Oxidation Number

Full Notes

Oxidation State (Oxidation Number)

Oxidation states help track electron transfer in reactions. It is straightforward to see how atoms have lost or gained electrons when ions get formed, however it can be harder to see how atoms have lost or gained electron density when dealing with molecules.

For example, carbon is oxidised to form carbon dioxide when combusted. However, no ions get formed, meaning it isn’t immediately clear how electrons are involved!

To help, we consider each atom to have an ‘imaginary’ charge, described as its oxidation number (or state).

Rules for assigning oxidation states:

1. Uncombined elements (e.g., O₂, N₂, Fe) have an oxidation state of 0.
2. Group 1 metals = +1, Group 2 metals = +2.
3. Oxygen is −2, except:
   • In peroxides (O₂²⁻), oxygen is −1.
   • With fluorine (OF₂), oxygen is +2.
4. Hydrogen is +1, except in metal hydrides (e.g., NaH), where it is −1.
5. In a neutral compound, the sum of oxidation states = 0.
6. In polyatomic ions, the sum of oxidation states = charge of the ion.

Using these rules, we can see now how carbon gets oxidised from an oxidation state of 0 in C(s) to +4 in CO₂(g).

An increase in oxidation number (gets more positive) means oxidation has occurred. A decrease in oxidation number (gets more negative) means reduction has occurred.

Rules for Assigning Oxidation Numbers

Example Assign oxidation states in H₂SO₄

H = +1 (there are 2 H, total +2).
O = −2 (there are 4 O, total −8).
The total charge must be 0, so:
S must be +6 to balance the equation: 2(+1) + S + 4(−2) = 0 → S = +6.

Roman Numerals in Names

Oxidation numbers are shown in Roman numerals in the names of compounds – particularly for transition metals and other elements with variable oxidation states.

Examples Variable oxidation states

• FeCl₂ → iron(II) chloride → Fe = +2
• FeCl₃ → iron(III) chloride → Fe = +3
• MnO₄⁻ → manganate(VII) ion → Mn = +7

Oxidising and Reducing Agents

In any redox reaction, one species donates electrons (reducing agent), and another accepts them (oxidising agent):

• Oxidising agent: Accepts electrons (is reduced). Causes another species to be oxidised.
• Reducing agent: Donates electrons (is oxidised). Causes another species to be reduced.

Example Zn + Cu²⁺ → Zn²⁺ + Cu

Zn is oxidised (loses electrons) = reducing agent.
Cu²⁺ is reduced (gains electrons) = oxidising agent.

Fractional Oxidation Numbers

Sometimes, we come across compounds where the oxidation number of an element appears to be a fraction, which might feel confusing at first.

For Example
In C₃O₂ (carbon suboxide), the oxidation number of carbon = 4/3

What’s really going on?

Fractional oxidation numbers don’t mean electrons are shared in fractions. Instead, they indicate an average oxidation number across multiple atoms of the same element – often because different atoms are in different oxidation states within the same molecule.

C₃O₂ (Carbon Suboxide)

Two terminal carbon atoms are in the +2 oxidation state. The central carbon (with asterisk) is in the 0 oxidation state. Average = (2 + 0 + 2)/3 = 4/3

Types of Redox Reactions

Redox reactions can be classified into four main types depending on how the oxidation and reduction occur. Let’s explore each one with examples and reasoning.

Combination Reactions

In a combination reaction, two substances combine to form a single product.

General form: A + B → C

For it to be a redox reaction, at least one element must be in its elemental form. Most combination reactions involving oxygen are redox reactions.

Examples Combination redox

• C(s) + O₂(g) → CO₂(g)
   C is oxidised (0 → +4) and O is reduced (0 → −2)
• 3Mg(s) + N₂(g) → Mg₃N₂(s)
   Mg is oxidised (0 → +2) and N is reduced (0 → −3)
• CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
   C is oxidised and O is reduced

Decomposition Reactions

These are the opposite of combination reactions. A compound breaks into two or more simpler substances, often involving redox changes.

General form: AB → A + B

At least one of the products should be in elemental form. Often requires heat (Δ) or light.

Examples Decomposition redox

• 2H₂O(l) → 2H₂(g) + O₂(g)
   H is reduced and O is oxidised
• 2NaH(s) → 2Na(s) + H₂(g) 
  H is oxidised (−1 → 0) and Na is reduced
• 2KClO₃(s) → 2KCl(s) + 3O₂(g)
   Cl is reduced (+5 → −1) and O is oxidised (−2 → 0)

Note: Not all decomposition reactions are redox – e.g., CaCO₃(s) → CaO(s) + CO₂(g) has no change in oxidation states.

Displacement Reactions

In these reactions, an atom or ion in a compound is displaced by another atom or ion.

General form: X + YZ → XZ + Y

Two types:

(a) Metal Displacement:
A more reactive metal displaces a less reactive metal from its salt solution. Based on the activity series of metals.

Examples Metal displacement

• CuSO₄(aq) + Zn(s) → ZnSO₄(aq) + Cu(s)
• V₂O₅(s) + 5Ca(s) → 2V(s) + 5CaO(s)
• Cr₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Cr(s)

The metal that loses electrons (oxidised) is the reducing agent.

