Reduction and Oxidation Reactions

Lessons

Senior 4 · Chemistry

Reduction and Oxidation Reactions

5 lessons

Lesson 2 of 5

Oxidation Numbers

30 min

Is it redox?

Hydrogen peroxide slowly decomposes into water and oxygen: $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$ . No metal is involved, no ions are formed, and nothing obviously "loses electrons". Is this a redox reaction? What about $\mathrm{AgNO_3 + NaCl \rightarrow AgCl + NaNO_3}$ , where a white precipitate appears? To answer quickly and reliably, chemists give every atom in a substance an oxidation number, a kind of electron bookkeeping. If any oxidation number changes, the reaction is redox. This lesson sets out the rules and shows how to use them.

Definition

Oxidation number (oxidation state): the charge an atom would have if all its bonds were fully ionic, with the shared electrons given to the more electronegative atom. Example: in $\mathrm{H_2O}$ , O is −2 and each H is +1.

Oxidation is an increase in oxidation number; reduction is a decrease.

Stock notation: a Roman numeral in brackets giving the oxidation number of an element in a compound's name. Example: iron(III) chloride, $\mathrm{FeCl_3}$ .

Disproportionation: a redox reaction in which the same element is both oxidised and reduced. Example: $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$ .

The rules

Apply the rules in order; a rule higher in the list wins over a lower one.

Rule	Oxidation number	Example
1. Uncombined element	0	Na, $\mathrm{O_2}$ , $\mathrm{P_4}$ , $\mathrm{S_8}$ : all 0
2. Simple (monatomic) ion	equal to its charge	$\mathrm{Fe^{3+}}$ : +3; $\mathrm{Cl^-}$ : −1
3. Sum in a neutral compound	0	$\mathrm{H_2SO_4}$ : $2 (+ 1) + x + 4 (- 2) = 0$
4. Sum in a polyatomic ion	equal to the charge of the ion	$\mathrm{SO_4^{2-}}$ : $x + 4 (- 2) = - 2$
5. Fluorine in compounds	always −1	$\mathrm{OF_2}$ : F −1, so O +2
6. Group 1 / Group 2 metals, Al	+1 / +2 / +3	$\mathrm{NaH}$ : Na +1, so H −1
7. Hydrogen	+1 (−1 in metal hydrides)	HCl: +1; $\mathrm{CaH_2}$ : −1
8. Oxygen	−2 (−1 in peroxides; +2 in $\mathrm{OF_2}$ )	$\mathrm{H_2O_2}$ : −1
9. Halogens Cl, Br, I	−1, except when bonded to O or to a more electronegative halogen	NaCl: −1; $\mathrm{ClO_3^-}$ : +5

Calculating an oxidation number

Call the unknown $x$ and use rule 3 or 4. For manganese in the manganate(VII) ion: $x + 4(-2) = -1 \quad\Rightarrow\quad x = +7$ For chromium in the dichromate(VI) ion: $2x + 7(-2) = -2 \quad\Rightarrow\quad 2x = +12,\ x = +6$ When one element appears in two different environments the formula gives only the average. In the thiosulfate ion, $\mathrm{S_2O_3^{2-}}$ , $2 x + 3 (- 2) = - 2$ gives $x = + 2$ on average; in iron(II,III) oxide, $\mathrm{Fe_3O_4}$ , $3 x = + 8$ gives $x = +\frac{8}{3}$ , because it contains one $\mathrm{Fe^{2+}}$ and two $\mathrm{Fe^{3+}}$ . In ammonium nitrate the two nitrogen atoms must be worked out separately, from the two ions: N is −3 in $\mathrm{NH_4^+}$ and +5 in $\mathrm{NO_3^-}$ .

Figure 16.2: The oxidation numbers of nitrogen, from −3 to +5. Moving up the ladder is oxidation; moving down is reduction. Nitrogen can lose its 5 outer electrons or gain 3.

