Module 2: Foundations in Chemistry · Year 12
2.1.5 Redox
Assign oxidation numbers, use them in formulae and names, then track electron transfer and oxidation-state changes through redox reactions.
What you need to know
Open a line for a quick recap. If it feels obvious, move straight to the linked practice.
2.1.5 a Apply oxidation number rules to atoms in elements, compounds and ions, including peroxides and metal hydrides. Quick revision
When you assign oxidation numbers, start with the rules you know and make the total add up to the overall charge. An element on its own is 0; a simple ion has the same oxidation number as its charge; oxygen is usually −2 and hydrogen is usually +1.
There are two OCR exceptions worth keeping visible: oxygen is −1 in peroxides such as H₂O₂, and hydrogen is −1 in metal hydrides such as NaH. I always check for those before applying the usual O and H rules.
- element in its standard elemental form: 0
- simple ion: oxidation number = ion charge
- oxygen usually −2; peroxide oxygen = −1
- hydrogen usually +1; metal-hydride hydrogen = −1
- sum of oxidation numbers = overall charge
Find the oxidation number of sulfur in SO₄²⁻.
- let sulfur = x
- x + 4(−2) = −2
- x = +6
Answer Sulfur is +6.
2.1.5 b Write formulae using oxidation numbers. Quick revision
Oxidation numbers can tell you the ratio of atoms needed in a neutral compound. Make the positive and negative contributions add to zero, using the smallest whole-number ratio.
For iron(III) oxide, Fe is +3 and O is −2. Two Fe atoms give +6 and three O atoms give −6, so the formula is Fe₂O₃. It is the same charge-balance idea you used for ionic formulae.
2.1.5 c Use Roman numerals to name and interpret compounds where an element has variable oxidation states. Quick revision
The Roman numeral in a name tells you the oxidation number of the named element in that compound. So iron(III) chloride contains Fe in oxidation state +3, while iron(II) chloride contains Fe in +2.
If you are given the formula, work the oxidation number out from the other atoms and then choose the numeral. For FeCl₃, three Cl atoms contribute −3 altogether, so Fe must be +3: iron(III) chloride.
2.1.5 d(i) Explain oxidation and reduction in terms of electron transfer. Quick revision
Track the electrons. Oxidation is loss of electrons; reduction is gain of electrons. If you write a half-equation, that tells you immediately which process you have.
For Mg → Mg²⁺ + 2e⁻, the electrons appear on the right, so magnesium has lost electrons and is oxidised. For Cu²⁺ + 2e⁻ → Cu, the electrons are on the left, so copper(II) ions gain electrons and are reduced.
2.1.5 d(ii) Explain oxidation and reduction in terms of changes in oxidation number. Quick revision
You can identify the same redox change without writing electrons. If an element’s oxidation number increases, it has been oxidised. If its oxidation number decreases, it has been reduced.
I find it useful to write the before and after numbers over the species. Fe²⁺ → Fe³⁺ is +2 to +3, so that is oxidation; Cl₂ → 2Cl⁻ is 0 to −1, so chlorine is reduced.
2.1.5 e Write full equations for metals reacting with acids to form salts and hydrogen. Quick revision
For the familiar metal–acid reactions in this part of OCR, use the pattern metal + acid → salt + hydrogen. The acid tells you the salt anion, so HCl gives a chloride and dilute H₂SO₄ gives a sulfate.
Write the correct salt formula first, then balance the equation. For magnesium and hydrochloric acid, MgCl₂ needs two chloride ions, so you need two HCl molecules; the two H atoms then form H₂.
2.1.5 f Interpret familiar and unfamiliar redox reactions to identify electron loss/gain and make predictions. Quick revision
When OCR gives you an unfamiliar redox reaction, do not rely on recognising the chemicals. Track what happens to the electrons or oxidation numbers. That lets you decide which species is oxidised and which is reduced from the information in front of you.
Then use the direction of those changes to make the prediction the question asks for. I’d annotate the changing element first; it is usually much quicker than trying to reason from the whole equation at once.
- oxidation: electrons lost / oxidation number rises
- reduction: electrons gained / oxidation number falls