Faculty of Chemistry and Technology in Split

First published: 2018/08/19

A simple redox reaction in which only one species has oxidized and only one reduced can be easily separated into oxidation and reduction half-reactions. However, when it comes to complex reactions where there are multiple redox couples, we must combine them in such a way that we end up with two half-reactions. It might seem logical to gather all of the atoms that have been oxidized into the half-reaction of oxidation and all atoms that have been reduced into the half-reaction of reduction, but that is not always the case. For example, if a species is made up of different redox atoms, all redox pairs in which it is found (regardless of oxidation or reduction) need to be combined into a single equation. The combined equation is then added to the oxidation or reduction partial reactions, depending on whether it receives or gives electrons.

In order to make the text more readable and easier to follow, we've singled out the most important terms we've used for describing the state of a specific member in a reaction:

*atom*is a member of the reaction defined by its type (element) and oxidation number,*redox atom*is an atom that has oxidized (is present on the left side of the equation with a greater oxidation number) or reduced (is present on the right side of the equation with a lower oxidation number),*redox couple*is a conjugated pair of a higher and lower (or lower and higher) degree of oxidation of a specific atom,*non-redox*is an atom that has the same oxidation number on both sides of the equation,*dual atom*is an atom that is simultaneously a redox and non-redox atom,*redox species*is a species which consists of one or more redox atoms,*non-redox species*is a species which doesn't consist of a single redox atom,*simple redox species*is a species which consists of only one redox atom,*partial reaction of oxidation (reduction)*is a synonym for a*half-reaction of oxidation (reduction)*when used in the Oxidation Number Change Method. However, a partial reaction is also every reaction produced by combining redox couples that are on the path to the partial reaction of oxidation (reduction).

There is one term we must explain in more detail: *combining equations* (redox pairs). When combining equations we differentiate between two cases: (1) all redox species from equations we're combining are simple redox species, and (2) in the equations we're combining there is a shared species with multiple redox atoms.

In the first case, combining is exactly the same as summing the equation, though in the second case that is no longer true. When combining equations that have a shared species with multiple redox atoms, we first must ensure that the stoichiometric factor in front of the species is the same for both equations. The equations are then combined by copying the shared species and summing the rest. For example, by combining the two oxidation equations for the reaction

we only get one molecule of Sb_{2}S_{3} instead of two. If you take a closer look at the example below you will notice that this is quite logical as only antimony is balanced in the first reaction, while only sulfur is balanced in the second reaction. Only in the combined reaction are the two redox atoms from Sb_{2}S_{3} balanced.

Sb+3_{2}S-2_{3}
→
Sb+5_{2}O-2_{5} + 4e^{-}

(Sb)

Sb+3_{2}S-2_{3}
→
3H+1S+6O-2_{4}^{-} + 24e^{-}

(S)

Sb_{2}S_{3} → Sb_{2}O_{5} + 3HSO_{4}^{-} + 28e^{-}

You can combine one equation with multiple others. If the equation is the only member of the oxidation reaction and after combining must join the reduction reactions, you can leave a copy untouched in the oxidation reaction. Naturally, the same applies in reverse.

In the first reaction, out of the five redox pairs, three redox pairs contain species with multiple redox atoms on the right side. We can solve this by combining the reduction of nitrogen with the oxidation of carbon and the reduction of carbonate. In the second equation the disproportionation of Na_{3}[Co(NO_{2})_{6}] produces three redox couples, so we must combine the reduction of CoCl_{2} with the remaining two redox pairs to ensure that cobalt is balanced in both partial reactions.

By following these five simple steps you can separate even the most complex of equations into partial reactions. Just so we're clear, it's not always necessary to go through all five steps. With simpler redox equations we can obtain the two partial reactions after just one step.

- Identify all redox couples
- Remove unnecessary redox couples
- Balance redox atoms that aren't a part of a redox couple
- Balance non-redox atoms that are only present in redox species
- Combine the remaining equations into two half-reactions

When you know the oxidation numbers of all atoms in an equation, it is simple to determine which atoms have been oxidized and which reduced:

- if the oxidation number of an atom on the right side of the equation is greater than on the left side, then the atom has been oxidized,
- if the oxidation number of an atom on the right side of the equation is lower than on the left side, then the atom has been reduced,
- the atom whose oxidation number has not changed does not participate in the electron transfer.

If you haven't already determined the oxidation numbers, and you don't know how to do so, you can find the rules on the Oxidation numbers calculator page. Write down all of the redox pairs you can with the above-determined redox atoms. A redox pair consists of two different oxidation states (oxidized and reduced form) of the same element that is found on both sides of the equation.

*Oxidation:* A_{Red} = A_{Ox} + ne^{-}

*Reduction:* B_{Ox} + ne^{-} = B_{Red}

Some of the redox couples we've constructed are either unlikely or simply unnecessary. Such redox couples can be constructed from atoms which, despite having different oxidation numbers on the left and right side of the equation, are not redox atoms, or from atoms that play a dual role by simultaneously behaving like redox and non-redox atoms. In some redox equations, however, it is impossible to determine to role of an atom without knowing the chemical reaction described by the equation.

