Redox reactions involve the simultaneous transfer of electrons — one species loses electrons (oxidation) and another gains electrons (reduction). The two half-processes always occur together, hence the name redox (reduction-oxidation).
Key Concepts
Oxidation: Loss of electrons, increase in oxidation number.
Reduction: Gain of electrons, decrease in oxidation number.
Oxidising Agent: The species that gets reduced (accepts electrons).
Reducing Agent: The species that gets oxidised (donates electrons). Memory tip: OIL RIG — Oxidation Is Loss, Reduction Is Gain.
- 1.Oxidation Number (Oxidation State): A formal charge assigned to an atom in a compound. Rules:
- 2.Oxidation number of free elements = 0.
- 3.For monoatomic ions, oxidation number = charge of ion.
- 4.Oxygen is usually -2 (except in peroxides where it is -1, and in OF2 where it is +2).
- 5.Hydrogen is +1 with non-metals and -1 with metals (metal hydrides).
- 6.Sum of oxidation numbers in a neutral compound = 0; in a polyatomic ion = ionic charge.
Balancing Redox Reactions: Two methods — Oxidation Number Method and Half-Reaction (Ion-Electron) Method.
- 1.In the half-reaction method:
- 2.Write separate oxidation and reduction half-equations.
- 3.Balance atoms (use H2O for O, H+ for H in acidic medium; OH- and H2O in basic medium).
- 4.Balance charges by adding electrons.
- 5.Multiply to equalise electrons transferred.
- 6.Add the half-equations and cancel common species.
Electrode Reactions and Standard Electrode Potential (E°): In electrochemistry, the tendency of an electrode to gain electrons is expressed as its standard reduction potential. The species with higher E° is a better oxidising agent.
EMF of cell: E°cell = E°cathode - E°anode
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Worked Examples
Find the oxidation number of Mn in KMnO4.
Let Mn = x. K is +1, O is -2 (four oxygens = -8).
+1 + x + (-8) = 0 → x = +7. Oxidation number of Mn = +7.
Find oxidation number of S in H2SO4.
H2 = +2, O4 = -8. +2 + x - 8 = 0 → x = +6.
Identify oxidising and reducing agents in:
2Mg + O2 → 2MgO
Mg goes from 0 to +2 (oxidised) → Mg is the reducing agent.
O2 goes from 0 to -2 (reduced) → O2 is the oxidising agent.
Balance by half-reaction method (acidic medium):
MnO4^- + Fe2+ → Mn2+ + Fe3+
Oxidation: Fe2+ → Fe3+ + e^-
Reduction: MnO4^- + 8H+ + 5e^- → Mn2+ + 4H2O
Multiply oxidation half × 5: 5Fe2+ → 5Fe3+ + 5e^-
Add: MnO4^- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
Balance by oxidation number method: Cr2O72- + SO2 → Cr3+ + SO42- (acidic)
Cr: +6 → +3 (gain of 3e- per Cr, total 6e- for 2Cr).
S: +4 → +6 (loss of 2e- per S). Multiply S term by 3 to equalise: 3SO2 oxidised.
After balancing atoms and charges: Cr2O72- + 3SO2 + 2H+ → 2Cr3+ + 3SO42- + H2O.
Disproportionation — Cl2 + NaOH → NaCl + NaOCl + H2O
Cl2 (0) → Cl^- (-1) in NaCl (reduction) and Cl^+ (+1) in NaOCl (oxidation). This is a disproportionation reaction where the same element is both oxidised and reduced.
Calculate E°cell for the cell Zn | Zn2+ || Cu2+ | Cu, given E°(Zn2+/Zn) = -0.76 V and E°(Cu2+/Cu) = +0.34 V.
E°cell = E°cathode - E°anode = 0.34 - (-0.76) = +1.10 V. Since E°cell > 0, the reaction is spontaneous.
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Key Formulas
Key formulas
Common mistakes
- Forgetting that in peroxides (like H2O2 or Na2O2), oxygen has oxidation number -1, not -2.
- Confusing the oxidising agent with the species being oxidised — the oxidising agent is reduced.
- When balancing in basic medium, adding H+ instead of OH- ions.
Summary
Redox reactions hinge on electron transfer, tracked through oxidation numbers. Balancing them requires equalising electrons transferred between the oxidation and reduction half-reactions. The E° values predict spontaneity of redox processes, foundational for electrochemistry.