Review of Oxidation and Reduction Reactions
- Page ID
- 9317
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 |
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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 |
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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