(b) Non-metal Displacement:
Mostly involves halogens or hydrogen. A more reactive halogen displaces a less reactive one from its salt.

Examples Non-metal displacement

• Cl₂(g) + 2KBr(aq) → 2KCl(aq) + Br₂(l)
• Br₂(l) + 2I⁻(aq) → 2Br⁻(aq) + I₂(s)

Disproportionation Reactions

In these, the same element is both oxidised and reduced.

Conditions:

• One element must have at least three possible oxidation states.
• Starts from an intermediate oxidation state.

Examples Disproportionation

• 2H₂O₂(aq) → 2H₂O(l) + O₂(g)
   O is oxidised and reduced (oxidation number goes from −1 to −2 and 0).
• Cl₂(g) + 2OH⁻(aq) → Cl⁻(aq) + ClO⁻(aq) + H₂O(l)
   Cl is oxidised and reduced (oxidation number goes from 0 to −1 and +1).
• P₄(s) + 3OH⁻(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂⁻(aq)
   P is oxidised and reduced (oxidation number goes from 0 to −3 and +1).

Balancing of Redox Reactions

Balancing redox reactions ensures conservation of both mass and charge. Two widely used methods are:

• Oxidation Number Method
• Half-Reaction Method

Each method has its own advantages, and the choice depends on the type of redox reaction.

Oxidation Number Method

This method balances redox equations by comparing changes in oxidation numbers.

Steps:

1. Write the correct formulae for all reactants and products.
2. Assign oxidation numbers to all atoms. Identify the atoms whose oxidation states change.
3. Calculate the increase and decrease in oxidation numbers. Multiply so that total increase = total decrease.
4. For ionic reactions, balance charges using H⁺ or OH⁻. Add H₂O as needed to balance O and H atoms.
5. Check and confirm that atoms and charges are balanced.

Example Problem 7.8 (acidic solution)

Balance the net ionic equation for the reaction between potassium dichromate and sodium sulphite in acidic solution:

Unbalanced:
K₂Cr₂O₇ + Na₂SO₃ + H⁺ → Cr³⁺ + SO₄²⁻

Stepwise Balanced Equation:
Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O

Half-Reaction Method

This method splits the redox equation into oxidation and reduction half-reactions, balances each, then combines them.

Steps:

1. Write the unbalanced ionic equation.
2. Separate into oxidation and reduction half-reactions.
3. Balance atoms other than O and H.
4. Balance O atoms using H₂O, and H atoms using H⁺ (in acid) or OH⁻ (in base).
5. Balance charge by adding electrons.
6. Multiply half-reactions so electrons cancel.
7. Add the half-reactions and simplify.

Example (acidic medium)

Balance the equation: Fe²⁺ + Cr₂O₇²⁻ → Fe³⁺ + Cr³⁺ in acidic medium.

Stepwise Solution:
Oxidation half: Fe²⁺ → Fe³⁺ + e⁻
Reduction half: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Balance electrons:
Multiply oxidation half by 6: 6Fe²⁺ → 6Fe³⁺ + 6e⁻

Add both:
6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

Example (basic medium)

Balance the reaction: MnO₄⁻ + I⁻ → MnO₂ + I₂ in basic solution.

Stepwise Solution:
Half-reactions: I⁻ → I₂ (oxidation); MnO₄⁻ → MnO₂ (reduction)

Balance O with H₂O and H with H⁺: MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O

Since basic, add 4OH⁻ to both sides: MnO₄⁻ + 4H⁺ + 4OH⁻ → MnO₂ + 2H₂O + 4OH⁻

Simplify: MnO₄⁻ + 2H₂O → MnO₂ + 2OH⁻

Final equation: 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + 2OH⁻ + I₂

Redox Reactions as the Basis for Titrations

Redox titrations are widely used in analytical chemistry to determine unknown concentrations.

• Common titrants: KMnO₄ (permanganometry), K₂Cr₂O₇ (dichrometry), I₂ (iodometry)
• Indicators: Sometimes self-indicating (e.g., KMnO₄ is purple but turns colorless upon reduction)

For Example Fe²⁺(aq) vs MnO₄⁻(aq)

In a redox titration involving Fe²⁺(aq) and MnO₄⁻(aq), the MnO₄⁻ ions get reduced to Mn²⁺ and there is a colour change from purple to colourless.

Endpoint = first permanent pink colour (excess MnO₄⁻)

Limitations of the Concept of Oxidation Number

• It does not depict actual electron sharing in covalent compounds.
• Often arbitrary in assigning oxidation states in molecules with delocalized electrons (e.g., resonance structures).
• Does not always represent real oxidation states in coordination compounds or organometallics.

Summary

• Oxidation number is a useful bookkeeping tool for electron transfer.
• Rules allow assignment of oxidation states in compounds and ions.
• Redox reactions include combination, decomposition, displacement and disproportionation.
• Balancing can be done by oxidation number or half-reaction methods.
• Redox titrations use visible changes to determine unknown concentrations.