Redox Reactions — Science Sage

Oxidation and Reduction — Three Definitions

All three definitions describe the same phenomenon — they are consistent with one another. Whichever definition you apply to a reaction, you will always get the same answer about what is oxidised and what is reduced.

In terms of	Oxidation	Reduction
Electrons	Loss of electrons (LEO)	Gain of electrons (GER)
Oxygen	Gain of oxygen	Loss of oxygen
Hydrogen	Loss of hydrogen	Gain of hydrogen

OIL RIG — Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). Oxidation and reduction always occur SIMULTANEOUSLY — you cannot have one without the other. The substance that causes oxidation is the oxidising agent; the one that causes reduction is the reducing agent.

⚠ Counterintuitive — memorise this: The reducing agent is itself oxidised; the oxidising agent is itself reduced. The agent does the opposite of what it is called. It donates or accepts electrons to cause the change in the other substance, and in doing so undergoes the reverse change itself.

Oxidation Numbers (Oxidation States)

The oxidation number (ON) is the charge an atom would have if the compound were fully ionic — all shared electrons assigned to the more electronegative atom. Oxidation numbers are a bookkeeping tool to track electron transfer in redox reactions.

Rules for assigning oxidation numbers:

Rule 1 — Pure element: ON = 0 (e.g. Fe, O₂, Cl₂, Na, H₂ — atoms in their elemental form)
Rule 2 — Monoatomic ion: ON = charge of the ion (e.g. Fe³⁺ → ON = +3; Cl⁻ → ON = −1; Ca²⁺ → ON = +2)
Rule 3 — Oxygen: usually −2 (exceptions: peroxides e.g. H₂O₂, ON(O) = −1; in OF₂, ON(O) = +2 because F is more electronegative)
Rule 4 — Hydrogen: usually +1 (exception: in metal hydrides e.g. NaH, CaH₂, ON(H) = −1)
Rule 5 — Fluorine: always −1 (most electronegative element; never positive)
Rule 6 — Sum rule: The sum of all oxidation numbers in a species = the overall charge of that species (0 for neutral compounds; equal to the ionic charge for polyatomic ions)

Worked examples:

H₂O: H(+1)×2 + O(?) = 0 → O = −2 ✓
SO₄²⁻: S(?) + O(−2)×4 = −2 → S + (−8) = −2 → S = +6
MnO₄⁻: Mn(?) + O(−2)×4 = −1 → Mn + (−8) = −1 → Mn = +7
Cr₂O₇²⁻: 2×Cr(?) + O(−2)×7 = −2 → 2Cr − 14 = −2 → Cr = +6
Na₂O₂: Na(+1)×2 + O(?)×2 = 0 → 2 + 2O = 0 → O = −1 (peroxide!)

If ON increases → the species is OXIDISED → it is the REDUCING AGENT (gives away electrons, causing its own ON to rise). If ON decreases → the species is REDUCED → it is the OXIDISING AGENT (receives electrons, causing its own ON to fall).

Balancing Redox Equations — Half-Reaction Method

Step-by-step procedure for acidic solutions:

Step 1: Write the unbalanced ionic equation and identify which atoms change oxidation number (those are the atoms being oxidised or reduced).
Step 2: Separate into two half-reactions — one for oxidation, one for reduction.
Step 3: Balance all atoms except O and H in each half-reaction.
Step 4: Balance O atoms by adding H₂O to whichever side is deficient in oxygen.
Step 5: Balance H atoms by adding H⁺ to whichever side is deficient in hydrogen.
Step 6: Balance charge by adding electrons (e⁻) — electrons appear on the LEFT of the reduction half-reaction (being gained) and on the RIGHT of the oxidation half-reaction (being released).
Step 7: Make the number of electrons equal in both half-reactions by multiplying each half-reaction by appropriate integers.
Step 8: Add the two half-reactions together; cancel the electrons and any species that appear identically on both sides.
Step 9: Simplify if possible (reduce coefficients if they all share a common factor).

