JEE Chemistry Redox Reactions Notes Set 04

Oxidation and Reduction

Oxidation is loss of electrons, gain of oxygen or loss of hydrogen; reduction is the opposite. The electron picture is the most general: an oxidation number rises when a species is oxidised and falls when it is reduced. Both processes always occur together, so such a reaction is called a redox reaction.

Oxidation Number and the Rules for Finding It

The oxidation number (state) of an atom is the charge it would carry if all bonding electron pairs were assigned to the more electronegative atom. It is a bookkeeping device, not always a real charge. The usual rules are:

1. An atom in its elemental form has oxidation number 0 (\( \ce{Na} \), \( \ce{O2} \), \( \ce{P4} \), \( \ce{S8} \)).
2. A monatomic ion has an oxidation number equal to its charge.
3. Fluorine is always -1 in compounds. Oxygen is -2, except in peroxides (-1), superoxides (-1/2) and in \( \ce{OF2} \) (+2).
4. Hydrogen is +1 with non-metals and -1 in metal hydrides such as NaH.
5. Alkali metals are +1 and alkaline earth metals are +2 in their compounds.
6. The sum of the oxidation numbers equals zero in a neutral compound and equals the charge in an ion.

Examples: in \( \ce{Mg2P2O7} \), \( 2(+2) + 2x + 7(-2) = 0 \) gives P = +5. In \( \ce{Fe3O4} \), \( 3x = 8 \), so the average oxidation state of Fe is +8/3. In \( \ce{Na2O2} \), oxygen is -1. In \( \ce{S4O6^{2-}} \), the average oxidation state of S is \( (-2 + 12)/4 = +2.5 \).

Oxidation states of sulphur in some oxo-anions: \( \ce{S2O4^{2-}} \) +3, \( \ce{SO3^{2-}} \) +4, \( \ce{S2O6^{2-}} \) +5, \( \ce{SO4^{2-}} \) +6. Oxidation states of nitrogen range from -3 (\( \ce{NH3} \)) to +5 (\( \ce{HNO3} \)), with -1 in \( \ce{N2H2} \) and -1/3 in the azide ion.

Oxidising and Reducing Agents

An oxidising agent (oxidant) accepts electrons and is itself reduced; a reducing agent (reductant) donates electrons and is itself oxidised. A species whose element is in its highest oxidation state can only act as an oxidant (for example \( \ce{KMnO4} \), \( \ce{K2Cr2O7} \), \( \ce{HNO3} \), \( \ce{H2SO4} \)); one in its lowest state can only be a reductant (for example \( \ce{H2S} \), \( \ce{I^-} \), \( \ce{NH3} \)). A species in an intermediate state can act as either, as hydrogen peroxide and sulphur dioxide do.

Hydrogen peroxide is an oxidant when oxygen goes from -1 to -2 and a reductant when oxygen goes from -1 to 0. In the reaction \( 2\ce{Fe^{2+}} + \ce{H2O2} + 2\ce{H+} \rightarrow 2\ce{Fe^{3+}} + 2\ce{H2O} \) it is the oxidant; in the reaction with acidified permanganate it is the reductant.

Types of Redox Reactions

Combination, decomposition and displacement reactions are redox if any element changes its oxidation number. Disproportionation is a reaction in which the same element in one species is both oxidised and reduced; the element must be present in an intermediate oxidation state, which can go up and down. Examples: \( \ce{Cl2} + 2\ce{OH^-} \rightarrow \ce{Cl^-} + \ce{ClO^-} + \ce{H2O} \); \( 2\ce{H2O2} \rightarrow 2\ce{H2O} + \ce{O2} \); \( 4\ce{KClO3} \rightarrow 3\ce{KClO4} + \ce{KCl} \); \( 3\ce{HNO2} \rightarrow \ce{HNO3} + 2\ce{NO} + \ce{H2O} \). Fluorine never disproportionates because it has no positive oxidation state.

The reverse process, where two species containing the same element in different oxidation states give a single product, is comproportionation (for example \( \ce{I^-} + \ce{IO3^-} \) in acid giving \( \ce{I2} \)). A decomposition such as \( 2\ce{KClO3} \rightarrow 2\ce{KCl} + 3\ce{O2} \) is not a disproportionation because chlorine only goes down while oxygen goes up.

