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:
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 Sb2S3 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 Sb2S3 balanced.
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 Na3[Co(NO2)6] produces three redox couples, so we must combine the reduction of CoCl2 with the remaining two redox pairs to ensure that cobalt is balanced in both partial reactions.
Regardless of whether we're balancing the redox reaction with the ion-electron method, oxidation number change method or the ARS method, it is necessary to divide it into reactions of oxidation and reduction. Since textbooks mostly use simple redox reactions, we will describe in detail, and support with examples, the whole process of dividing redox reactions into partial 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 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: ARed = AOx + ne-
Reduction: BOx + ne- = BRed
Only a small number of redox reactions really transfers electrons from one molecule (or atom) to another. In most cases the "transfer of electrons" applies only to simply changing the oxidation state of a redox atom. Defining oxidation as a "increase in the oxidation state" is applicable even to organic compounds.
If we applied the above-mentioned rules to organic molecules with numerous carbon atoms with different oxidation numbers, we could end up with hundreds of redox pairs. We would end up at a point where we couldn't see the forest from the trees. As such, when working with organic compounds, we can a) use molecular formulas and average oxidation numbers instead of structural formulas; or b) use the general formula of a organic compound while replacing everything that doesn't change until the first C-C bond with the abbreviation R.
a) Usage of average oxidation numbers
In order to simplify the task, we will temporarily write molecular formulas and calculate oxidation numbers instead of using structural formulas. Depending on the compound, the oxidation number of carbon can be any decimal number between -4 and +4. How this looks like in practice, we can see through the oxidation of xylidine to benzoquinone, or through the reduction of linoleic acid to stearic acid.
Be careful! You cannot simply write a molecular formula and determine the oxidation numbers within it by following the rules. In order to get the correct average oxidation numbers, you have to know the structural formula of the compound. You can only determine the oxidation number of every atom from the structural formula. For example, in the thermal degradation of di-tert-butyl peroxide and in the periodate cleavage of 1,2-octanediol, the same molecular formula (C8H18O2) has different oxidation numbers (-2,+1,-1 or -1.75,+1,-2 respectively).
How are average oxidation numbers calculated? Simple, divide the sum of oxidation numbers of all atoms of the same element by the number of atoms. For example, the average oxidation numbers in the molecular formula of di-tert-butyl peroxide can be calculated as follows:
C: (-3)*3*1 + (+1)*1 + (+1)*1 + (-3)*3*1 = (-16)/8 = -2
H: (+1)*3*3 + (+1)*3*3 = (+18)/18 = +1
O: (-1)*1 + (-1)*1 = (-2)/2 = -1
b) Usage of general formulas
We can simplify large organic molecules by replacing everything that doesn't change until the first C-C bond with the abbreviation R. Unlike radicals in organic chemistry, this R cannot be hydrogen, nor can you write, for example, alcohols by the general formula ROH (the correct way is to write R-CH2OH). Since electrons are equally split between two atoms of carbon, the R group doesn't change the oxidation number of the carbon atom to which it is bound. Since we are interested only in the change of oxidation numbers, we can join any oxidation number we wish to the R group. Using the general formulas, let's write Tollens' test for aldehydes and the reduction of carboxylic acids to corresponding primary alcohols.
After we balance our equation, we'll replace the temporary molecular formulas (or general formulas) we used with structural formulas from the task.
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 appears 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 K2SO4 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.
Depending on the number of species containing a redox atom, this rule can have two distinct cases:
a) Redox atom with a specific oxidation number is present in more redox species. 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 HNO3 because it contains only one redox atom, unlike Fe(NO3)2. If we change our selection and choose a more complex species ((NH4)2S2O8) in the second reaction, the program won't be able to balance the equation.
b) Redox atom with a specific oxidation number is present in only one redox specie. In this case redox and non-redox atoms are summed in the same species. Since this species is the only source of redox and non-redox atoms, the transfer of electrons has to be written for every redox atom that has really oxidized or reduced, and not for every atom present in the species.
Since I still haven't written an algorithm for solving this 'rule', I'll leave the two examples for you to puzzle over. Both equations are fully solvable.
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 H2S 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:
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:
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.
Following these five simple steps you will be able to separate even the most complex reaction into partial reactions. That said, it is not always necessary to follow all five steps. Sometimes, with simpler redox reactions, two partial reactions can be obtained after just the first step.
Citing this page:
Generalic, Eni. "Divide the redox reaction into two half-reactions." EniG. Periodic Table of the Elements. KTF-Split, 18 Jan. 2024. Web. {Date of access}. <https://www.periodni.com/divide_redox_reaction.html>.
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