Redox Reactions

In the older, classical picture, oxidation meant addition of oxygen or any electronegative element to a substance, or removal of hydrogen or any electropositive element from it. Reduction meant exactly the opposite — addition of hydrogen/electropositive element, or removal of oxygen/electronegative element.

Since one substance can be oxidised only if another is simultaneously reduced, the two changes always occur together. A reaction in which oxidation and reduction take place side by side is called a redox reaction.

The modern, more general definition is based on electron transfer. Oxidation is the loss of one or more electrons by a species, and reduction is the gain of one or more electrons by a species. The species that loses electrons is the reducing agent (reductant) — it is itself oxidised. The species that gains electrons is the oxidising agent (oxidant) — it is itself reduced.

Worked example: Consider zinc metal reacting with copper sulphate solution:

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

In ionic form: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). This can be split into two half reactions:

• Oxidation half: Zn(s) → Zn²⁺(aq) + 2e^- (zinc loses 2 electrons)
• Reduction half: Cu²⁺(aq) + 2e^- → Cu(s) (copper(II) ion gains 2 electrons)

Here Zn is the reducing agent (it gets oxidised) and Cu²⁺ is the oxidising agent (it gets reduced). The 2 electrons lost by zinc are exactly the 2 electrons gained by the copper ion — this equality of electrons lost and gained is the basis for balancing redox equations.

The oxidation number (oxidation state) of an atom is the formal charge it appears to have when the electrons in each bond are assigned to the more electronegative atom. It lets us track electron transfer even in reactions where the bonding is largely covalent.

Oxidation numbers are assigned using the following rules:

1. The oxidation number of an atom in its free (elemental) form is always zero, e.g. Na(s), O₂, P₄.
2. For a monatomic ion, the oxidation number equals the charge on the ion, e.g. O.N. of Na in Na⁺ is +1, of Cl in Cl^- is -1.
3. Fluorine, the most electronegative element, always has an oxidation number of -1 in every compound it forms.
4. The other halogens (Cl, Br, I) are usually -1 in their compounds, but show positive oxidation numbers when bonded to a more electronegative atom such as oxygen or fluorine, e.g. Cl is +5 in ClO₃^-, +1 in OCl^-, and +7 in ClO₄^-.
5. Oxygen is usually -2 in its compounds. Exceptions: in peroxides (H₂O₂, Na₂O₂) it is -1; in superoxides (KO₂) it is -1/2; and in OF₂, where it is bonded to the more electronegative fluorine, it is +2.
6. Hydrogen is usually +1 in its compounds, except in metal hydrides (NaH, CaH₂), where it is -1.
7. Alkali metals are always +1 and alkaline earth metals are always +2 in their compounds.
8. The algebraic sum of the oxidation numbers of all atoms in a neutral molecule is zero; in a polyatomic ion, the sum equals the charge on the ion.

Worked example: Find the oxidation number of Mn in KMnO₄ and of Cr in K₂Cr₂O₇.

In KMnO₄, let O.N. of Mn be x: (+1) + x + 4(-2) = 0, so x - 7 = 0, giving x = +7.

In K₂Cr₂O₇, let O.N. of each Cr be x: 2(+1) + 2x + 7(-2) = 0, so 2 + 2x - 14 = 0, 2x = 12, giving x = +6.

So Mn is +7 in KMnO₄, and each Cr atom is +6 in K₂Cr₂O₇.

Combination reactions: two or more substances combine to form one product. This is a redox reaction only when at least one reactant is in elemental form, so oxidation numbers actually change.

Example: C(s) + O₂(g) → CO₂(g); here C goes from 0 to +4 and O goes from 0 to -2. Not every combination reaction is redox — in CaO(s) + CO₂(g) → CaCO₃(s), no atom changes oxidation number, so it is not a redox reaction.

Decomposition reactions: a single compound breaks into simpler substances; these are usually redox, being the reverse of combination. (Some decomposition reactions, such as CaCO₃(s) → CaO(s) + CO₂(g), are not redox, since no atom changes oxidation number.)

Example: 2KClO₃(s) → 2KCl(s) + 3O₂(g); Cl goes from +5 to -1 (reduced) and O goes from -2 to 0 (oxidised).

Displacement reactions: an atom/ion in a compound is replaced by an atom/ion of another element.

• Metal displacement: CuSO₄(aq) + Zn(s) → ZnSO₄(aq) + Cu(s) — zinc, a stronger reducing agent, displaces copper.
• Non-metal (halogen) displacement: Cl₂(g) + 2NaBr(aq) → 2NaCl(aq) + Br₂(l) — chlorine, a stronger oxidising agent, displaces bromine.

Disproportionation reactions: the same element, present in one intermediate oxidation state, is simultaneously oxidised and reduced to two different oxidation states.

Example: 2H₂O₂(aq) → 2H₂O(l) + O₂(g); oxygen in H₂O₂ is -1, and appears as -2 in H₂O (reduced) and 0 in O₂ (oxidised).