Many reactions are not redox: precipitation, neutralisation, and the conversion of chromate to dichromate in acid \( 2\ce{CrO4^{2-}} + 2\ce{H+} \rightarrow \ce{Cr2O7^{2-}} + \ce{H2O} \), since chromium stays +6.

Balancing Redox Equations

Oxidation-number method. (1) Write the skeleton equation and assign oxidation numbers. (2) Find the change in oxidation number per atom of each changing element and multiply by the number of such atoms in the formula. (3) Multiply the formulas by factors that make the total increase equal the total decrease. (4) Balance the remaining atoms, usually H and O last, with the help of \( \ce{H2O} \), \( \ce{H+} \) or \( \ce{OH^-} \) as the medium requires.

Ion-electron (half-reaction) method. (1) Split the reaction into oxidation and reduction half-reactions. (2) Balance atoms other than H and O. (3) Balance O with \( \ce{H2O} \). (4) Balance H with \( \ce{H+} \) (acidic medium). In basic medium add one \( \ce{OH^-} \) to both sides for every \( \ce{H+} \) and combine \( \ce{H+} + \ce{OH^-} \) into \( \ce{H2O} \). (5) Balance the charge with electrons. (6) Multiply the half-reactions so that the electrons cancel, then add them.

Worked example 1 (acidic). \( \ce{MnO4^-} + \ce{C2O4^{2-}} \rightarrow \ce{Mn^{2+}} + \ce{CO2} \). Reduction: \( \ce{MnO4^-} + 8\ce{H+} + 5e^- \rightarrow \ce{Mn^{2+}} + 4\ce{H2O} \). Oxidation: \( \ce{C2O4^{2-}} \rightarrow 2\ce{CO2} + 2e^- \). Multiply by 2 and 5 and add: \( 2\ce{MnO4^-} + 5\ce{C2O4^{2-}} + 16\ce{H+} \rightarrow 2\ce{Mn^{2+}} + 10\ce{CO2} + 8\ce{H2O} \).

Worked example 2 (oxidation-number method). \( \ce{HI} + \ce{HNO3} \rightarrow \ce{NO} + \ce{I2} + \ce{H2O} \). Each I rises by 1 (two I atoms in \( \ce{I2} \) give a rise of 2); N falls by 3. The factors are 3 for \( \ce{I2} \) and 2 for NO, giving \( 6\ce{HI} + 2\ce{HNO3} \rightarrow 2\ce{NO} + 3\ce{I2} + 4\ce{H2O} \).

Worked example 3 (basic). \( \ce{Mn(OH)2} + \ce{MnO4^-} \rightarrow \ce{MnO2} \). Mn rises from +2 to +4 in the first species (2 electrons) and falls from +7 to +4 in the second (3 electrons), so the ratio is 3 : 2: \( 3\ce{Mn(OH)2} + 2\ce{MnO4^-} \rightarrow 5\ce{MnO2} + 2\ce{OH^-} + 2\ce{H2O} \).

Worked example 4. \( \ce{P4} + \ce{HNO3} \rightarrow \ce{H3PO4} + \ce{NO2} + \ce{H2O} \). One \( \ce{P4} \) loses 20 electrons (each P from 0 to +5); each N gains 1 electron (+5 to +4). So \( \ce{P4} + 20\ce{HNO3} \rightarrow 4\ce{H3PO4} + 20\ce{NO2} + 4\ce{H2O} \).

Worked example 5. \( \ce{Cu3P} \) with dichromate in acid. In \( \ce{Cu3P} \), Cu is +1 and P is -3. Three Cu go to +2 (3 electrons) and P goes to +5 (8 electrons): 11 electrons per formula unit. Dichromate gains 6 electrons, so \( 6\ce{Cu3P} \) pairs with \( 11\ce{Cr2O7^{2-}} \).

Points to Remember

Peroxides and superoxides are exceptions to the oxygen rule. Average oxidation numbers in mixed species such as \( \ce{Fe3O4} \) and \( \ce{S4O6^{2-}} \) can be fractional. In ion-electron balancing, check both atoms and charges at the end. A disproportionation always has the same element oxidised and reduced, and at least three oxidation states of that element must exist.