Oxidation States and Redox Reactions
Oxidation numbers, identification of redox processes, and balancing redox reactions by the electron method in acidic and basic media.
Complete Theory
4The oxidation state (oxidation number, NO) is the formal charge an atom would have if all bonds were ionic, i.e., if electrons were assigned to the more electronegative atom. It is an accounting tool for tracking electron transfers in chemical reactions.
Main rules for assignment:
Main rules for assignment:
- Elements in their free state (O₂, Fe, S₈, Cl₂): NO = 0
- Monoatomic ions: NO = ionic charge (e.g. Fe³⁺ → NO = +3)
- Fluorine: always −1 (the most electronegative element)
- Oxygen: usually −2, except in peroxides (H₂O₂, NO = −1), superoxides (KO₂, NO = −½), and OF₂ (NO = +2)
- Hydrogen: +1 except in metal hydrides (NaH, CaH₂) where NO = −1
- The algebraic sum of NOs in a neutral compound = 0; in a polyatomic ion = the ion charge
A redox (reduction-oxidation) reaction involves the transfer of electrons between two chemical species. The fundamental concepts are complementary and inseparable:
The activity series of metals ranks metals by their tendency to be oxidized (lose electrons). A metal higher in the series can reduce the ion of a metal lower in the series (e.g. Zn + Cu²⁺ → Zn²⁺ + Cu).
Real-world applications: iron corrosion (rust) is a redox process: (oxidation) and (reduction). Photosynthesis converts CO₂ to glucose (carbon reduction) using solar energy.
- Oxidation: loss of electrons → the oxidation number increases (e.g. : NO from +2 to +3). The species that is oxidized is a reducing agent (donates electrons).
- Reduction: gain of electrons → the oxidation number decreases (e.g. : NO of Mn from +7 to +2). The species that is reduced is an oxidizing agent (accepts electrons).
The activity series of metals ranks metals by their tendency to be oxidized (lose electrons). A metal higher in the series can reduce the ion of a metal lower in the series (e.g. Zn + Cu²⁺ → Zn²⁺ + Cu).
Real-world applications: iron corrosion (rust) is a redox process: (oxidation) and (reduction). Photosynthesis converts CO₂ to glucose (carbon reduction) using solar energy.
The electron method (half-reaction method) for acidic medium follows systematic steps:
For complex reactions (e.g. with polyatomic ions), the ion-electron method can be more convenient, directly balancing the species as they appear in solution.
Real-world applications: redox balancing is essential for analytical chemistry titrations (permanganometry), designing industrial electrochemical processes (chlor-alkali production, metal electrorefining), and understanding biological electron transport chains (mitochondria).
- 1. Assign oxidation numbers to all atoms and identify the two half-reactions (oxidation and reduction)
- 2. Balance atoms other than H and O in each half-reaction
- 3. Add H₂O to balance oxygen atoms, then H⁺ to balance hydrogen atoms
- 4. Add electrons (e⁻) to the appropriate side to balance charge in each half-reaction
- 5. Multiply the half-reactions by the least common multiple (LCM) of the electrons exchanged, so that electrons lost in oxidation equal those gained in reduction
- 6. Sum the two half-reactions, canceling species that appear on both sides
For complex reactions (e.g. with polyatomic ions), the ion-electron method can be more convenient, directly balancing the species as they appear in solution.
Real-world applications: redox balancing is essential for analytical chemistry titrations (permanganometry), designing industrial electrochemical processes (chlor-alkali production, metal electrorefining), and understanding biological electron transport chains (mitochondria).
An element's tendency to be oxidized (lose electrons) or reduced (gain electrons) is closely related to its position in the periodic table and its electronegativity:
Real-world applications: the electrochemical series predicts metal corrosion, enables battery design (e.g. Daniell cell: V), and helps choose oxidizing/reducing agents for chemical synthesis.
- Alkali metals (group 1) and alkaline earth metals (group 2): very low electronegativity, easily donate valence electrons → strong reducing agents (e.g. Na → Na⁺ + e⁻, Mg → Mg²⁺ + 2e⁻). They oxidize spontaneously in air.
- Halogens (group 17): high electronegativity, easily accept electrons → strong oxidizing agents (e.g. Cl₂ + 2e⁻ → 2Cl⁻). Fluorine is the most powerful oxidizer.
- Transition metals: can assume multiple oxidation states thanks to partially filled d orbitals. Examples: Mn (from +2 to +7 in MnO₄⁻), Cr (from +2 to +6 in Cr₂O₇²⁻), Fe (+2 and +3). This versatility makes them widely used redox catalysts in industry and biochemistry.
