Redox

Define oxidation in terms of electron transfer.

Oxidation is the loss of electrons.

Define reduction in terms of electron transfer.

Reduction is the gain of electrons.

State the mnemonic for electron transfer in redox.

OIL RIG - Oxidation Is Loss of electrons, Reduction Is Gain of electrons.

Define oxidation in terms of oxidation number.

Oxidation is an increase in oxidation number.

Define reduction in terms of oxidation number.

Reduction is a decrease in oxidation number.

Define an oxidising agent.

An oxidising agent is a reagent that oxidises another species (takes electrons from it) and is itself reduced.

Define a reducing agent.

A reducing agent is a reagent that reduces another species (adds electrons to it) and is itself oxidised.

State the rule for oxidation numbers in an uncombined element.

The oxidation number of an uncombined element is 0 (zero).

State the rule for oxidation numbers of Group 1 metals.

Group 1 metals have an oxidation number of +1.

State the rule for oxidation numbers of Group 2 metals.

Group 2 metals have an oxidation number of +2.

State the rule for oxidation numbers of aluminium.

Aluminium has an oxidation number of +3.

State the rule for oxidation numbers of hydrogen in compounds.

Hydrogen has an oxidation number of +1 in compounds (except in metal hydrides where it is -1).

State the rule for oxidation numbers of oxygen in compounds.

Oxygen has an oxidation number of -2 in compounds (except in peroxides where it is -1 and in F₂O where it is +2).

State the rule for oxidation numbers of fluorine.

Fluorine always has an oxidation number of -1.

State the rule for the sum of oxidation numbers in a neutral compound.

The sum of all oxidation numbers in a neutral compound is 0 (zero).

State the rule for the sum of oxidation numbers in a polyatomic ion.

The sum of all oxidation numbers in a polyatomic ion equals the charge on the ion.

State the exception to the oxidation number of oxygen.

Oxygen is -2 except in peroxides (where it is -1) and in F₂O (where it is +2).

State the exception to the oxidation number of hydrogen.

Hydrogen is +1 except in metal hydrides (where it is -1).

Define a disproportionation reaction.

A disproportionation reaction is a redox reaction in which the same element is both oxidised and reduced.

Define an oxyanion.

An oxyanion is a negative ion containing oxygen and one or more other elements. The names of oxyanions usually end in -ate.

Explain why Roman numerals are used in the names of oxyanions.

Roman numerals indicate the oxidation state of the element in the oxyanion (e.g. sulfate(VI) means sulfur is in the +6 oxidation state).

Calculate the oxidation number of sulfur in SO₂.

Oxygen is -2. S + 2(-2) = 0. S = +4.

Calculate the oxidation number of sulfur in SO₄²⁻.

Oxygen is -2. S + 4(-2) = -2. S = +6.

Calculate the oxidation number of sulfur in SO₃²⁻.

Oxygen is -2. S + 3(-2) = -2. S = +4.

Calculate the oxidation number of sulfur in H₂S.

Hydrogen is +1. 2(+1) + S = 0. S = -2.

Calculate the oxidation number of sulfur in S.

S is uncombined, so oxidation number = 0.

Calculate the oxidation number of sulfur in S₂O₃²⁻.

Oxygen is -2. 2S + 3(-2) = -2. 2S - 6 = -2. 2S = +4. S = +2.

Calculate the oxidation number of nitrogen in NO₃⁻.

Oxygen is -2. N + 3(-2) = -1. N - 6 = -1. N = +5.

Calculate the oxidation number of nitrogen in NO₂⁻.

Oxygen is -2. N + 2(-2) = -1. N - 4 = -1. N = +3.

Calculate the oxidation number of nitrogen in N₂O₃.

Oxygen is -2. 2N + 3(-2) = 0. 2N - 6 = 0. N = +3.

Calculate the oxidation number of nitrogen in NO.

Oxygen is -2. N + (-2) = 0. N = +2.

Calculate the oxidation number of nitrogen in NO₂.

Oxygen is -2. N + 2(-2) = 0. N - 4 = 0. N = +4.

Calculate the oxidation number of nitrogen in NH₄⁺.

Hydrogen is +1. N + 4(+1) = +1. N + 4 = +1. N = -3.

Calculate the oxidation number of nitrogen in Mg(NO₃)₂.

In NO₃⁻, N = +5 (as calculated above).

Calculate the oxidation number of nitrogen in NH₃.

Hydrogen is +1. N + 3(+1) = 0. N = -3.

Calculate the oxidation number of nitrogen in N₂H₄.

Hydrogen is +1. 2N + 4(+1) = 0. 2N = -4. N = -2.

Calculate the oxidation number of chlorine in Cl₂.

Uncombined element, oxidation number = 0.

Calculate the oxidation number of chlorine in HCl.

Hydrogen is +1. H + Cl = 0. Cl = -1.

Calculate the oxidation number of chlorine in Cl⁻.

Monatomic ion, oxidation number = charge = -1.

Calculate the oxidation number of chlorine in ClO⁻.

Oxygen is -2. Cl + (-2) = -1. Cl = +1.

Calculate the oxidation number of chlorine in ClO₃⁻.

Oxygen is -2. Cl + 3(-2) = -1. Cl - 6 = -1. Cl = +5.

Calculate the oxidation number of chlorine in ClO₄⁻.

Oxygen is -2. Cl + 4(-2) = -1. Cl - 8 = -1. Cl = +7.

Calculate the oxidation number of chlorine in HClO₄.

Hydrogen +1, Oxygen -2. +1 + Cl + 4(-2) = 0. Cl - 7 = 0. Cl = +7.

Calculate the oxidation number of manganese in MnO₄⁻.

Oxygen is -2. Mn + 4(-2) = -1. Mn - 8 = -1. Mn = +7.

Calculate the oxidation number of manganese in MnO₂.

Oxygen is -2. Mn + 2(-2) = 0. Mn = +4.

Calculate the oxidation number of manganese in Mn²⁺.

Monatomic ion, oxidation number = charge = +2.

Calculate the oxidation number of manganese in K₂MnO₄.

Potassium +1, Oxygen -2. 2(+1) + Mn + 4(-2) = 0. 2 + Mn - 8 = 0. Mn = +6.

Calculate the oxidation number of chromium in Cr₂O₇²⁻.

Oxygen is -2. 2Cr + 7(-2) = -2. 2Cr - 14 = -2. 2Cr = +12. Cr = +6.

Calculate the oxidation number of chromium in CrO₄²⁻.

Oxygen is -2. Cr + 4(-2) = -2. Cr - 8 = -2. Cr = +6.

Calculate the oxidation number of chromium in Cr³⁺.

Monatomic ion, oxidation number = +3.