Redox reducing agents are electron donors oxidising agents are electron acceptors oxidation is the process of electron loss: Zn Zn2+ + 2eIt involves an increase in oxidation number reduction is the process of electron gain: Cl2 + 2e- 2ClIt involves a decrease in oxidation number Redox equations and half equations Br2 (aq) + 2I- (aq) I 2 (aq) + 2 Br- (aq) Br2 (aq) + 2e- + 2 Br- (aq) 2I- (aq) I 2 (aq) + 2 eBr has reduced as it has gained electrons I has oxidised as it has lost electrons When naming oxidising and reducing agents always refer to full name of substance and not just name of element The oxidising agent is Bromine water . It is an electron acceptor The reducing agent is the Iodide ion. It is an electron donor An oxidising agent (or oxidant) is the species that causes another element to oxidise. It is itself reduced in the reaction A reducing agent (or reductant) is the species that causes another element reduce. It is itself oxidised in the reaction. A reduction half equation only shows the parts of a chemical equation involved in reduction The electrons are on the left An oxidation half equation only shows the parts of a chemical equation involved in oxidation The electrons are on the right. Balancing Redox equations Writing half equations 1. Work out oxidation numbers for element being oxidised/ reduced Zn Zn2+ Zn changes from 0 to +2 2. Add electrons equal to the change in oxidation number For reduction add e’s to reactants For oxidation add e’s to products Zn Zn2+ + 2e- 3. check to see that the sum of the charges on the reactant side equals the sum of the charges on the product side 0 +2 –2 =0 -1 + 8 -5 = +2 More complex Half equations If the substance that is being oxidised or reduced contains a varying amount of O (eg MnO4 – Mn2+ ) then the half equations are balanced by adding H+ , OHions and H2O. In acidic conditions use H+ and H2O Example: Write the half equation for the change MnO4 – Mn2+ 1. Balance the change in O.N. with electrons MnO4 – + 5e- Mn Mn changes from +7 to +2 2+ Add 5 electrons to reactants 2. Add H2O in products to balance O’s in MnO4 – MnO4 – + 5e- Mn2+ + 4H2O 3. Add H+ in reactants to balance H’s in H2O MnO4 – + 8H+ + 5e- Mn2+ + 4H2O 4. check to see that the sum of the charges on the reactant side equals the sum of the charges on the product side Combining half equations To make a full redox equation combine a reduction half equation with a oxidation half equation To combine two half equations there must be equal numbers of electrons in the two half equations so that the electrons cancel out Reduction MnO4 – + 8 H+ + 5 e- → Mn2+ + 4 H2O Oxidation C2O4 2- → 2 CO2 + 2 ex2 x5 Multiply the half equations to get equal electrons 2MnO4 – + 16 H+ + 5C2O4 2- → 2Mn2+ + 10 CO2 + 8 H2O Add half equations together and cancel electrons -4 + 4 = 0 Example: Write the half equation for the change SO4 2- SO2 1. Balance the change in O.N. with electrons SO4 2- + 2e- SO2 S changes from +6 to +4 Add 2 electrons to reactants 2. Add H2O in products to balance O’s in SO4 2- 3. Add H+ in reactants to balance H’s in H2O SO4 2- + 4H+ + 2e- SO2 + 2H2O 4. check to see that the sum of the charges on the reactant side equals the sum of the charges on the product side 0 SO4 2- + 2e- SO2 + 2H2O Reduction SO4 2- + 10H+ + 8e- H2S+ 4H2O Oxidation 2I- → I 2 + 2 ex4 Multiply the half equations to get equal electrons 8I- + SO4 2- + 10H+ H2S+ 4I2 + 4H2O Add half equations together and cancel electrons

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5.2.3 Redox and electrode potentials

Redox (a) explanation and use of the terms oxidising agent and reducing agent (see also 2.1.5 Redox) (b) construction of redox equations using halfequations and oxidation numbers M0.2 (c) interpretation and prediction of reactions involving electron transfer