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.)

Elements in elemental form are 0 (e.g. \(Fe(s)\), \(O_2 (g)\), etc.)
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)
All group I metals and Ag are +1
F is -1
All group II metals and Zn, Cd are +2
H is +1
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 7) Now add them up. \[ Cr_2O_7^{-2} + 8 H^+ + 3 H_2O_2 \rightarrow 2 Cr^{+3} + 7 H_2O + 3 O_2 \] 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.

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

7) Now add them up.

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

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 \[ BrO_3^- \rightarrow BrO_4^-\] The bromines are already balanced. 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 \[BrO_3- + 2 OH^- \rightarrow BrO_4^- + H_2O + 2 e^-\]

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

\[ BrO_3^- \rightarrow BrO_4^-\]

The bromines are already balanced.

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