Worked example: White phosphorus disproportionates in hot concentrated alkali:

P₄(s) + 3OH^-(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^-(aq)

Phosphorus (O.N. = 0 in P₄) is reduced to -3 (in PH₃) and oxidised to +1 (in H₂PO₂^-), so phosphorus is simultaneously the oxidising agent and the reducing agent.

The oxidation number method ensures that the total increase in oxidation number (oxidation) equals the total decrease in oxidation number (reduction). Steps:

1. Write the skeletal (unbalanced) equation.
2. Identify atoms whose oxidation number changes.
3. Find the increase/decrease per atom, and multiply by the number of such atoms in the formula to get the total change.
4. Choose integer multiplying factors for the oxidant and reductant so the total increase equals the total decrease.
5. Balance atoms other than O and H by inspection.
6. Balance O by adding H₂O, then balance H by adding H⁺ (acidic medium).

Worked example: Balance the reaction of potassium permanganate with potassium iodide in sulphuric acid medium:

KMnO₄ + KI + H₂SO₄ → MnSO₄ + I₂ + K₂SO₄ + H₂O

Mn goes from +7 (in KMnO₄) to +2 (in MnSO₄), a decrease of 5 units per Mn atom. I goes from -1 (in KI) to 0 (in I₂), an increase of 1 unit per I atom.

To equalise total change, 2 KMnO₄ (total decrease = 2 × 5 = 10) must balance 10 KI (total increase = 10 × 1 = 10), i.e. 5 I₂. Balancing the remaining K, S, O and H atoms by inspection gives:

2KMnO₄ + 10KI + 8H₂SO₄ → 2MnSO₄ + 5I₂ + 6K₂SO₄ + 8H₂O

Check: K (12 = 12), Mn (2 = 2), S (8 = 8), I (10 = 10), O (40 = 40), H (16 = 16) — balanced.

The ion-electron (half-reaction) method splits the overall change into an oxidation half-equation and a reduction half-equation, balances each independently, then combines them after equalising electrons. Steps in acidic medium:

1. Write separate skeletal half-equations for oxidation and reduction.
2. Balance atoms other than O and H in each half-equation.
3. Balance oxygen by adding H₂O to the side deficient in oxygen.
4. Balance hydrogen by adding H⁺ to the side deficient in hydrogen.
5. Balance charge on each half-equation by adding electrons (e^-).
6. Multiply each half-equation so electrons lost equal electrons gained, then add and cancel common terms.

In basic medium, the equation is first balanced as in acid, then an equal number of OH^- ions is added to both sides to convert every H⁺ into H₂O (H⁺ + OH^- → H₂O), and common water molecules are simplified.

Worked example: Balance the titration reaction between dichromate ion and iron(II) ion in acidic medium:

Cr₂O₇^2-(aq) + Fe²⁺(aq) → Cr³⁺(aq) + Fe³⁺(aq)

Oxidation half-equation: Fe²⁺ → Fe³⁺ + e^-

Reduction half-equation: balance Cr (2 each side), balance O with 7H₂O, balance H with 14H⁺, balance charge with 6 electrons:

Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

Multiply the oxidation half-equation by 6: 6Fe²⁺ → 6Fe³⁺ + 6e^-. Adding both half-equations and cancelling the 6 electrons gives the final balanced ionic equation:

Cr₂O₇^2-(aq) + 14H⁺(aq) + 6Fe²⁺(aq) → 2Cr³⁺(aq) + 6Fe³⁺(aq) + 7H₂O(l)

Check: Cr (2 = 2), Fe (6 = 6), O (7 = 7), H (14 = 14); charge (left: -2 + 14 + 12 = +24, right: +6 + 18 = +24) — balanced.

Redox reactions form the chemical basis of electrode processes and electrochemical cells. When a spontaneous redox reaction is carried out by physically separating its oxidation and reduction half-reactions at two different electrodes, the electrons released at one electrode must travel through an external wire to the other electrode, and this flow of electrons constitutes an electric current.

The electrode at which oxidation occurs (electrons are released into the circuit) is called the anode; the electrode at which reduction occurs (electrons are taken up from the circuit) is called the cathode.

Example (Daniell cell): a zinc rod in ZnSO₄ solution is connected through an external circuit and a salt bridge to a copper rod in CuSO₄ solution.

• At the anode (zinc electrode), oxidation occurs: Zn(s) → Zn²⁺(aq) + 2e^-
• At the cathode (copper electrode), reduction occurs: Cu²⁺(aq) + 2e^- → Cu(s)

The overall cell reaction is the same redox change as the direct reaction between zinc metal and copper sulphate solution:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Because the two half-reactions are physically separated, the chemical energy of this spontaneous redox reaction is converted into electrical energy instead of being released simply as heat. This link between separated oxidation/reduction half-reactions and useful electrical current is the essential connection between redox reactions and electrode processes; the detailed study of cell potentials is developed further in electrochemistry.