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Oxidation–reduction (redox) reactions

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What is an oxidation–reduction reaction?

Sunlight shining on bright green plant leaves.
Plants use photosynthesis, a redox process, to derive energy from the sun. Image credit: Eschtar M. on Pixabay, Pixabay Licence.
An oxidation–reduction or redox reaction is a reaction that involves the transfer of electrons between chemical species (the atoms, ions, or molecules involved in the reaction). Redox reactions are all around us: the burning of fuels, the corrosion of metals, and even the processes of photosynthesis and cellular respiration involve oxidation and reduction. Some examples of common redox reactions are shown below.
CHX4(g)+2OX2(g)COX2(g)+2HX2O(g)(combustion of methane)\small{\ce{CH4}(g) + \ce{2O2}(g) \rightarrow \ce{CO2}(g) + \ce{2H2O}(g)\kern0.75em(\text{combustion of methane})}
2Cu(s)+OX2(g)2CuO(s)(oxidation of copper)\small{\ce{2Cu}(s) + \ce{O2}(g) \rightarrow \ce{2CuO}(s)\kern7.65em (\text{oxidation of copper})}
6COX2(g)+6HX2O(l)CX6HX12OX6(s)+6OX2(g)(photosynthesis)\small{\ce{6CO2}(g) + \ce{6H2O}(l) \rightarrow \ce{C6H12O6}(s) + \ce{6O2}(g)\kern1.90em(\text{photosynthesis})}
During a redox reaction, some species undergo oxidation, or the loss of electrons, while others undergo reduction, or the gain of electrons. For example, consider the reaction between iron and oxygen to form rust:
4Fe(s)+3OX2(g)2FeX2OX3(s)(rusting of iron)\small{\ce{4Fe}(s) + \ce{3O2}(g) \rightarrow \ce{2Fe2O3}(s)\kern8.65em(\text{rusting of iron})}
In this reaction, neutral F, e loses electrons to form FeX3+\ce{Fe^3+} ions and neutral OX2\ce{O2} gains electrons to form OX2\ce{O^2-} ions. In other words, iron is oxidized and oxygen is reduced. Importantly, oxidation and reduction don’t occur only between metals and nonmetals. Electrons can also move between nonmetals, as indicated by the combustion and photosynthesis examples above.

Oxidation numbers

How can we determine if a particular reaction is a redox reaction? In some cases, it is possible to tell by visual inspection. For example, we could have determined that the rusting of iron is a redox process by simply noting that it involves the formation of ions (FeX3+\ce{Fe^3+} and OX2\ce{O^2-}) from free elements (F, e and OX2\ce{O2}). In other cases, however, it is not as obvious, particularly when the reaction in question involves only nonmetal substances.
To help identify these less obvious redox reactions, chemists have developed the concept of oxidation numbers, which provides a way to track electrons before and after a reaction. An atom’s oxidation number (or oxidation state) is the imaginary charge that the atom would have if all of the bonds to the atom were completely ionic. Oxidation numbers can be assigned to the atoms in a reaction using the following guidelines:
  1. An atom of a free element has an oxidation number of 0. For example, each C, l atom in ClX2\ce{Cl2} has an oxidation number of 0. The same is true for each H atom in HX2\ce{H2}, each S atom in SX8\ce{S8}, and so on.
  2. A monatomic ion has an oxidation number equal to its charge. For example, the oxidation number of CuX2+\ce{Cu^2+} is plus, 2, and the oxidation number of BrX\ce{Br-} is minus, 1.
  3. When combined with other elements, alkali metals (Group 1, start text, A, end text) always have an oxidation number of plus, 1, while alkaline earth metals (Group 2, start text, A, end text) always have an oxidation number of plus, 2.
  4. Fluorine has an oxidation number of minus, 1 in all compounds.
  5. Hydrogen has an oxidation number of plus, 1 in most compounds. The major exception is when hydrogen is combined with metals, as in N, a, H or LiAlHX4\ce{LiAlH4}. In these cases, the oxidation number of hydrogen is minus, 1.
  6. Oxygen has an oxidation number of minus, 2 in most compounds. The major exception is in peroxides (compounds containing OX2X2\ce{O2^2-}), where oxygen has an oxidation number of minus, 1. Examples of common peroxides include HX2OX2\ce{H2O2} and NaX2OX2\ce{Na2O2}.
  7. The other halogens (C, l, B, r, and I) have an oxidation number of minus, 1 in compounds, unless combined with oxygen or fluorine. For example, the oxidation number of C, l in the ion ClOX4X\ce{ClO4-} is plus, 7 (since O has an oxidation number of minus, 2 and the overall charge on the ion is minus, 1).
  8. The sum of the oxidation numbers for all atoms in a neutral compound is equal to zero, while the sum for all atoms in a polyatomic ion is equal to the charge on the ion. Consider the polyatomic ion NOX3X\ce{NO3-}. Each O atom has an oxidation number of minus, 2 (for a total of minus, 2, times, 3, equals, minus, 6). Since the overall charge on the ion is minus, 1, the oxidation number of the N atom must be plus, 5.
One thing to note is that oxidation numbers are written with the sign (plus or minus) before the number. This is in contrast to the charges on ions, which are written with the sign after the number. Now, let’s see some examples of assigning oxidation numbers!