- Noble gases (group 18): stable filled-shell configuration, neither oxidized nor reduced under normal conditions.
Real-world applications: the electrochemical series predicts metal corrosion, enables battery design (e.g. Daniell cell: V), and helps choose oxidizing/reducing agents for chemical synthesis.
Worked Examples
2Example 1Assigning oxidation numbers in
Given
Find
Oxidation number of each element
Step-by-step solution
1Potassium is an alkali metal (group 1), so its oxidation number is always +1 in compounds: NO(K) = +1. Two K atoms give a total of +2.
2Oxygen follows the general rule with NO = −2. Seven O atoms give a total of −14.
3The sum of oxidation numbers in a neutral compound must be zero: . Solving: , so , . Cr in dichromate has its maximum NO (+6), making Cr₂O₇²⁻ a powerful oxidizing agent used in analytical chemistry (alcohol testing, redox titrations).
✓ Final result: K: +1, Cr: +6, O: −2
Example 2Balancing in acidic medium
Given
(acidic)
: Mn +7
Find
Balanced half-reactions
Overall balanced reaction
Step-by-step solution
1Write the reduction half-reaction: Mn goes from NO +7 (in MnO₄⁻) to +2 (in Mn²⁺), gaining 5 electrons. In acidic medium, balance oxygen with H₂O and hydrogen with H⁺: .
2Write the oxidation half-reaction: Fe²⁺ loses one electron to become Fe³⁺: .
3Calculate the least common multiple of the electrons exchanged: 5 (reduction) and 1 (oxidation) → LCM = 5. Multiply the oxidation half-reaction by 5: .
4Sum the two half-reactions, canceling electrons: . Verify: left charge = −1 + 8 + 10 = +17; right charge = +2 + 0 + 15 = +17. Balanced!
✓ Final result:
Exercises with Solutions
3Exercise 1Oxidation numbersMedium
Problem to solve
Assign the oxidation number of each atom in (phosphoric acid) and (sulfuric acid).
Given data
H₃PO₄: H +1, O −2H₂SO₄: H +1, O −2
Step-by-step solution
1H₃PO₄: H has NO +1, O has NO −2. For a neutral compound the sum is zero: . Phosphorus has NO +5, its maximum oxidation state.
2H₂SO₄: . Sulfur has NO +6, also its maximum. Both acids are strong oxidizers when hot.
✓ Final answer: P = +5 (in H₃PO₄); S = +6 (in H₂SO₄)
Exercise 2Redox balancingMedium
Problem to solve
Balance the redox reaction in acidic medium.
Given data
Cr₂O₇²⁻ (Cr +6 → +3)SO₃²⁻ (S +4 → +6)
Step-by-step solution
1Reduction half-reaction (Cr from +6 to +3, gains 3 e⁻ per atom, 6 e⁻ total): . Cr₂O₇²⁻ (dichromate) is a strong oxidizer in acidic medium.
2Oxidation half-reaction (S from +4 to +6, loses 2 e⁻): . The sulfite ion SO₃²⁻ is oxidized to sulfate SO₄²⁻.
3LCM of electrons = 6 (3 × 2e⁻ from oxidation). Multiply oxidation by 3 and sum: .
4Cancel H₂O, H⁺, and e⁻ on both sides: . Charge check: left −2 + 8 − 6 = 0; right +6 + 0 − 6 = 0.
✓ Final answer:
Exercise 3Redox in basic mediumHard
Problem to solve
Balance in basic medium: (chlorine disproportionation).
Given data
Cl₂ (Cl 0) → Cl⁻ (−1) + ClO⁻ (Cl +1)
Step-by-step solution
1This is a disproportionation reaction: the same species (Cl₂) is simultaneously oxidized and reduced. NO of Cl in Cl₂ = 0; in Cl⁻ = −1 (reduction); in ClO⁻ = +1 (oxidation).
2Reduction half-reaction: (Cl₂ gains 2 e⁻).
3Oxidation half-reaction (in basic medium, use OH⁻ directly): (Cl₂ loses 2 e⁻).
4LCM = 2 → electrons are already balanced. Sum the half-reactions: .
5Cancel electrons and divide by 2: . This is the bleach-making reaction: chlorine gas in basic solution forms hypochlorite (ClO⁻), the active bleaching and disinfecting agent.
✓ Final answer:
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