Substitute coefficients of the ionic equation into the general equation
Finally, the stoichiometric coefficients from the previous equation are transferred to the first equation:
Cu (s) + 2AgNO 3 (aq) → Cu (NO 3) 2 + 2Ag (s)
Note that 2 was positioned with AgNO 3 because in this salt silver is as Ag +, and the same happens with Cu (NO 3) 2. If this equation is not balanced at the end, we proceed to carry out the trial.
The equation proposed in the previous steps could have been balanced directly by trial and error. However, there are redox reactions that need an acidic (H +) or basic (OH -) medium to take place. When this happens, it cannot be balanced assuming that the medium is neutral; as just shown (neither H + nor OH - was added).
On the other hand, it is convenient to know that the atoms, ions or compounds (mostly oxides) in which the changes in oxidation numbers occur are written in the half-reactions. This will be highlighted in the exercises section.
- Balance in acid medium
When the medium is acid, it is necessary to stop at the two half-reactions. This time when balancing we ignore the oxygen and hydrogen atoms, and also the electrons. The electrons will balance in the end.
Then, on the side of the reaction with fewer oxygen atoms, we add water molecules to make up for it. On the other side, we balance the hydrogens with H + ions. And finally, we add the electrons and proceed by following the general steps already outlined.
- Balance in basic medium
When the medium is basic, one proceeds in the same way as in the acidic medium with a small difference: this time on the side where there are more oxygen, a number of water molecules equal to this excess oxygen will be located; and on the other side, OH ions - to compensate for hydrogens.
Finally, the electrons are balanced, the two half-reactions are added, and the coefficients of the global ionic equation are substituted into the general equation.
Examples
The following balanced and unbalanced redox equations serve as examples to see how much they change after applying this balancing method:
P 4 + ClO - → PO 4 3- + Cl - (unbalanced)
P 4 + 10 ClO - + 6 H 2 O → 4 PO 4 3- + 10 Cl - + 12 H + (balanced acid medium)
P 4 + 10 ClO - + 12 OH - → 4 PO 4 3- + 10 Cl - + 6 H 2 O (balanced basic medium)
I 2 + KNO 3 → I - + KIO 3 + NO 3 - (unbalanced)
3I 2 + KNO 3 + 3H 2 O → 5I - + KIO 3 + NO 3 - + 6H + (balanced acid medium)
Cr 2 O 2 7- + HNO 2 → Cr 3+ + NO 3 - (unbalanced)
3HNO 2 + 5H + + Cr 2 O 2 7- → 3NO 3 - + 2Cr 3+ + 4H 2 O (balanced acid medium)
Exercises
Exercise 1
Balance the following equation in basic medium:
I 2 + KNO 3 → I - + KIO 3 + NO 3 -
General steps
We begin by writing down the oxidation numbers of the species that we suspect have been oxidized or reduced; in this case, the iodine atoms:
I 2 0 + KNO 3 → I - + KI 5+ O 3 + NO 3 -
Note that iodine is oxidized and at the same time reduced, so we proceed to write their two respective half-reactions:
I 2 → I - (reduction, for every I - 1 electron is consumed)
I 2 → IO 3 - (oxidation, for every IO 3 - 5 electrons are released)
In the oxidation half-reaction we place the anion IO 3 -, and not the iodine atom as I 5+. We balance the iodine atoms:
I 2 → 2I -
I 2 → 2IO 3 -
Balance in basic medium
Now we focus on balancing the oxidation semi-reaction in a basic medium, since it has an oxygenated species. We add on the product side the same number of water molecules as there are oxygen atoms:
I 2 → 2IO 3 - + 6H 2 O
And on the left side we balance the hydrogens with OH -:
I 2 + 12OH - → 2IO 3 - + 6H 2 O
We write the two half-reactions and add the missing electrons to balance the negative charges:
I 2 + 2e - → 2I -
I 2 + 12OH - → 2IO 3 - + 6H 2 O + 10e -
We equalize the numbers of the electrons in both half-reactions and add them:
(I 2 + 2e - → 2I -) x 10
(I 2 + 12OH - → 2IO 3 - + 6H 2 O + 10e -) x 2
12I 2 + 24 OH - + 20e - → 20I - + 4IO 3 - + 12H 2 O + 20e -
The electrons cancel out and we divide all coefficients by four to simplify the global ionic equation:
(12I 2 + 24 OH - → 20I - + 4IO 3 - + 12H 2 O) x ¼
3I 2 + 6OH - → 5I - + IO 3 - + 3H 2 O
And finally, we substitute the coefficients of the ionic equation in the first equation:
3I 2 + 6OH - + KNO 3 → 5I - + KIO 3 + NO 3 - + 3H 2 O
The equation is already balanced. Compare this result with the balancing in acid medium in Example 2.
