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2.6.2 Balancing equations using Oxidation Numbers

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Presentation on theme: "2.6.2 Balancing equations using Oxidation Numbers"— Presentation transcript:

1 2.6.2 Balancing equations using Oxidation Numbers

2 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

3 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]

4 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 5 8 5 4 +7 +2 +2 +3 Oxidation Reduction 5. Write electron gains and losses 6. Balance electron transfers 5 Fe = Fe3+ - 1 e- 5 5 Mn = Mn2+ + 5 e- 7. Put numbers up into the equation 8. Balance equation by inspection - leave H+ till last

5 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]

6 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 + Cr2O H+ = CH3COOH + Cr H2O 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 C = C2 e- 2 x2 Cr2O = Cr3+ 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

7 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 = Fe Cl- MnO41- + H2O2 + H+ = Mn2+ + O2 +H2O Cr2O72- + CH3OH + H+ = Cr3+ + HCHO + H2O Cr2O72- + CH3CHO + H+ = Cr3+ + CH3COOH + H2O

8 Answers 5C2H5OH + 2MnO41- + 6H+ = 5CH3CHO + 2Mn2+ + 8H2O
5C2H5OH + 4MnO H+ = 5CH3COOH + 4Mn H2O Cr2O Fe H+ = 2Cr3+ + 6Fe3+ + 7H2O 2MnO C2O H+ = 10CO2 + 2Mn2+ + 8H2O 2Fe2+ + Cl2 = 2Fe Cl- 2MnO H2O2 + 8H+ = 2Mn2+ + 5O2 +8H2O Cr2O72- + CH3OH + H+ = Cr3+ + HCHO + H2O Cr2O72- + CH3CHO + H+ = Cr3+ + CH3COOH + H2O


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