While developing the ARS method we encountered a couple of rules that will help you decide which redox pairs to keep and which to discard. In order to make the rules a bit more understandable we have also illustrated each one with an example. However, these rules are not laws. Their only goal is to help you separate an equation into partial reactions of oxidation and reduction. As such, regardless of how you combine or erase redox pairs, there is one rule that must be satisfied:

**Rule 0:** There must always remain at least one oxidation reaction and at least one reduction reaction.

**Rule 1:** All atoms of a certain element are non-redox atoms.

If an element has the same oxidation numbers on both sides of the equation, then it is a non-redox element. All redox couples in which its atoms participate as redox atoms have to be removed.

In the first equation sulfur has the same oxidation numbers on both sides of the equation (+6 and -2). Similarly, nitrogen in the second equation has the same oxidation numbers on both sides of the equation (+5 and -3).

**Rule 2:** Some of the redox atoms are actually non-redox atoms.

If an element has different oxidation numbers on the left and right side of the equation, and we know that its atom of a certain degree of oxidation does not participate in the electron exchange (it's a non-redox atom), we have to remove all redox pairs in which that non-redox atom appears as a redox atom.

In these reactions the cyanide ion and ammonium are spectators. We know that nitrogen from the cyanide ion and ammonium does not participate in the electron transfer, so we must remove all redox couples in which it appear as a redox atom.

**Rule 3:** The redox atom has a non-redox 'twin' on the left side.

If a redox atom with a specific oxidation number present in one or more redox species on the right side of the skeleton equation simultaneously appears in a non-redox species on the left side, all redox couples leading to that oxidation number, except the one present in the simplest redox species, have to be removed. The simplest redox species is the one that has the least redox atoms. This redox atom does not need to be balanced in the redox pair. The number of redox atoms on the left side determines how many electrons have been exchanged, while the right side can have more or less of these atoms. For example:

In the first reaction it doesn't matter which species with chloride we'll choose because neither contains other redox atoms. In the second reaction we will only retain K_{2}SO_{4} because it's the only species that contains sulfur(VI) without other redox atoms.

**Rule 4:** The redox atom has a non-redox 'twin' on the right side.

If multiple redox pairs on the left side contain redox atoms with the same oxidation number as the species on the right side, all reactions besides the simplest ones should be removed. Why should they be removed? Given that we can choose which species is going to be oxidized/reduced and which will be the non-redox species, we will always choose the simplest species as the redox species.

Since we can choose which species with a nitrogen(V) atom is going to get reduced, we will choose HNO_{3} because it contains only one redox atom, unlike Fe(NO_{3})_{2}.

**Rule 5:** A redox atom is simultaneously the product and the reactant.

If a redox atom with the same oxidation number is found on the left side in one redox couple and on the right side in another - one of the redox couples must be removed (they are nullified). It's necessary to know the chemical reaction in order to choose which redox couple should be removed.

In the first reaction it is obvious which redox couple we need to remove - the reduction of chlorate to chloride.

In the case of oxidation of hydrogen sulfide by potassium permanganate, the situation is much more complex. In redox processes with weaker oxidizing agents H_{2}S is oxidized to elemental sulfur. However, with stronger oxidizing agents the oxidation of the sulfide proceeds all the way until the +4 or +6 oxidation state. Depending on the conditions, this reaction can be run in two ways:

a) The sulfide ion oxidizes into elemental sulfur. In this case the sulfate ion is a spectator ion so we use Rule 2. The balanced equation for this reaction is:

2KMnO_{4} + 5H_{2}S + 3H_{2}SO_{4}
→
2MnSO_{4} + K_{2}SO_{4} + 8H_{2}O + 5S

b) The sulfide ion is partially oxidized into elemental sulfur and partially into sulfate ions. In this case sulfuric acid is added to make the solution acidic so we use Rule 3. Since we have a sulfate on the left side of one and on the right side of another redox couple, we will also use Rule 5 - the reduction of the sulfate is unlikely and as such should be removed. The balanced equation for this reaction is:

2KMnO_{4} + 2H_{2}S + 2H_{2}SO_{4}
→
2MnSO_{4} + K_{2}SO_{4} + 4H_{2}O + S

When multiple atoms that change their oxidation number are present in a species, we must calculate the transfer of electrons for the entire species. Because of this it is necessary to combine all redox couples that have species with multiple different redox atoms into one equation. The combined equation is then added to either the oxidation or reduction reactions depending on whether it is losing or gaining electrons.

a) The species with multiple redox atoms is on the left side

Disproportionation is the special case of a species with multiple redox atoms on the left side.

b) The species with multiple redox atoms is on the right side

When the non-redox atom is found only in redox species it must be balanced before the electron transfer is equalized. Since the redox atoms are already balanced, non-redox atoms can only be balanced by multiplying the entire equation (redox couple) by some factor and then summing it with another oxidation or reduction equation. If possible, all equations that have a different number of non-redox atoms on the left and right side should be paired up in either the oxidation or reduction reactions.

Non-redox atoms that are already balanced do not affect the result and should not be touched.

The remaining reactions are summed into two half-reactions: one for oxidation and the other for reduction. In special cases, when we have a species with multiple redox atoms that we could not balance in Step 3, the partial reactions should be combined and not summed.

- E. Generalić, N. Vladislavić, Aggregate Redox Species Method - An Improved Oxidation Number Change Method for Balancing Redox Equations,
*Chemistry Journal*, Vol.**4**, No. 3, 43-49 (2018)

Citing this page:

Generalic, Eni. "Divide the redox reaction into two half-reactions." *EniG. Periodic Table of the Elements*. KTF-Split, 19 Aug. 2018. Web. {Date of access}. <https://www.periodni.com/divide_redox_reaction.html>.

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