# Review of Oxidation and Reduction Reactions


## Oxidation State

Oxidation state (also known as oxidation number or valence state), is a quick way of assessing the oxidation or reduction of particular atoms according to a prescribed set of rules. Oxidation state is not the same as formal charge, which tries to estimate the real charge distribution of a molecule or ion among its constituent atoms. Formal charge uses information about single bonds, multiple bonds and octet and non-octet structures. Dipole moment, the actual distribution of charges in a molecule or ion, must be experimentally measured.

Empirical Method of Determining Oxidation State: These rules must be memorized and applied in order, the first rule has the highest priority, the last has the least priority.)

1. Elements in elemental form are 0 (e.g. $$Fe(s)$$, $$O_2 (g)$$, etc.)
2. The sum of the oxidation states = the overall charge of the molecule/polyatomic ion. (e.g. $$ClO_4^-$$, Cl is +7, O is -2, 7+(4x-2) = -1)
3. All group I metals and Ag are +1
4. F is -1
5. All group II metals and Zn, Cd are +2
6. H is +1
7. O is -2

## Some Definitions

• Reduction: The charge (oxidation state) is reduced by gaining electrons (e.g., in $$Fe^{+2} + 2e^- \rightarrow Fe(s)$$, the charge goes from +2 to 0 and so the $$Fe^{+2}$$ is reduced to $$Fe(s)$$.)
• Oxidation: The charge (oxidation state) increases by losing electrons (eg., $$Ag(s) \rightarrow Ag^+ + 1e^-$$ ).
• Reducing agent: The reducing agent is the compound which contains the element that is oxidized. When reducing another element or compound, the reducing agent must give electrons away so it becomes oxidized.
• Oxidizing agent: The oxidizing agent is the compound which contains the element that is reduced.
• Redox Reaction: A balanced chemical reaction consisting of both an oxidation and a reduction (i.e., the sum of an oxidation half-reaction and a reduction half-reaction). These are electron transfer reactions.

## Balancing Redox Reactions in Acids

The "HElOHEN" six step method:

 H E O H E N Half rxn's: Elements Oxygen Hydrogen Electrons Number of electrons break up into Half-reactions balance all Elements except O and H balance O with $$H_2O$$ balance H with $$H^+$$ add the $$e^-$$ to balance charge multiply so the Number of $$e^-$$ is the same in both

Example

Balance the following reaction in acid:

$Cr_2O_7^{-2} + H_2O_2 \rightarrow Cr^{+3} + O_2$

1) H: 1/2 rxn's

Cr2O7-2 \rightarrow Cr+3 red (+6 \rightarrow +3)

H2O2 \rightarrow O2 ox (-1 \rightarrow 0)

2) El: elements

Cr2O7-2 \rightarrow 2 Cr+3 red

H2O2 \rightarrow O2 ox

3) O: oxygen

$Cr_2O_7^{-2} \rightarrow 2 Cr^{+3} + 7 H_2O$ red

$H_2O_2 \rightarrow O_2$ ox

4) $Cr_2O_7^{-2} + 14 H^+ \rightarrow 2 Cr^{+3} + 7 H_2O red$

$H_2O_2 \rightarrow O_2 + 2 H^+ ox$

5) E: electrons Be sure to use the net (total) charge on each sides. Multiply coefficients by charges to get the net.

Cr2O7-2 + 14 H+ + 6e- \rightarrow 2 Cr+3 + 7 H2O red

H2O2 \rightarrow 2 e- + O2 + 2 H+ ox

6) N: number

$Cr_2O_7^{-2} + 14 H^+ + 6e^- \rightarrow 2 Cr^{+3} + 7 H_2O$ red

$3H_2O_2 \rightarrow 6 e^- + 3 O_2 + 6 H^+$ ox

$Cr_2O_7^{-2} + 8 H^+ + 3 H_2O_2 \rightarrow 2 Cr^{+3} + 7 H_2O + 3 O_2$

8) Always check that in your answer all electrons cancel and atoms and charges balance! Also check to see if you can divide by a common divisor (see if you have the lowest possible coefficients) and if you can collect terms.

## Balancing in Aqueous Bases

The procedure is the same as with acid except that after you add the $$H^+$$ to balance the hydrogens, add the same number of OH- to both sides and form water on one side.

Example

Balance the following 1/2 rxn in base: $$BrO_3^- \rightarrow BrO_4^-$$

1) H: (The 1/2 rxn is given in this case)

$BrO_3^- \rightarrow BrO_4^-$

2) El: elements

$BrO_3^- \rightarrow BrO_4^-$

3) O: oxygen

$BrO_3^- + H_2O \rightarrow BrO_4^-$

4) H: hydrogen (but because its really in base, after adding H+, then add OH- to both sides.)

$BrO_3^- + H_2O + 2 OH^- \rightarrow BrO_4^- + 2H^+ + 2 OH^-$

If you put H+ and OH- together, they will form water.

$BrO_3^- + H_2O + 2 OH^- \rightarrow BrO_4^- + (2H^+ + 2 OH^- \rightarrow 2 H_2O)$

Cancel the waters and continue.

$BrO_3^- + 2 OH^- \rightarrow BrO_4^- + H_2O$

5) E: electrons

$BrO_3- + 2 OH^- \rightarrow BrO_4^- + H_2O + 2 e^-$

## Contributors

• Jim Hollister, SASC
• Rolf Unterleitner, SASC

Review of Oxidation and Reduction Reactions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.