Electrochemistry Revision Notes — 9701 Paper 1

Oxidation numbers: the rules, and how to use them

Syllabus requirement · §6.1

“

calculate oxidation numbers of elements in compounds and ions … use a Roman numeral to indicate the magnitude of the oxidation number of an element

”

What an oxidation number tracks

In an ionic compound it is easy to see who has the electrons: in $\text{Na}^+\text{Cl}^-$ , sodium has lost one and chlorine has gained one. In a covalent molecule the electrons are shared, so nothing has an obvious charge.

An oxidation number fixes this. It is the charge an atom would have if every bond it forms were fully ionic: for each shared pair of electrons, give both electrons to the more electronegative atom (see “Electronegativity” in the AS Chemical Bonding note). In $\text{CO}_2$ , for example, oxygen is more electronegative, so each O takes the electrons of its double bond and counts as $-2$ , which leaves carbon at $+4$ .

No bond in $\text{CO}_2$ is really ionic, so this is only bookkeeping. But it gives every atom in every substance a number that behaves like a charge, and that is all you need to follow electrons through a reaction.

You do not need to draw bonds each time. The rules below give the same answer much faster.

Rule	Value	Exceptions
A free element (uncombined, any allotrope)	$0$	none
A simple (monatomic) ion	equals the ion's charge	none — e.g. $\text{Na}^+ = +1$ , $\text{O}^{2-} = -2$
Fluorine, combined	always $-1$	none — the most electronegative element there is
Oxygen, combined	usually $-2$	peroxides ( $\text{O}_2^{2-}$ , e.g. $\text{H}_2\text{O}_2$ ) $= -1$ ; $\text{OF}_2 = +2$ (F outranks O)
Hydrogen, combined	usually $+1$	metal hydrides (e.g. $\text{NaH}$ ) $= -1$ , since H is more electronegative than a reactive metal there
Group 1 and Group 2 metals, combined	Group 1 $= +1$ , Group 2 $= +2$	none
Cl, Br or I, combined	usually $-1$	positive when bonded to O, F or a more electronegative halogen (e.g. Cl in $\text{ClO}^-$ $= +1$ , I in $\text{ICl}$ $= +1$ )
Sum in a neutral species	all oxidation numbers add to $0$	—
Sum in an ion	all oxidation numbers add to the ion's overall charge	—

Use the fixed values first (free element, simple ion, F, O, H, Group 1 and 2 metals, halogens). Find the element that is left over last, by making the total come out right.

The order to work in: fixed values first, then solve for whatever is left so the total is zero (or the ion's charge).

When two rules clash

The rules follow electronegativity, so the more electronegative atom always wins the negative value. That is why O is $+2$ in $\text{OF}_2$ (F beats O), H is $-1$ in $\text{NaH}$ (H beats Na), and in a compound of two non-metals with no F, O or H, such as $\text{ICl}$ or $\text{SCl}_2$ , the more electronegative atom (Cl) takes its usual negative value.

Finding an oxidation number in a molecule and in an ion

Find the oxidation number of (a) sulfur in $\text{H}_2\text{SO}_4$ and (b) manganese in $\text{MnO}_4^-$ .

Show full working

1
(a) Fixed values first: H is $+1$ and O is $-2$ .
Check the exceptions before using a usual value: this is not a metal hydride and not a peroxide, so the usual values hold.
2
Let the oxidation number of S be $x$ . Count the atoms: $2$ H, $1$ S, $4$ O. The molecule is neutral, so the total is $0$ : $2(+1) + x + 4(-2) = 0$
Multiply each value by the number of atoms of that element. Forgetting the 4 on oxygen is the most common slip.
3
Multiply out: $2 + x - 8 = 0$
4
Solve: $x = +6$
Write the sign every time. It is optional for a positive value, but leaving it off becomes a habit that costs marks on negative values.
5
(b) Fixed value first: O is $-2$ . Let Mn be $y$ . There are $4$ O atoms.
6
This is an ion with charge $-1$ , so the total is $-1$ , not $0$ : $y + 4(-2) = -1$
For an ion, the oxidation numbers add up to the charge on the ion. Setting the total to 0 here would give +8, which is wrong.
7
Multiply out and solve: $y - 8 = -1$ $y = +7$

Answer

(a) S = +6 in H₂SO₄. (b) Mn = +7 in MnO₄⁻.

