Redox Reactions Revision Guide (AQA)

1.7 Redox Oxidation is the process of electron loss: Reduction is the process of electron gain: Zn Zn2+ + 2e- Cl2 + 2e-  2Cl- It involves an increase in oxidation number It involves a decrease...

1.7 Redox Oxidation is the process of electron loss: Reduction is the process of electron gain: Zn Zn2+ + 2e- Cl2 + 2e-  2Cl- It involves an increase in oxidation number It involves a decrease in 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 In NaCl Na= +1 Cl= -1 up to zero Sum = +1 -1 = 0 3. The oxidation number of a monoatomic ion is equal to the e.g. Zn2+ = +2 Cl- = -1 ionic charge 4. In a polyatomic ion (CO32-) the sum of the individual e.g. in CO32- C = +4 and O = -2 oxidation numbers of the elements adds up to the charge sum = +4 + (3 x -2) = -2 on the ion 5. Several elements have invariable oxidation numbers in their common compounds. Group 1 metals = +1 Group 2 metals = +2 Al = +3 We use these rules to H = +1 (except in metal hydrides where it is –1 eg NaH) identify the oxidation numbers of elements that F = -1 have variable oxidation Cl, Br, I = –1 except in compounds with oxygen and fluorine numbers. O = -2 except in peroxides (H2O2 ) where it is –1 and in compounds with fluorine. Note the oxidation number of Cl What is the oxidation number of Fe in FeCl 3 in CaCl2 = -1 and not -2 because Using rule 5, Cl has an O.N. of –1 there are two Cl’s Using rule 2, the O.N. of the elements must add up to 0 Always work out the oxidation per one atom of the element. Fe must have an O.N. of +3 in order to cancel out 3 x –1 = -3 of the Cl’s Redox equations and half equations Br2 (aq) + 2I- (aq)  I2 (aq) + 2 Br- (aq) Br2 (aq) + 2e-  + 2 Br- (aq) 2I- (aq)  I2 (aq) + 2 e- Br has reduced as it has gained electrons I has oxidised as it has lost electrons A reduction half equation only shows the parts An oxidation half equation only shows the reducing agents are of a chemical equation involved in reduction. parts of a chemical equation involved in electron donors The electrons are on the left oxidation. The electrons are on the right oxidising agents are electron acceptors The oxidising agent is Bromine The reducing agent is the Iodide water. It is an electron acceptor ion. It is an electron donor When naming oxidising and reducing agents An oxidising agent (or oxidant) is the A reducing agent (or reductant) is the species that causes another element to species that causes another element always refer to full name oxidise. It is itself reduced in the reaction. reduce. It is itself oxidised in the reaction.. of substance and not just name of element or ion N Goalby chemrevise.org 1 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 Zn  Zn2+ + 2e- For oxidation add e’s to products 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 More complex half-equations If the substance that is being oxidised or reduced contains a varying amount of O (eg In acidic conditions MnO4-  Mn2+ ) then the half equations are balanced by adding H+, OH- ions and H2O. use H+ and H2O Example: Write the half equation for the change MnO4-  Mn2+ Mn changes from +7 to +2 1. Balance the change in O.N. with electrons Add 5 electrons to reactants MnO4- + 5e-  Mn2+ 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 -1 + 8 -5 = +2 +2 Example: Write the half equation for the change SO42-  SO2 S changes from +6 to +4 1. Balance the change in O.N. with electrons Add 2 electrons to reactants SO42- + 2e-  SO2 2. Add H2O in products to balance O’s in SO42- SO42- + 2e-  SO2 + 2H2O 3. Add H+ in reactants to balance H’s in H2O SO42- + 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 -4 + 4 = 0 0 Combining half-equations To combine two half equations there must be To make a full redox equation combine a reduction equal numbers of electrons in the two half half equation with a oxidation half equation equations so that the electrons cancel out Example 1 Reduction MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O x2 Multiply the half equations to get Oxidation C2O42- 2 CO2 + 2 e- x5 equal electrons 2MnO4- + 16 H+ + 5C2O42- 2Mn2+ + 10 CO2 + 8 H2O Add half equations together and cancel electrons Example 2 Reduction SO42- + 10H+ + 8e- H2S+ 4H2O Multiply the half equations to get equal electrons Oxidation 2I- I2 + 2 e- x4 Add half equations together and 8I- + SO42- + 10H+  H2S+ 4I2 + 4H2O cancel electrons N Goalby chemrevise.org 2