JEE Chemistry Redox Reactions Notes Set 02

Redox Reactions: Key Ideas

A redox reaction is one in which electrons are transferred, so oxidation and reduction always occur together. Oxidation is loss of electrons (rise in oxidation number); reduction is gain of electrons (fall in oxidation number). The species that is reduced is the oxidising agent (oxidant); the species that is oxidised is the reducing agent (reductant).

Oxidation Number: Rules

1. Atoms of a free element, in any allotrope, have oxidation number 0; so do all atoms in an alloy or amalgam.
2. A monoatomic ion has an oxidation number equal to its charge.
3. The oxidation numbers in a neutral molecule add up to 0; in a polyatomic ion they add up to the charge on the ion.
4. Fluorine is always -1 in compounds. Group 1 metals are +1 and group 2 metals are +2.
5. Hydrogen is +1, except in metal hydrides (NaH, CaH2), where it is -1.
6. Oxygen is -2, except in peroxides (-1), superoxides (-1/2), \( \ce{OF2} \) (+2) and \( \ce{O2F2} \) (+1).
7. Neutral ligands such as CO, NO, \( \ce{NH3} \) and \( \ce{H2O} \) count as zero in a complex.
8. In a covalent bond the shared pair is assigned to the more electronegative atom. In a coordinate bond the donated pair is also assigned to the more electronegative atom.

Fractional values (for example \( \ce{Fe3O4} \): 8/3, \( \ce{C3O2} \): 4/3, \( \ce{Na2S4O6} \): 2.5) are averages. Real atoms have whole-number states, so look at the structure (a peroxide link or an S-S bond) when you need the actual values. Examples worth remembering: \( \ce{CrO5} \) has Cr = +6 (one oxide and two peroxide groups); \( \ce{H2SO5} \) and \( \ce{H2S2O8} \) have S = +6; in \( \ce{NH4NO3} \) the two nitrogens are -3 and +5.

Range of Oxidation States

A compound whose key element is in its highest state can act only as an oxidant (\( \ce{KMnO4} \), \( \ce{K2Cr2O7} \), \( \ce{H2SO4} \), \( \ce{HNO3} \), \( \ce{HClO4} \)). One whose key element is in its lowest state acts only as a reductant (\( \ce{H2S} \), \( \ce{NH3} \), \( \ce{PH3} \), \( \ce{CH4} \), \( \ce{I^-} \)). One in an intermediate state can act as both (\( \ce{H2O2} \), \( \ce{SO2} \), \( \ce{HNO2} \), \( \ce{H3PO3} \)).

Uses of Oxidation Number

Acid strength of oxoacids of one element rises with the oxidation number of the central atom: \( \ce{HClO} \lt \ce{HClO2} \lt \ce{HClO3} \lt \ce{HClO4} \). Oxidation numbers also let you test a proposed formula (the total must be zero), identify the oxidant and reductant, and count electrons exchanged.

Types of Redox Reactions

1. Combination: for example \( \ce{C + O2 -> CO2} \).
2. Decomposition: for example \( \ce{2KClO3 -> 2KCl + 3O2} \). Not every decomposition is redox: \( \ce{CaCO3 -> CaO + CO2} \) is not.
3. Displacement: a more reactive metal or halogen replaces a less reactive one, for example \( \ce{Zn + Cu^2+ -> Zn^2+ + Cu} \).
4. Disproportionation: one element in one species is oxidised and reduced at the same time, for example \( \ce{Cl2 + 2OH^- -> Cl^- + ClO^- + H2O} \) (cold, dilute) and \( \ce{3Cl2 + 6OH^- -> 5Cl^- + ClO3^- + 3H2O} \) (hot, concentrated). The element must be in an intermediate state to disproportionate; \( \ce{F2} \) cannot.
5. Comproportionation is the reverse: two states of one element give a single intermediate state, as in \( \ce{NH4NO3 -> N2O + 2H2O} \).
6. Intermolecular redox involves two different species; intramolecular redox involves two different atoms in the same molecule, as in \( \ce{2KClO3 -> 2KCl + 3O2} \).

Balancing Redox Equations

Oxidation number method: find the atoms whose oxidation numbers change, equalise the total increase and decrease by choosing multipliers, then balance the remaining atoms, and finally the charge.