Exercise 2
Balance the following equation in an acid medium:
Fe 2 O 3 + CO → Fe + CO 2
General steps
We look at the oxidation numbers of iron and carbon to find out which of the two has been oxidized or reduced:
Fe 2 3+ O 3 + C 2+ O → Fe 0 + C 4+ O 2
Iron has been reduced, making it the oxidizing species. Meanwhile, the carbon has been oxidized, behaving as the reducing species. The half-reactions for oxidation and reduction concerned are:
Fe 2 3+ O 3 → Fe 0 (reduction, for each Fe 3 electrons are consumed)
CO → CO 2 (oxidation, for every CO 2 2 electrons are released)
Note that we write the oxide, Fe 2 O 3, because it contains Fe 3+, rather than just placing Fe 3+. We balance the atoms that are needed except those of oxygen:
Fe 2 O 3 → 2Fe
CO → CO 2
And we proceed to carry out the balancing in an acid medium in both semi-reactions, since there are oxygenated species in between.
Balance in acid medium
We add water to balance the oxygens, and then H + to balance the hydrogens:
Fe 2 O 3 → 2Fe + 3H 2 O
6H + + Fe 2 O 3 → 2Fe + 3H 2 O
CO + H 2 O → CO 2
CO + H 2 O → CO 2 + 2H +
Now we balance the charges by placing the electrons involved in the half-reactions:
6H + + 6e - + Fe 2 O 3 → 2Fe + 3H 2 O
CO + H 2 O → CO 2 + 2H + + 2e -
We equalize the number of electrons in both half-reactions and add them:
(6H + + 6e - + Fe 2 O 3 → 2Fe + 3H 2 O) x 2
(CO + H 2 O → CO 2 + 2H + + 2e -) x 6
12 H + + 12e - + 2Fe 2 O 3 + 6CO + 6H 2 O → 4Fe + 6H 2 O + 6CO 2 + 12H + + 12e -
We cancel electrons, H + ions and water molecules:
2Fe 2 O 3 + 6CO → 4Fe + 6CO 2
But these coefficients can be divided by two to simplify the equation even more, having:
Fe 2 O 3 + 3CO → 2Fe + 3CO 2
This question arises: was redox balancing necessary for this equation? By trial and error it would have been much faster. This shows that this reaction proceeds regardless of the pH of the medium.
References
- Whitten, Davis, Peck & Stanley. (2008). Chemistry (8th ed.). CENGAGE Learning.
- Helmenstine, Anne Marie, Ph.D. (September 22, 2019). How to Balance Redox Reactions. Recovered from: thoughtco.com
- Ann Nguyen & Luvleen Brar. (June 05, 2019). Balancing Redox Reactions. Chemistry LibreTexts. Recovered from: chem.libretexts.org
- Quimitube. (2012). Exercise 19: Adjustment of a redox reaction in basic medium with two oxidation half-reactions. Recovered from: quimitube.com
- Washington University in St. Louis. (sf). Practice Problems: Redox Reactions. Recovered from: chemistry.wustl.edu
- John Wiley & Sons. (2020). How to Balance Redox Equations. Recovered from: dummies.com
- Rubén Darío OG (2015). Balancing chemical equations. Recovered from: aprendeenlinea.udea.edu.co