Using oxidation numbers

Spotting redox. Write the oxidation number over every atom. In $\mathrm{AgNO_3 + NaCl \rightarrow AgCl + NaNO_3}$ : Ag +1 → +1, N +5 → +5, O −2 → −2, Na +1 → +1, Cl −1 → −1. Nothing changes: this is not a redox reaction, only an exchange of ions (a precipitation).

Disproportionation. In $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$ , oxygen goes from −1 to −2 in water (reduced) and from −1 to 0 in $\mathrm{O_2}$ (oxidised). Chlorine in cold alkali does the same: $\overset{0}{\mathrm{Cl_2}} + \mathrm{2NaOH \rightarrow Na}\overset{-1}{\mathrm{Cl}} + \mathrm{Na}\overset{+1}{\mathrm{Cl}}\mathrm{O + H_2O}$

The reverse, two oxidation states of an element meeting in the middle, happens in volcanic gases: $\mathrm{2H_2\overset{-2}{S} + \overset{+4}{S}O_2 \rightarrow 3\overset{0}{S} + 2H_2O}$ .

Naming. Stock names state the oxidation number: $\mathrm{FeSO_4}$ iron(II) sulfate; $\mathrm{Fe_2(SO_4)_3}$ iron(III) sulfate; $\mathrm{Cu_2O}$ copper(I) oxide; $\mathrm{KMnO_4}$ potassium manganate(VII); NaClO sodium chlorate(I).

Worked example 1: a set of oxidation numbers

Find the oxidation number of the underlined element: (a) S in $\mathrm{Na_2SO_3}$ ; (b) P in $\mathrm{H_2PO_4^-}$ ; (c) N in $\mathrm{N_2O_5}$ ; (d) C in $\mathrm{C_2O_4^{2-}}$ ; (e) O in $\mathrm{BaO_2}$ .

Solution. (a) $2(+1) + x + 3(-2) = 0 \Rightarrow x = +4$ . (b) $2(+1) + x + 4(-2) = -1 \Rightarrow x = +5$ . (c) $2x + 5(-2) = 0 \Rightarrow x = +5$ . (d) $2x + 4(-2) = -2 \Rightarrow x = +3$ . (e) Ba is +2 (Group 2), so $+2 + 2x = 0 \Rightarrow x = -1$ : a peroxide.

Worked example 2: copper and nitric acid

Copper reacts with dilute nitric acid: $\mathrm{3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O}$ . Use oxidation numbers to identify what is oxidised and what is reduced, and show that the electron changes balance.

Solution. Cu: 0 → +2 (oxidised; each Cu loses 2). N: +5 in $\mathrm{HNO_3}$ → +2 in NO (reduced; each N gains 3). Only 2 of the 8 nitrogen atoms change; the other 6 stay at +5 in the nitrate ions of $\mathrm{Cu(NO_3)_2}$ . Electrons lost $= 3 \times 2 = 6$ ; electrons gained $= 2 \times 3 = 6$ . They balance. H (+1) and O (−2) do not change.

Warning

Common mistakes. (1) Writing oxidation numbers as "2+" instead of "+2": the sign goes first for oxidation numbers, after for ionic charges. (2) Forgetting rule priority: in $\mathrm{H_2O_2}$ oxygen is −1, because hydrogen (+1) wins. (3) Thinking an average such as +2.5 is wrong: in $\mathrm{S_4O_6^{2-}}$ the four S atoms really differ, and +2.5 is their average. (4) Calling every reaction with a colour change or a precipitate "redox": check the oxidation numbers.

Safety. Copper with nitric acid releases toxic brown nitrogen dioxide; it is done only by a teacher in a fume cupboard. Concentrated hydrogen peroxide burns the skin; school work uses dilute (about 3–6%) solutions with goggles.

Tip

Up is oxidation, down is reduction. Write the numbers above every atom, look for the ones that change, and remember the fixed points: F −1, O −2, H +1, Group 1 + 1, Group 2 + 2.