Ion-electron (half-reaction) method:
1. Write the two half-reactions separately.
2. Balance all atoms except H and O.
3. Balance O by adding \( \ce{H2O} \).
4. Balance H: in acid add \( \ce{H+} \); in base add \( \ce{H2O} \) on the side that needs H and the same number of \( \ce{OH-} \) on the other side.
5. Balance the charge by adding electrons.
6. Multiply the half-reactions so that electrons cancel, then add.

Standard results: \( \ce{MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O} \) and \( \ce{Cr2O7^2- + 14H+ + 6e^- -> 2Cr^3+ + 7H2O} \). The ionic equation \( \ce{2MnO4^- + 5C2O4^2- + 16H+ -> 2Mn^2+ + 10CO2 + 8H2O} \) is the model for oxalate titrations.

n-factor and Equivalent Weight

The n-factor (valency factor) of an oxidant or reductant is the number of electrons gained or lost per formula unit. Equivalent weight = molar mass / n-factor, and normality = molarity x n-factor. Because the n-factor depends on the reaction, a single compound can have several equivalent weights.

For a species with several atoms changing, add the changes of every atom: for example \( \ce{Cu2S} \) going to \( \ce{Cu^2+} \) and \( \ce{SO2} \) has n = 2 x 1 + 6 = 8. For a disproportionation the n-factor is calculated from the total electrons lost (or gained) by the portion of the species that is oxidised (or reduced).

Law of Equivalence and Titration

In any redox reaction, equivalents of oxidant = equivalents of reductant. In terms of volumes: \( N_1V_1 = N_2V_2 \). When normality is not used, equate moles x n-factor: \( n_1 M_1 V_1 = n_2 M_2 V_2 \). Meq of a substance = mass (mg) / equivalent weight.

Worked example 1: the moles of \( \ce{KMnO4} \) needed to oxidise one mole of ferrous oxalate in acid. One mole loses 1 + 2 = 3 mol of electrons, and each permanganate takes 5, so the answer is 3/5 = 0.6 mol.

Worked example 2: the volume of 6 M \( \ce{HNO3} \) (reduced to NO) to oxidise 8 g of \( \ce{Fe^2+} \). Meq of Fe^2+ = (8/56) x 1000 = 142.9. For HNO3, n = 3 (N: +5 to +2), so N = 18. Volume = 142.9/18 = 7.94 mL.

Worked example 3: the weight of 85 percent pure \( \ce{Na2CO3} \) equivalent to 45.6 mL of 0.235 N \( \ce{H2SO4} \). Meq = 45.6 x 0.235 = 10.716. Pure mass = 10.716 x 53/1000 = 0.5679 g; sample mass = 0.5679/0.85 = 0.668 g.

Worked example 4: normality of 15.8 g of \( \ce{KMnO4} \) in 50 mL of acid. E = 158/5 = 31.6; N = (15.8 x 1000)/(31.6 x 50) = 10 N.

Oxidising and Reducing Agents: Quick Facts

1. Oxidants: \( \ce{KMnO4} \), \( \ce{K2Cr2O7} \) (acid only), \( \ce{Cl2} \), \( \ce{O3} \), \( \ce{HNO3} \), hot concentrated \( \ce{H2SO4} \), \( \ce{H2O2} \) (acid or base).
2. Reductants: \( \ce{H2S} \), \( \ce{SO2} \), \( \ce{Fe^2+} \), \( \ce{Sn^2+} \), \( \ce{I^-} \), \( \ce{C2O4^2-} \), \( \ce{H2} \), metals.
3. Order of oxidising power in acid: \( \ce{MnO4^-} \gt \ce{Cr2O7^2-} \gt \ce{VO2^+} \).
4. \( \ce{KMnO4} \) is reduced to colourless \( \ce{Mn^2+} \) in acid, brown \( \ce{MnO2} \) in neutral or weakly basic medium, and green \( \ce{MnO4^2-} \) in strong alkali.
5. \( \ce{H2O2} \) with permanganate in acid releases \( \ce{O2} \): \( \ce{2MnO4^- + 5H2O2 + 6H+ -> 2Mn^2+ + 5O2 + 8H2O} \).
6. Copper(II) oxidises iodide: \( \ce{2Cu^2+ + 4I^- -> 2CuI + I2} \); the liberated iodine is titrated with thiosulphate.