Two oxidation numbers for the same element in two related compounds

9701/22 F/M 2025 Q1(a)(i)2 marks

Phosphorus and chlorine are elements in Period 3 of the Periodic Table. Chlorine forms three different compounds with phosphorus. The most common compounds are $\text{PCl}_3$ and $\text{PCl}_5$ .

Complete Table 1.1.

Table 1.1

compound	oxidation number of P	oxidation number of Cl
$\text{PCl}_3$
$\text{PCl}_5$

Show full working

1
Chlorine is more electronegative than phosphorus and there is no F, O or H here, so Cl takes its usual value: $\text{Cl} = -1$ in both compounds.
In a compound of two non-metals, the more electronegative atom takes the negative value.
2
$\text{PCl}_3$ is neutral, with three Cl: $\text{P} + 3(-1) = 0$ $\text{P} = +3$
3
$\text{PCl}_5$ is neutral, with five Cl: $\text{P} + 5(-1) = 0$ $\text{P} = +5$
Each row of the table needs both numbers for its mark, so fill in Cl as well as P.

Answer

PCl₃: P = +3, Cl = −1. PCl₅: P = +5, Cl = −1.

The same element can have different oxidation numbers in different compounds. The names say it: phosphorus(III) chloride and phosphorus(V) chloride.

Comparing two oxidation numbers across a period, and explaining why they differ

9701/21 M/J 2025 Q1(c)2 marks

State the oxidation number of the Period 3 elements bonded to $\text{Cl}$ in $\text{NaCl}$ and $\text{PCl}_5$ .

Explain the difference in the oxidation number.

Show full working

1
$\text{NaCl}$ : Na is a Group 1 metal, so $\text{Na} = +1$ (and Cl is $-1$ , which checks: $+1 - 1 = 0$ ).
2
$\text{PCl}_5$ : Cl is $-1$ , five of them. $\text{P} + 5(-1) = 0$ $\text{P} = +5$
M1 needs both values together: NaCl +1 and PCl₅ +5.
3
Explain using outer-shell (valence) electrons. Sodium, $[\text{Ne}]\,3\text{s}^1$ , has only one valence electron, so its oxidation number can be at most $+1$ . Phosphorus, $[\text{Ne}]\,3\text{s}^2 3\text{p}^3$ , has five valence electrons, and in $\text{PCl}_5$ all five are used in bonding, giving $+5$ .
M2 needs both halves: Na has 1 valence electron AND P has 5. The configurations come from “Writing electron configurations” in the AS Atomic Structure note.

Answer

NaCl: Na = +1; PCl₅: P = +5. Na has only 1 valence (outer-shell) electron, while P has 5 valence electrons, all used in bonding in PCl₅.

An element's highest oxidation number is usually the number of its outer-shell electrons, which for a main-group element matches its group (Group 15 has 5).

Roman numerals: writing the oxidation number into a name

A Roman numeral in a name gives the oxidation number of the element just before it. You met this for metal ions: iron(III) chloride is $\text{FeCl}_3$ , with $\text{Fe} = +3$ .

The same idea names oxyanions (negative ions containing oxygen). The numeral gives the oxidation number of the central element:

name	formula	oxidation number of the central atom
sulfate(VI)	$\text{SO}_4^{2-}$	S $= +6$
sulfate(IV) (sulfite)	$\text{SO}_3^{2-}$	S $= +4$
nitrate(V)	$\text{NO}_3^-$	N $= +5$
nitrate(III) (nitrite)	$\text{NO}_2^-$	N $= +3$
chlorate(I)	$\text{ClO}^-$	Cl $= +1$
chlorate(V)	$\text{ClO}_3^-$	Cl $= +5$
manganate(VII)	$\text{MnO}_4^-$	Mn $= +7$
dichromate(VI)	$\text{Cr}_2\text{O}_7^{2-}$	Cr $= +6$

You do not need to learn this table: work the numeral out from the formula, or the formula's oxidation number from the numeral. The numeral gives the size of the oxidation number.