Worked Example 1: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic solution)

Oxidation half-reaction: Fe²⁺ → Fe³⁺ + e⁻
(ON of Fe: +2 → +3; loses 1 electron)

Reduction half-reaction: MnO₄⁻ → Mn²⁺
(ON of Mn: +7 → +2; gains 5 electrons)

Balance O: MnO₄⁻ → Mn²⁺ + 4H₂O
Balance H: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Balance charge: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Multiply oxidation × 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻

Add half-reactions:
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

Worked Example 2: Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ (acidic solution)

Oxidation: 2I⁻ → I₂ + 2e⁻
(ON of I: −1 → 0; each I loses 1 electron; 2 electrons total)

Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
(ON of Cr: +6 → +3; each Cr gains 3 electrons; 6 electrons total)

Multiply oxidation × 3: 6I⁻ → 3I₂ + 6e⁻

Add half-reactions:
Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂

Oxidising and Reducing Agents

Property	Oxidising Agent (OA)	Reducing Agent (RA)
Effect on other substance	Oxidises it	Reduces it
What happens to itself	Gets reduced	Gets oxidised
Electron movement	Gains electrons	Loses electrons
ON change	ON decreases	ON increases

Strong oxidising agents: MnO₄⁻, Cr₂O₇²⁻, F₂, Cl₂, H₂O₂, O₂, concentrated HNO₃
Strong reducing agents: Na, Li, K, Mg, Fe, Zn (active metals), H₂, I⁻, Fe²⁺, SO₂

The Electrochemical (Reduction Potential) Series

The electrochemical series lists standard reduction half-reactions alongside their standard reduction potentials (E°), measured in volts under standard conditions (298 K, 1 mol·dm⁻³, 101.3 kPa).

TOP of table: most positive E° → strongest oxidising agents (these species are most easily reduced)
BOTTOM of table: most negative E° → strongest reducing agents (these species are most easily oxidised)
A species in the series can oxidise any species that appears below it on the right-hand (reducing) side

Half-reaction (as written: reduction)	E° (V)
F₂ + 2e⁻ → 2F⁻	+2.87
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O	+1.51
Cl₂ + 2e⁻ → 2Cl⁻	+1.36
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O	+1.33
O₂ + 4H⁺ + 4e⁻ → 2H₂O	+1.23
Fe³⁺ + e⁻ → Fe²⁺	+0.77
I₂ + 2e⁻ → 2I⁻	+0.54
Fe²⁺ + 2e⁻ → Fe	−0.44
Zn²⁺ + 2e⁻ → Zn	−0.76
Na⁺ + e⁻ → Na	−2.71
Li⁺ + e⁻ → Li	−3.04

Predicting spontaneous reactions: The species with the higher (more positive) E° acts as the oxidising agent and gets reduced. Calculate E°cell = E°cathode − E°anode. If E°cell > 0, the reaction is spontaneous. If E°cell < 0, the reaction is non-spontaneous under standard conditions.

Corrosion of Iron (Rusting)

Iron is oxidised in the presence of water and dissolved oxygen — rusting is an electrochemical process that requires both an oxidant (O₂) and an electrolyte (water with dissolved ions).

Anode (oxidation): Fe → Fe²⁺ + 2e⁻
Cathode (reduction): O₂ + 4H⁺ + 4e⁻ → 2H₂O
or (neutral/basic): O₂ + 2H₂O + 4e⁻ → 4OH⁻

Overall: 4Fe + 3O₂ + xH₂O → 2Fe₂O₃·xH₂O (rust — hydrated iron(III) oxide)

Water acts as the electrolyte; dissolved salt (e.g. sea spray) speeds up rusting by greatly increasing the solution's conductivity.

Painting / coating: physical barrier excluding O₂ and water; only effective while the coating is intact
Galvanising (zinc coating): Zn has a more negative E° than Fe (−0.76 V vs −0.44 V), so Zn corrodes preferentially — it acts as a sacrificial anode. Even if the coating is scratched, Zn continues to protect the underlying Fe.
Tin plating (tin cans): Sn is below Fe in the activity series (less negative E°) — only protective while intact. If scratched, Fe acts as the anode and corrodes FASTER (galvanic acceleration).
Cathodic protection: Connect iron structure to a block of more active metal (Mg or Zn). The active metal is oxidised (sacrificial anode); iron acts as the cathode and is protected.
Electroplating with noble metals: Coating with Cr, Ni provides a hard, corrosion-resistant surface barrier.

