JEE Chemistry Redox Reactions Notes Set 05

Redox Reactions and Oxidation Number

1. A redox reaction involves oxidation (loss of electrons, rise in oxidation number) and reduction (gain of electrons, fall in oxidation number) at the same time.
2. Oxidation number is the real or hypothetical charge an atom would carry if the shared electrons in each bond went to the more electronegative atom.
3. It can be positive, negative, zero or fractional (for example N in HN₃ is \( -\frac13 \), and S in S₄O₆²⁻ averages +2.5).
4. The same element can show different oxidation numbers in different compounds, for example S in H₂S (−2), H₂SO₃ (+4) and H₂SO₄ (+6).

Rules for Finding Oxidation Numbers

1. Atoms of a free element, in any allotrope or in an alloy such as an amalgam, have oxidation number 0.
2. Fluorine is always −1. Group 1 metals are +1 and group 2 metals are +2.
3. Hydrogen is +1, except in metal hydrides such as NaH and CaH₂ where it is −1.
4. Oxygen is −2, except in peroxides (−1), superoxides (−1/2), ozonides (−1/3), OF₂ (+2) and O₂F₂ (+1).
5. A monatomic ion has oxidation number equal to its charge. Neutral ligands such as NH₃, H₂O and CO count as 0.
6. The sum of oxidation numbers is 0 in a neutral molecule and equals the charge in an ion.
7. Example: for H₂S₂O₈, which has one peroxide linkage, 2(+1) + 2x + 6(−2) + 2(−1) = 0, so S = +6.
8. Example: in [Co(NH₃)₆]Cl₂Br, x + 0 − 2 − 1 = 0, so Co = +3.

Applications of Oxidation Number

1. For oxoacids of the same element, acid strength increases with oxidation number: HClO₄ > HClO₃ > HClO₂ > HClO.
2. A species with the element in its highest oxidation state can only act as an oxidising agent (KMnO₄, K₂Cr₂O₇, H₂SO₄, HNO₃, HClO₄).
3. A species with the element in its lowest oxidation state can only act as a reducing agent (NH₃, PH₃, CH₄).
4. A species with the element in an intermediate state can act as both (HNO₂, H₃PO₃, SO₂, H₂O₂).
5. The oxidation numbers in a proposed formula must add up to zero, which tests whether the formula is possible.

n-factor and Equivalent Weight

1. The n-factor of an oxidant or reductant is the number of electrons gained or lost per formula unit, equal to the total change in oxidation number of the atoms that change.
2. Equivalent weight = molar mass divided by n-factor.
3. KMnO₄: n = 5 in acidic medium (to Mn²⁺), 3 in neutral or weakly basic medium (to MnO₂), 1 in strongly basic medium (to MnO₄²⁻).
4. K₂Cr₂O₇ acts as an oxidant in acid only, with n = 6 (2 Cr from +6 to +3), so its equivalent weight is 294/6 = 49.
5. H₂O₂ has n = 2 both as oxidant (to H₂O) and as reductant (to O₂). MnO₂ has n = 2 in acid. Oxalate (to CO₂) has n = 2 and Fe²⁺ to Fe³⁺ has n = 1.
6. At equivalence, equivalents of oxidant equal equivalents of reductant. For example, 1 mol MnO₄⁻ (5 equivalents) oxidises 5/2 mol of oxalate.

Balancing Redox Equations (Ion-Electron Method)

1. Write the two half reactions, balance atoms other than O and H, then balance O using H₂O.
2. In acid, balance H by adding H⁺. In base, balance as in acid and then add the same number of OH⁻ to both sides to remove H⁺.
3. Balance the charge with electrons, multiply the half reactions so electrons cancel, then add.
4. Example: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
5. Example in base: 4Zn + NO₃⁻ + 7H₂O → 4Zn²⁺ + NH₄⁺ + 10OH⁻.