Example 1: Assigning oxidation numbers

What is the oxidation number of each atom in (a) SFX6\ce{SF6}, (b) HX3POX4\ce{H3PO4} and (c) IOX3X\ce{IO3-}?
To assign the oxidation numbers to the atoms in each compound, let’s follow the guidelines outlined above.
(a) We know that the oxidation number of F is minus, 1 (guideline 4). Because the sum of the oxidation numbers of the six F atoms is minus, 6 and SFX6\ce{SF6} is a neutral compound, the oxidation number of S must be plus, 6:
SF6+61\begin{aligned} &\kern0.75em\blueD{\text{S}}\maroonD{\text{F}}_6 \\ &\footnotesize\blueD{{+6}}\kern0.35em\maroonD{{-1}} \end{aligned}
(b) The oxidation number of H is plus, 1 (guideline 5) and the oxidation number of O is minus, 2 (guideline 6). The sum of these oxidation numbers is 3, left parenthesis, plus, 1, right parenthesis, plus, 4, left parenthesis, minus, 2, right parenthesis, equals, minus, 5. Since HX3POX4\ce{H3PO4} has no net charge, the oxidation number of P must be plus, 5:
H3PO4+1+52\begin{aligned} &\kern0.60em\blueD{\text{H}}_3\maroonD{\text{P}}\greenD{\text{O}}_4 \\ &\footnotesize\blueD{{+1}}\kern0.35em\maroonD{{+5}}\kern0.35em\greenD{{-2}} \end{aligned}
(c) The oxidation number of O is minus, 2 (guideline 6), so the sum of the oxidation numbers of the three O atoms is minus, 6. Since the net charge on IOX3X\ce{IO3-} is minus, 1, the oxidation number of I must be plus, 5:
IO3+52\begin{aligned} &\kern0.75em\blueD{\text{I}}\maroonD{\text{O}}_3^- \\ &\footnotesize\blueD{{+5}}\kern0.35em\maroonD{{-2}} \end{aligned}
Concept check: What is the oxidation number of the carbon atom in COX3X2\ce{CO3^2-}?

Recognizing redox reactions

How do we actually use oxidation numbers to identify redox reactions? To find out, let’s revisit the reaction between iron and oxygen, this time assigning oxidation numbers to each atom in the equation:
4Fe(s)+3OX2(g)2FeX2OX3(s)00+32\begin{aligned} &\small{\ce{4Fe}(s) + \ce{3O2}(g) \rightarrow \ce{2Fe2O3}(s)} \\ &\footnotesize\kern0.95em\blueD{{0}}\kern3.50em\blueD{{0}}\kern3.75em\blueD{{+3}}\kern0.50em\blueD{{-2}} \end{aligned}
Notice how iron (which we already know is oxidized in this reaction) changes from an oxidation number of 0 to an oxidation number of plus, 3. Similarly, oxygen (which we know is reduced) changes from an oxidation number of 0 to an oxidation number of minus, 2. From this, we can conclude that oxidation involves an increase in oxidation number, while reduction involves a decrease in oxidation number.
So, we can identify redox reactions by looking for changes in oxidation numbers over the course of a reaction. Let’s explore this idea more in the next example.

Example 2: Using oxidation numbers to identify oxidation and reduction

Consider the following reaction:
4NHX3(g)+5OX2(g)4NO(g)+6HX2O(g)\small{\ce{4NH3}(g) + \ce{5O2}(g) \rightarrow \ce{4NO}(g) + \ce{6H2O}(g)}
Is this reaction a redox reaction? If so, which element in the reaction is oxidized and which element is reduced?
Considering this is an article about redox reactions, the reaction probably is a redox reaction! However, let’s prove it by assigning oxidation numbers to the atoms of each element in the equation:
4NHX3(g)+5OX2(g)4NO(g)+6HX2O(g)3+10+22+12\begin{aligned} &\small{\ce{4NH3}(g) + \ce{5O2}(g) \rightarrow \ce{4NO}(g) + \ce{6H2O}(g)} \\ &\footnotesize\blueD{{-3}}\kern0.35em\blueD{{+1}}\kern3.25em\blueD{{0}}\kern3.20em\blueD{{+2}}\kern0.35em\blueD{{-2}}\kern2.15em\blueD{{+1}}\kern0.50em\blueD{{-2}} \end{aligned}
The oxidation numbers of N and O are different on either side of the equation, so this is definitely a redox reaction! The oxidation number of N increases from minus, 3 to plus, 2, which means that N loses electrons and is oxidized during the reaction. The oxidation number of O decreases from 0 to minus, 2, which means that O gains electrons and is reduced during the reaction.

