half reaction method for balancing in an acid or base

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Presentation transcript:

half reaction method for balancing in an acid or base electrochem part 2 half reaction method for balancing in an acid or base

MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (acid) Half-Reaction Method : MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (acid)

Half-Reaction Method (in acid) : MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) Step 1: Divide into two half-reactions, MnO4-  Mn2+ C2O42-  CO2 Step 2: Balance main element MnO4-  Mn2+ C2O42-  2CO2

MnO4-  Mn2+ + 4H2O C2O42-  2CO2 MnO4-  Mn2+ + 4H2O 8H+ + Step 3: Balance the O atoms by adding H2O : : MnO4-  Mn2+ + 4H2O C2O42-  2CO2 Step 4: Balance the H atoms by adding H+ 8H+ + MnO4-  Mn2+ + 4H2O C2O42-  2CO2 Step 5: Balance charge with electrons +7 +2 5e- + 8H++ MnO4-  Mn2+ + 4H2O -2 C2O42-  2CO2 + 2e-

5e- + 8H++ MnO4-  Mn2+ + 4H2O C2O42-  2CO2 + 2e- Step 6: Make the electrons balance : 10 16 2 2 8 5e- + 8H++ MnO4-  Mn2+ + 4H2O 5 10 10 C2O42-  2CO2 + 2e- Step 7: Cancel and add 10e- + 16H++ 2MnO4-  2Mn2+ + 8H2O + 5C2O42-  10CO2 +10e- 16H+ + 2MnO4- + 5C2O42-  2Mn2+ + 10CO2 + 8H2O

CN- + MnO4-  CNO- + MnO2 -1 +1 H2O + CN-  CNO- + 2H+ + 2e- 3e-+ Balancing in base CN- + MnO4-  CNO- + MnO2 -1 +1 3 3 6 3 6 H2O + CN-  CNO- + 2H+ + 2e- 6 2 8 2 4 3e-+ 4H+ + MnO4-  MnO2 + 2H2O +3 2H+ + 3CN- + 2MnO4-  3CNO- + 2MnO2 + H2O + 2OH- 2OH- 2H2O 1

Cr(OH)3 + ClO  CrO4-2 + Cl2 Have at it Balancing in base 4Cr(OH)3 + 6ClO + 8OH- 4CrO4-2 + 3Cl2 + 10H2O