IEB Extension: Electrochemical Cells & Electrolysis

Galvanic (voltaic) cells convert spontaneous chemical energy into electrical energy.

Anode: oxidation occurs; negative electrode in a galvanic cell
Cathode: reduction occurs; positive electrode in a galvanic cell
Salt bridge: maintains electrical neutrality between the two half-cells by allowing ion flow (typically KNO₃ or KCl in agar)
Cell notation: Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s) — left of || is anode; right of || is cathode; single | = phase boundary
E°cell = E°cathode − E°anode; spontaneous when E°cell > 0

Electrolytic cells use an external power source to drive non-spontaneous redox reactions.

Anode: oxidation; connected to + terminal of external supply
Cathode: reduction; connected to − terminal of external supply
Electrolysis of brine (concentrated NaCl):

Cathode: 2H₂O + 2e⁻ → H₂↑ + 2OH⁻ (hydrogen gas)
Anode: 2Cl⁻ → Cl₂↑ + 2e⁻ (chlorine gas)
Net products: H₂, Cl₂, NaOH(aq) — the chlor-alkali process

Simulation Mode

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e⁻ transfer

—e⁻

Question 1 · NSC

In the reaction 2Mg + O₂ → 2MgO, which statement is correct?

Mg goes from ON = 0 to ON = +2 in MgO → ON increases → Mg is OXIDISED → Mg is the REDUCING AGENT. O₂ goes from ON = 0 to ON = −2 → ON decreases → O₂ is REDUCED → O₂ is the OXIDISING AGENT.

Question 2 · NSC

What is the oxidation number of Cr in Cr₂O₇²⁻?

Let ON(Cr) = x. Two Cr atoms + seven O atoms (each −2) = total charge −2. So 2x + 7(−2) = −2 → 2x − 14 = −2 → 2x = 12 → x = +6.

Question 3 · NSC

In the redox reaction MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺, the oxidising agent is:

MnO₄⁻ is reduced (Mn goes from +7 to +2; ON decreases) → MnO₄⁻ is the oxidising agent. Fe²⁺ is oxidised (Fe goes from +2 to +3) → Fe²⁺ is the reducing agent.

Question 4 · NSC

When balancing the half-reaction MnO₄⁻ → Mn²⁺ in acidic solution, what must be added to balance the oxygen atoms?

MnO₄⁻ has 4 oxygen atoms; Mn²⁺ has none. To balance O, add 4H₂O to the right (product) side: MnO₄⁻ → Mn²⁺ + 4H₂O. Then balance the 8 H atoms by adding 8H⁺ to the left.

Question 5 · NSC

Iron rusts faster in sea water than in fresh water because:

Rusting is an electrochemical process — it requires an electrolyte to conduct ions. Dissolved salts (NaCl) ionise in solution, greatly increasing the conductivity of sea water → faster ion flow → faster corrosion.

Question 6 · NSC

Zinc is used to galvanise iron. If the zinc coating is scratched, the iron underneath does NOT rust because:

Zinc has a more negative reduction potential than iron (E°(Zn²⁺/Zn) = −0.76 V vs E°(Fe²⁺/Fe) = −0.44 V). So Zn is preferentially oxidised (acts as the sacrificial anode), protecting the iron from corrosion even when the coating is broken.

Question 7 · NSC

Chlorine gas is bubbled into cold, dilute NaOH solution to make household bleach: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. What has happened to the chlorine in this reaction?

In Cl₂, chlorine has ON = 0. In the product Cl⁻, ON = −1 (a decrease → reduction). In the product ClO⁻, ON = +1 (an increase → oxidation). The SAME element, starting from the SAME oxidation state, ends up both oxidised and reduced in different product species — this is called a disproportionation reaction. It shows that redox analysis must be applied per species, not assumed to be a single simple oxidation-or-reduction change.

Question 8 · NSC

In the balanced redox reaction Cr₂O₇²⁻ + 3H₂S + 8H⁺ → 2Cr³⁺ + 3S + 7H₂O, how many electrons are transferred per Cr₂O₇²⁻ ion reduced?