Naming from a formula, and a formula from a name

(a) Name $\text{NaClO}_3$ using a Roman numeral. (b) Potassium manganate(VI) contains the $\text{MnO}_4^{\,n-}$ ion. Find $n$ .

Show full working

1
(a) Na is $+1$ and O is $-2$ . Let Cl be $x$ : $(+1) + x + 3(-2) = 0$
2
Solve: $x = +5$ So the anion is chlorate(V), and the compound is sodium chlorate(V).
The numeral goes on the part that contains the element with the variable oxidation number, here the chlorate ion.
3
(b) The name tells you Mn $= +6$ . O is $-2$ , four of them. Add them up to get the charge on the ion: $+6 + 4(-2) = -2$
Here you run the calculation backwards: the name gives you the oxidation number, and the sum gives the charge.
4
So the ion is $\text{MnO}_4^{2-}$ and $n = 2$ . Compare manganate(VII), $\text{MnO}_4^-$ : same atoms, different charge, different oxidation number.

Answer

(a) Sodium chlorate(V). (b) n = 2 (MnO₄²⁻).

Average oxidation numbers

Sometimes the sum rule gives a fraction. In the tetrathionate ion, $\text{S}_4\text{O}_6^{2-}$ , with O at $-2$ :

4x + 6(-2) = -2

4x = 10

x = +2.5

No single atom has an oxidation number of $+2.5$ . The four sulfur atoms are not all in the same situation, and $+2.5$ is their average. A question that says "use the average oxidation number" wants exactly this number; use it like any other value when you compare reactants and products. The same happens in $\text{Fe}_3\text{O}_4$ (average Fe $= +\tfrac{8}{3}$ ).

