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Redox Reactions

Written by tutor Joshua K.

Oxidation and Reduction (redox) reactions are complementary reactions that involve the transfer of electrons from one molecule or ion (reductant) to another (oxidant). They always work in concert with one-another. That is, whenever an oxidation reaction occurs, so does a reduction reaction. Redox reactions can occur through all previously mentioned reactions schemes: synthesis, decomposition, single-displacement, double-displacement, and combustion. Aside from combustion, none of these reactions necessarily involve redox, so when can a reaction be classified as redox?

(a) CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)
(b) NaOH(aq) + HCl(aq) --> H2O(l) + NaCl(aq)

Figure 1: Reaction (a) is a redox reaction and reaction (b) is not.

Chemists use a conceptual tool called the oxidation number (oxidation state) as a means of keeping track of the number of electrons borne by each atom in a molecule or ion. Each atom in each molecule or ion has its own oxidation state. A tell-tale sign that a redox reaction has taken place is that the oxidation number of a particular atom has changed over the course of a reaction. If each of these numbers remains invariant over the course of the reaction, then we may conclude that no redox has taken place. So how do we determine a particular atom’s oxidation number? Here is a list of guidelines for determining an atom’s oxidation number:

1
For neutral molecules, the sum of the oxidation numbers of each atom must equal zero.
For ions, the sum of the oxidation numbers of each atom must equal the charge of the ion.
2
The oxidation state of each atom in its pure elemental form is zero.
a In F2, N2, O2, etc. the oxidation state for F, N, O is zero.
3
For monatomic ions, the oxidation number is just the charge of the ion.
a In F-, Cl-, etc. the oxidation state for F and Cl is -1.
b In O-2, S-2, etc. the oxidation state for O and S is -2.
c In Fe+3, Cr+3, etc. the oxidation state for Fe and Cr is +3.

4
Fluorine’s oxidation state is always -1 in compounds with other elements, even oxygen.

5
Oxygen’s oxidation state is typically -2 (unless bonded to fluorine) and hydrogen’s oxidation state is typically +1 (except in the case of metal hydrides in hydrogen’s oxidation state is -1).
6
Unless covalently bonded to fluorine or oxygen, the oxidation state of chloride, bromide, and iodide is -1.

Using these rules and a bit of algebra, we can easily determine the oxidation state of each atom in most compounds.

Determine the oxidation state of each atom in the indicated molecule or ion. List your answers in the order in which the atoms appear, separated by a comma.

CH4

Solution: We know that each of the four hydrogens has an oxidation state of +1 and the overall charge of the molecule is zero. So the oxidation state of carbon plus four times the oxidation state of hydrogen (to account for each hydrogen) must equal zero. This gives carbon an oxidation state of -4.

{-4, +1|-4, 1}
MnO4-

Solution: The overall charge of this ion is -1, so all of the oxidation states of each atom must sum to -1. Each oxygen has a charge of -2, so if we call x the oxidation state of manganese, then 4(-2) + x = -1. Solving for x, we see that manganese has an oxidation state of +7.

{+7, -2|7, -2}
CO2

Solution: Each oxygen has an oxidation state of -2 and the overall charge of the molecule is zero. If x is the oxidation state of carbon, then x + 2(-2)=0. x = +4

{+4, -2|4, -2}

Equations describing oxidation and reduction reactions may be more difficult to balance by inspection using the methods stated in previous help sections. Carefully keeping track of oxidation numbers and the method of half-reactions provide a systematic way to balance complex redox equations. Many aqueous phase redox reactions will not proceed at neutral pH, so either an acid or base must be added. Let’s look at the reaction between potassium oxalate and aqueous potassium permanganate in sulfuric acid. Start by writing down the unbalanced equation, identifying all species present in the reactants and products.

Balance the following chemical equation

K2C2O4(aq) + KMnO4(aq) + H2SO4(aq) --> CO2(g) + H2O(l) + MnSO4(aq) + K2SO4(aq)

Solution: At first glance, this seems like a daunting equation to balance, but from the net ionic equation, there are a lot of species we may ignore because their oxidation state remains the same of the course of the reaction.

C2O4-2(aq) + MnO4-(aq) + H+(aq) --> CO2(g) + H2O(l) + Mn+2(aq)

Start by identifying the oxidation states of carbon and manganese before and after the reaction. Carbon goes from +3 in oxalate to +4 in carbon dioxide. Manganese goes from +7 in permanganate to +2 in manganese sulfate. Let’s write down these two steps separately and represent the transfer of an electron with e-.

Oxidation half-reaction:

C2O4-2(aq) --> 2CO2(g) +2 e-

Reduction half-reaction:

MnO4-(aq) + 8 H+(aq) + 5 e- --> Mn+2(aq) + 4 H2O(l)

Make sure that each half-reaction is individually balanced in terms of mass and charge. To balance oxygen, add water, and to balance hydrogens, add protons (H+). Notice that in the oxidation half-reaction, two electrons are donated and in the reduction reaction, five electrons are accepted. We need to make it so the number of electrons donated and accepted is the same. To do this, multiply each stoichiometric coefficient in the oxidation reaction by five and each stoichiometric coefficient in the reduction reaction by two. This results in a total of ten electrons being transferred. So let’s combine the now-balanced half-reactions.

5 C2O4-2(aq) + 2 MnO4-(aq) + 16 H+(aq) + 10 e- --> 10 CO2(g) + 2 Mn+2(aq) + 8 H2O(l) + 10 e-

Notice that 10 electrons appear on both sides of the equation, so they cancel out. Therefore, the balanced net ionic equation is

5 C2O4-2(aq) + 2 MnO4-(aq) + 16 H+(aq) --> 10 CO2(g) + 2 Mn+2(aq) + 8 H2O(l)

Works Cited

Gilbert, T. R., R. V. Kirss, N. Foster, and G. Davies. Chemistry, The Science in Context. Second. New York: W. W. Norton & Company, 2008. Print.

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