Wayne Breslyn: How to Find Oxidation Numbers, Rules and Examples (7 min). Watch the rules applied one by one. Note the presenter's own correction: N in N₂O is +1.

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Practice

Give the oxidation number of: (a) Cl in $\mathrm{ClO_4^-}$ ; (b) N in $\mathrm{NO_2^-}$ ; (c) Mn in $\mathrm{MnO_2}$ ; (d) H in LiH; (e) S in $\mathrm{H_2S}$ .
Calculate the average oxidation number of S in the tetrathionate ion, $\mathrm{S_4O_6^{2-}}$ .
Give the Stock names of $\mathrm{CuCl}$ , $\mathrm{FeCl_3}$ , $\mathrm{KClO_3}$ and $\mathrm{PbO_2}$ .
Use oxidation numbers to decide whether each is a redox reaction: (a) $\mathrm{CaCO_3 \rightarrow CaO + CO_2}$ ; (b) $\mathrm{2KI + Cl_2 \rightarrow 2KCl + I_2}$ ; (c) $\mathrm{NaOH + HCl \rightarrow NaCl + H_2O}$ .
Show that $\mathrm{3NO_2 + H_2O \rightarrow 2HNO_3 + NO}$ is a disproportionation.
In $\mathrm{MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O}$ , identify the element oxidised and the element reduced, with their changes in oxidation number, and explain why only half of the chlorine atoms are oxidised.
Explain why oxygen has a positive oxidation number in $\mathrm{OF_2}$ .
Iron in rust can be written $\mathrm{Fe_2O_3 \cdot xH_2O}$ . What is the oxidation number of iron, and is the rusting of iron an oxidation or a reduction of iron?

Show answersHide answers

(a) +7; (b) +3; (c) +4; (d) −1; (e) −2.
$4x + 6(-2) = -2 \Rightarrow 4x = +10 \Rightarrow x = +2.5$ .
Copper(I) chloride; iron(III) chloride; potassium chlorate(V); lead(IV) oxide.
(a) Not redox (Ca +2, C +4, O −2 throughout). (b) Redox: I −1 → 0 (oxidised), Cl 0 → −1 (reduced). (c) Not redox (neutralisation).
N is +4 in $\mathrm{NO_2}$ ; it becomes +5 in $\mathrm{HNO_3}$ (oxidised) and +2 in NO (reduced): the same element is both oxidised and reduced.
Mn: +4 → +2 (reduced). Cl: −1 → 0 in $\mathrm{Cl_2}$ (oxidised). Of the 4 Cl atoms, 2 become $\mathrm{Cl_2}$ and 2 stay as $\mathrm{Cl^-}$ in $\mathrm{MnCl_2}$ : Mn gains 2 electrons, so only 2 Cl need to lose one each.
Fluorine is more electronegative than oxygen and is always −1, so O must be +2 for the sum to be zero.
+3; iron goes from 0 to +3, so it is oxidised.

Quick check

Key points

Oxidation number is the charge an atom would carry if all its bonds were ionic. An increase is oxidation; a decrease is reduction.
Rules in priority order: elements 0; simple ions = charge; sum = 0 (or the ion's charge); F −1; Group 1 + 1, Group 2 + 2, Al +3; H +1 (−1 in metal hydrides); O −2 (−1 in peroxides).
A reaction is redox only if some oxidation number changes. In disproportionation one element is both oxidised and reduced.
Stock names give oxidation numbers in Roman numerals: iron(III) chloride, potassium manganate(VII).

Sources and credits

Image: Nitrogen dioxide.jpg by Izmaelt, via Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: How to Find Oxidation Numbers (Rules and Examples), Wayne Breslyn (Dr. B.), embedded from YouTube.
Written by Ganzaa: the text, rules table, diagram, worked examples and Practice items.

1Is it redox?

2The rules

3Calculating an oxidation number

4Using oxidation numbers

Is it redox?

The rules

Calculating an oxidation number

Using oxidation numbers