Oxidation and Reduction Chapter 14

Oxidation and Reduction Oxidation - addition of Oxygen. Reduction - removal of Oxygen. In terms of electrons Oxidation is the loss of electrons. (OILRIG) Reduction is the gain of electrons

Oxidising and reducing agents An oxidising agent causes another substance to become oxidised and, in the process, the oxidising agent is reduced. Common oxidising agents include O2, Cl2, ClO-, KMnO4 and H2O2 A reducing agent causes another substance to become reduced and, in the process, the reducing agent is oxidised. Common reducing agents include CO, SO2, SO32- Oxidising agents are useful in Sterilising and Bleaching.

Experiments: (a) Halogens as Oxidising agents Experiments: (a) Halogens as Oxidising agents (b) Displacement reactions of Metals. (See Book Pages 184-186)

Oxidation Numbers The oxidation number of an atom is the charge that the atom has or appears to have when electrons are distributed according to certain rule. Rules The ON of an element is 0. The ON of an ion is the charge that it has. The ON of F is always -1. The ON of O IS -2 except OF2 (+2) and H2O2 (-1). The ON of H is +1 except with metal hydrides (-1)

Oxidation and reduction in terms of Oxidation numbers Oxidation is an increase in Oxidation number. Reduction is a decrease in Oxidation number. Balancing Redox reactions (See Book Page 187-188) + HANDOUT

2.6.2 Balancing equations using Oxidation Numbers

Chemical reactions are all about the movement of electrons When an equation is balanced the electron movements must balance Sometimes it is possible to balance an equation in more than one way It is therefore possible to make the numbers of atoms balance but not the electron movement Using oxidation numbers prevents this

Sequence of events Write out the formulae of all reactants and products Assign oxidation numbers to all atoms Pick out the atoms whose oxidation numbers change Write down the electron movements Balance the electron movements Enter the numbers from balanced electron movement into the overall equation Balance equation by inspection - [do H last]

MnO4- + Fe2+ + H+ = Mn2+ + Fe3+ + H2O +7 -2 +2 +1 +2 +3 +1 -2 5 8 5 4 1. Assign oxidation numbers (a) Assign Oxidation – increase in Ox. No (b) Assign Reduction – decrease in Ox. No 2. Select those which change MnO4- + Fe2+ + H+ = Mn2+ + Fe3+ + H2O +7 -2 +2 +1 +2 +3 +1 -2 5 8 5 4 +7 +2 +2 +3 Oxidation Reduction 5. Write electron gains and losses 6. Balance electron transfers 5 Fe2+ = Fe3+ +2 +3 - 1 e- 5 5 Mn = Mn2+ +7 +2 + 5 e- 7. Put numbers up into the equation 8. Balance equation by inspection - leave H+ till last

Sequence of events Write down reactants and products Assign oxidation numbers Identify oxidation and reduction Write down changes and electrons involves Balance e- change Put changes into equation Balance equation by inspection [H last]

x2 3 C2H5OH + Cr2O72-+ H+ = CH3COOH + Cr3++ H2O 2 16 3 4 11 -2 +6 +3 1. Assign oxidation numbers (a) Assign Oxidation – increase in Ox. No (b) Assign Reduction – decrease in Ox. No 2. Select those which change 3 C2H5OH + Cr2O72-+ H+ = CH3COOH + Cr3++ H2O -2 +1-2+1 +6 -2 +1 0 +1 0 -2-2+1 +3 +1- 2 2 16 3 4 11 -2 +6 +3 Reduction Oxidation 5. Write electron gains and losses 6. Balance electron transfers 3 12 4 3 x3 C2 = C2 -2 0 - 2 e- 2 Cr2O72- = Cr3+ +6 +3 x2 + 3 e- 12 6 2 4 7. Put numbers up into the equation Balance equation by inspection - leave H+ till last :- O 3+(2*7) = (3*2) + x  x=11 H  3(6*1) + X = (3*4) + 11(2*1)  x= 16

Examples C2H5OH + MnO41- + H+ = CH3CHO + Mn2+ + H2O C2H5OH + MnO41- + H+ = CH3COOH + Mn2+ + H2O Cr2O72- + Fe2+ + H+ = Cr3+ + Fe3+ + H2O MnO41- + C2O42- + H+ = CO2 + Mn2+ + H2O Fe2+ + Cl2 = Fe3+ + Cl- MnO41- + H2O2 + H+ = Mn2+ + O2 +H2O Cr2O72- + CH3OH + H+ = Cr3+ + HCHO + H2O Cr2O72- + CH3CHO + H+ = Cr3+ + CH3COOH + H2O

Answers 5C2H5OH + 2MnO41- + 6H+ = 5CH3CHO + 2Mn2+ + 8H2O 5C2H5OH + 4MnO41- + 12H+ = 5CH3COOH + 4Mn2+ + 11H2O Cr2O72- + 6Fe2+ + 14H+ = 2Cr3+ + 6Fe3+ + 7H2O 2MnO41- + 5C2O42- + 16H+ = 10CO2 + 2Mn2+ + 8H2O 2Fe2+ + Cl2 = 2Fe3+ + 2Cl- 2MnO41- + 5H2O2 + 8H+ = 2Mn2+ + 5O2 +8H2O Cr2O72- + CH3OH + H+ = Cr3+ + HCHO + H2O Cr2O72- + CH3CHO + H+ = Cr3+ + CH3COOH + H2O