Summary

The image shows four test tubes, each containing a solution of a different color. From left to right, the solution colors are yellow, blue, green, and purple.
The most common oxidation numbers of vanadium are plus, 5 (yellow), plus, 4 (blue), plus, 3 (green), and plus, 2 (purple). Image credit: "Vanadium oxidation states" by W. Oelen on Wikimedia Commons, CC BY-SA 3.0.
Oxidation–reduction reactions, commonly known as redox reactions, are reactions that involve the transfer of electrons from one species to another. The species that loses electrons is said to be oxidized, while the species that gains electrons is said to be reduced. We can identify redox reactions using oxidation numbers, which are assigned to atoms in molecules by assuming that all bonds to the atoms are ionic. An increase in oxidation number during a reaction corresponds to oxidation, while a decreases corresponds to reduction.

Want to join the conversation?

  • blobby green style avatar for user maks.berlec
    Shouldn’t equation H2 + O2 -> 2 H2O be balanced to 2 H2 + O2 -> 2 H2O?
    (62 votes)
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  • leaf blue style avatar for user bennetd
    In the last paragraph, it states that there is a transfer of electrons. Is there not always a transfer of electrons?
    (26 votes)
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    • blobby green style avatar for user yuki
      There is not always a transfer of electrons (not all reactions are redox reactions). An example of a reaction that is not a redox reaction might be a neutralization reaction:
      H3O+ + OH- ---> 2 H2O
      The oxidation number of H is +1 and the oxidation number of O is -2 for for both the reactants and products, so it is not a redox reaction.
      (44 votes)
  • leafers tree style avatar for user Quinn
    Is it possible to have reaction where only oxidation or reduction happens, or does the occurrence of one result in the other? Thank you.
    (18 votes)
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    • piceratops ultimate style avatar for user Just Keith
      They must both occur. The sum of all the oxidation states cannot change unless there is a change in the overall charge of the ion/molecule. Even with a change in the charge, there must be somewhere else that the change in charge (and thus oxidation states) is exactly offset. That is because of conservation of electrical charge.
      (13 votes)
  • piceratops seed style avatar for user anishvpalli
    What is the difference between a monatomic ion and an atom in its elemental state?
    (11 votes)
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    • piceratops seed style avatar for user RogerP
      An ion is an atom that has gained or lost electrons. Atoms in their elemental state are not ionised.

      Also, atoms in their elemental state can be joined to other atoms. For example, H2 is the elemental state for hydrogen.
      (22 votes)
  • marcimus pink style avatar for user ikuko mukai-cheh
    In the example of determining the oxidation state in H2 and H2O, it reads:
    Rule 4 tells us that all the oxidation numbers in a compound have to add up to charge on the compound, and rule 3 says that oxygen usually has an oxidation number of +2.
    The "+2" should be "-2".
    (13 votes)
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  • leaf green style avatar for user sg60847
    What is the use of knowing about oxidation numbers ?
    (4 votes)
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  • winston baby style avatar for user AJ
    Above it says, "oxygen is usually assigned a −2 oxidation number (except in peroxide compounds where it is −1, and in binary compounds with fluorine where it is positive);"

    The part where it says "...with fluorine where it is positive", did they mean +1 or +2? Or did they mean, it varies depending on the binary compound with fluorine but it'll always be positive?
    (6 votes)
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  • leaf green style avatar for user Nicholas Dymov
    In the section "Determining the oxidation state in H2 and H2O"
    There is a line (oxidation # of O×# of O atoms)+(oxidation # of H×# of H atoms)=(−2×1)+(−1×2)=−3
    But the result should be -4
    (6 votes)
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  • piceratops ultimate style avatar for user christian
    Does anyone know what the OH radical is? It is in one of my compounds for my experiment
    (3 votes)
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  • female robot ada style avatar for user Angelica Chen
    I encountered the following question in my Chemistry textbook, and because it is not assigned as homework, I wanted to ask it here...

    In a nickel-cadmium battery, the relevant redox reaction is:

    2NiO(OH) + Cd + 2H²O → 2Ni(OH)² + Cd(OH)²

    Does this agree with the EMF series? What are the oxidation states of nickel before and after the reaction?

    My answer is that although the correct metal is releasing its electrons (Cd→Ni), it is not valid because nickel's oxidation state of 3+ in the reactant is not a common ionic charge. The product Ni(II) is, but not Ni(III).

    Is my answer correct? Thank you for the help!
    (2 votes)
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    • leaf red style avatar for user Richard
      So it's a battery meaning it should produce a spontaneous redox reaction with a positive standard cell potential. Nickel goes from an oxidation state of +3 to +2 (so it's being reduced) and cadmium goes from 0 to +2 (so it's being oxidized). Looking at the standard electrode potentials (or standard reduction potentials or EMF series I suppose is how your book is referring to it as) of the half reactions. We can find that the cadmium half reaction has a value of -0.4 V while the nickel one has a value of +0.8 V. Being more positive means that Nickel is a stronger oxidizing agent (more likely to cause oxidation) and itself more likely to be reduced as compared to cadmium which is a stronger reducing agent (more likely to cause reduction).

      So seeing that nickel is being reduced and cadmium is being oxidized would agree with their standard electrode values. Additionally having the metals in that order in the redox reaction generates a positive standard cell potential which is necessary for a battery.

      Hope that helps.
      (4 votes)