## Assigning oxidation number

Rules for assigning oxidation numbers 1. All uncombined elements have an oxidation number of zero eg . Zn, Cl2, O2, Ar all have oxidation numbers of zero 2. The oxidation numbers of the elements in a compound add up to zero In NaCl Na= +1 Cl= -1 Sum = +1 -1 = 0 3. The oxidation number of a monoatomic ion is equal to the ionic charge e.g. Zn2+ = +2 Cl- = -1 4. In a polyatomic ion (CO3 2-) the sum of the individual oxidation numbers of the elements adds up to the charge on the ion e.g. in CO3 2- C = +4 and O = -2 sum = +4 + (3 x -2) = -2 5. Several elements have invariable oxidation numbers in their common compounds. Group 1 metals = +1 Group 2 metals = +2 Al = +3 H = +1 (except in metal hydrides where it is –1 eg NaH) F = -1 Cl, Br, I = –1 except in compounds with oxygen and fluorine O = -2 except in peroxides (H2O2 ) where it is –1 and in compounds with fluorine. We use these rules to identify the oxidation numbers of elements that have variable oxidation numbers. Note the oxidation number of Cl in CaCl2 = -1 and not -2 because there are two Cl’s Always work out the oxidation for one atom of the element What is the oxidation number of Fe in FeCl3 Using rule 5, Cl has an O.N. of –1 Using rule 2, the O.N. of the elements must add up to 0 Fe must have an O.N. of +3 in order to cancel out 3 x –1 = -3 of the Cl’s

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3.1.7 Oxidation, reduction and redox equations

The rules for assigning oxidation states.

Students should be able to:

• work out the oxidation state of an element in a compound or ion from the formula

## Balancing redox reactions

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 N Goalby chemrevise.org 2 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 -1 + 8 -5 = +2 +2 –2 =0 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 0 +2 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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3.1.7 Oxidation, reduction and redox equations

Oxidation is the process of electron loss and oxidising agents are electron acceptors.

Reduction is the process of electron gain and reducing agents are electron donors.

Students should be able to:

• write half-equations identifying the oxidation and reduction processes in redox reactions

• combine half-equations to give an overall redox equation.

Credits: Neil Goalby