Each Cr atom changes from ON = +6 (in Cr₂O₇²⁻) to ON = +3 (in Cr³⁺), a gain of 3 electrons per Cr atom. Since there are 2 Cr atoms per Cr₂O₇²⁻ ion, a total of 2 × 3 = 6 electrons are gained per Cr₂O₇²⁻ ion reduced. This matches the electrons lost by the reducing agent: each S atom changes from ON = −2 (in H₂S) to ON = 0 (elemental S), losing 2 electrons per atom; with 3 S atoms oxidised, 3 × 2 = 6 electrons are lost — confirming the electron balance.

IEB Extension Questions

Question 7 · IEB

In a galvanic cell, Zn|Zn²⁺||Cu²⁺|Cu: E°(Zn²⁺/Zn) = −0.76 V; E°(Cu²⁺/Cu) = +0.34 V. What is E°cell?

E°cell = E°cathode − E°anode = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn) = +0.34 − (−0.76) = +0.34 + 0.76 = +1.10 V. Positive E°cell confirms the reaction is spontaneous.

Question 8 · IEB

During the electrolysis of brine (concentrated NaCl solution), which gas is produced at the cathode?

At the cathode (reduction), water is preferentially reduced: 2H₂O + 2e⁻ → H₂ + 2OH⁻. Hydrogen gas is produced. At the anode, Cl⁻ is oxidised: 2Cl⁻ → Cl₂ + 2e⁻ (chlorine gas).

Answer all questions in full. Show all working where required. This workbook may be completed and submitted as a classwork or homework assignment.

Question 1 [6 marks]

For the following species, assign the oxidation number of the indicated element and state whether any change from a previous state represents oxidation or reduction:

(a) S in H₂SO₄ (b) S in H₂S (c) N in NO₃⁻ (d) N in NH₃

Using your answers, identify which of the following represents oxidation: H₂S → H₂SO₄ or NO₃⁻ → NH₃? Explain your answer in terms of oxidation numbers.

Question 2 [7 marks]

Balance the following redox equation in acidic solution using the half-reaction method. Show ALL steps clearly:

Cr₂O₇²⁻(aq) + Fe²⁺(aq) → Cr³⁺(aq) + Fe³⁺(aq)

Your answer must include: (i) the unbalanced half-reactions, (ii) balancing of atoms, (iii) balancing of charge with electrons, (iv) the final balanced overall equation.

Question 3 [4 marks]

Identify the oxidising agent and the reducing agent in each of the following reactions. Explain your choice by referring to oxidation numbers:

(a) Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂

(b) Zn + 2HCl → ZnCl₂ + H₂

Question 4 [5 marks]

A steel (iron) structure near the coast is corroding rapidly.

(a) Write the half-reactions for the corrosion of iron, clearly indicating which is oxidation and which is reduction. (2 marks)

(b) Name ONE method to protect this structure from corrosion. Explain in terms of electrochemistry WHY this method works, referring to reduction potentials or the activity series where relevant. (3 marks)

Question 5 [3 marks]

Use the electrochemical series to predict whether the following reaction will occur spontaneously. Show your reasoning by comparing reduction potentials and calculating E°cell.

Given: E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(I₂/I⁻) = +0.54 V

Reaction: I₂(aq) + 2Fe²⁺(aq) → 2I⁻(aq) + 2Fe³⁺(aq)

Question 6 [7 marks]

Two identical iron nails (each starting at 5.00 g) are left to corrode: one in a beaker of seawater, one in a beaker of freshwater. The mass of each nail is measured every 2 weeks:

Time (weeks)	Mass in seawater (g)	Mass in freshwater (g)
0	5.00	5.00
2	4.82	4.96
4	4.64	4.92
6	4.46	4.88
8	4.28	4.84

(a) Plot mass (y-axis) against time (x-axis) for both nails on the same set of axes. (2 marks)
(b) Calculate the rate of mass loss (gradient, in g·week⁻¹) for the nail in seawater. (2 marks)
(c) Calculate the rate of mass loss for the nail in freshwater, and determine how many times faster the seawater corrosion rate is. (2 marks)
(d) Explain, in terms of electrolyte conductivity, why the corrosion rate differs between the two beakers. (1 mark)