Your turn

1
Find the oxidation number of chromium in $\text{Cr}_2\text{O}_7^{2-}$ (dichromate).
Stuck? Show hint
There are two chromium atoms. Find their total first, then share it between them.
Show solution
1. 1
  O is $-2$ , and there are seven: $7 \times (-2) = -14$
2. 2
  Let each Cr be $x$ . The ion's charge is $-2$ : $2x + (-14) = -2$
  Write 2x, not x: both Cr atoms carry the same oxidation number.
3. 3
  Add 14 to both sides: $2x = 12$
4. 4
  Divide by 2: $x = +6$
  Forgetting to divide by 2 gives +12, which is impossible: chromium has only six electrons available for bonding.
Answer
+6 (hence the name dichromate(VI))
2
Find the oxidation number of sulfur in $\text{H}_2\text{SO}_3$ and in $\text{H}_2\text{S}$ . Why is it positive in one and negative in the other?
Show solution
1. 1
  $\text{H}_2\text{SO}_3$ : H is $+1$ (two of them), O is $-2$ (three of them). $2(+1) + x + 3(-2) = 0$ $x = +4$
2. 2
  $\text{H}_2\text{S}$ : only H is fixed, at $+1$ (two of them). $2(+1) + x = 0$ $x = -2$
3. 3
  In $\text{H}_2\text{SO}_3$ , S is bonded to O, which is more electronegative, so S "loses" electrons and is positive. In $\text{H}_2\text{S}$ , S is bonded only to H, which is less electronegative, so S "gains" electrons and is negative.
  The same element's oxidation number depends on what it is bonded to.
Answer
H₂SO₃: S = +4. H₂S: S = −2. S is positive next to more electronegative O, negative next to less electronegative H.
3
Find the oxidation number of: (a) O in $\text{H}_2\text{O}_2$ ; (b) H in $\text{CaH}_2$ ; (c) Cl in $\text{Cl}_2\text{O}$ ; (d) the average oxidation number of Fe in $\text{Fe}_3\text{O}_4$ .
Stuck? Show hint
Parts (a) to (c) each involve an exception. Decide first which atom is more electronegative.
Show solution
1. 1
  (a) H is $+1$ (two of them) and the molecule is neutral: $2(+1) + 2x = 0$ $x = -1$
  H₂O₂ is a peroxide, so O is −1, not −2. Trust the sum rule: it gives −1 automatically.
2. 2
  (b) Ca is a Group 2 metal, so $+2$ : $(+2) + 2x = 0$ $x = -1$
  CaH₂ is a metal hydride: H is more electronegative than Ca, so H is −1.
3. 3
  (c) O is more electronegative than Cl, so O keeps $-2$ : $2x + (-2) = 0$ $x = +1$
  Cl is only −1 when it is the more electronegative partner. Next to O it is positive.
4. 4
  (d) O is $-2$ (four of them); let the average Fe be $x$ (three of them): $3x + 4(-2) = 0$ $3x = 8$ $x = +\tfrac{8}{3} \approx +2.67$
  A fraction means an average. (In fact Fe₃O₄ contains one Fe²⁺ and two Fe³⁺: (2 + 3 + 3)/3 = 8/3.)
Answer
(a) −1 (b) −1 (c) +1 (d) +8/3 (about +2.67)
4
(a) Name $\text{NaClO}$ and $\text{KNO}_3$ using Roman numerals. (b) Write the formula of the sulfate(IV) ion.
Show solution
1. 1
  $\text{NaClO}$ : Na $+1$ , O $-2$ . $(+1) + x + (-2) = 0$ $x = +1$ So it is sodium chlorate(I).
2. 2
  $\text{KNO}_3$ : K $+1$ , O $-2$ (three of them). $(+1) + x + 3(-2) = 0$ $x = +5$ So it is potassium nitrate(V).
3. 3
  (b) Sulfate(IV) means S $= +4$ . Try $\text{SO}_3^{\,n-}$ : $+4 + 3(-2) = -2$ so the ion is $\text{SO}_3^{2-}$ .
  Sulfate(VI), SO₄²⁻, and sulfate(IV), SO₃²⁻, have the same charge but different numbers of O, so the numeral is what tells them apart.
Answer
(a) Sodium chlorate(I); potassium nitrate(V). (b) SO₃²⁻.

Paper 1 · MCQReal past-paper questions · Oxidation numbers and balancing Paper 2 · TheoryReal past-paper questions · Oxidation numbers and balancing

Checking your access…

Can you do all of these?

Apply the oxidation-number rules in the correct order: fixed values first (element, simple ion, F, O, H, Group 1/2, halogens), then solve for the rest
Calculate the oxidation number of any element in a compound or ion, including an average value such as +2.5 in S₄O₆²⁻
Read and write a Roman numeral as an oxidation number, including in oxyanion names such as chlorate(V)
Define oxidation and reduction both ways (electron transfer, oxidation-number change) and decide whether a reaction is redox
Identify the oxidising agent and reducing agent in a given reaction, and explain why
Define disproportionation, and prove it with three oxidation numbers: start, higher and lower
Balance a redox equation by matching the total increase and decrease in oxidation number, then balancing charge with H⁺ and H and O with H₂O
Use electron balance to find a reacting ratio, a mole amount or an unknown oxidation number without the full equation
Write a half-equation in acidic or alkaline solution, and combine two half-equations, cancelling electrons, H⁺ and H₂O

Paper 1 · MCQ

Real past-paper questions, sorted by difficulty, with mark schemes

Paper 2 · Theory

Real past-paper questions, sorted by difficulty, with mark schemes

Electrochemistry

Oxidation numbers: the rules, and how to use them

What an oxidation number tracks

Roman numerals: writing the oxidation number into a name

Average oxidation numbers

Redox, and oxidising and reducing agents

Disproportionation

Can you do all of these?

Oxidation numbers: the rules, and how to use them

What an oxidation number tracks

Roman numerals: writing the oxidation number into a name

Average oxidation numbers

The rest of this note

Can you do all of these?