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Writing complex redox half equations
Example 1: Permanagante ion (MnO4-) is reduced to form manganese ions(Mn 2+) •Oxidation numbers for Mn atoms: +7 +2 MnO4- → Mn 2+ •K already balanced for Mn atoms. MnO4- → Mn 2+ •O 4 atoms of oxygen = add 4 x H20 MnO4- → Mn 2+ + 4 H20 •H 8 atoms of hydrogen = add 8 x H+ MnO4- + 8H+ → Mn 2+ + 4 H20 •E charge on left = -1 + 8 x 1 = +7 charge on right = +2 difference in charge is 5 = 5 e- MnO4- + 8H+ + 5e- → Mn 2+ + 4 H20 •S include states MnO4-(aq) + 8H+(aq) + 5e- →Mn 2+(aq) + 4H20(l) Writing out a balanced equation 1.Write balanced half equations for oxidation and reduction. 2.Multiply each half equation by factors that will lead to the same number of electrons in each half equation. 3.Add the half equations and subtract terms that appear on both sides of the equation. (electrons will be cancelled out) Example: Permanagante ions (MnO4-) react with Chloride ions (Cl-) to form manganese ions(Mn 2+) and chlorine molecules (Cl2). MnO4- + Cl- → Mn 2+ + Cl2 Oxidation Half Equation: (2Cl- (aq) → Cl2 (aq) + 2e- ) x 5 Reduction Half Equation: (MnO4-(aq) + 8H+(aq) + 5e- →Mn 2+(aq) + 4H20(l)) x 2 Balanced Ionic Equation: 10Cl- (aq) + 2MnO4-(aq) + 16H+(aq) + 10e- → 5Cl2 (aq) + 2Mn 2+(aq) + 8H20(l)) + 10e- Cancel electrons. 10Cl- (aq) + 2MnO4-(aq) + 16H+(aq) → 5Cl2 (aq) + 2Mn 2+(aq) + 8H20(l) |
Example 2:
Chloride ions (Cl-) is oxidised to form chlorine molecules (Cl2) •Oxidation numbers for Cl atoms: -1 0 Cl- → Cl2 •K balance Cl atoms by adding 1. 2Cl- → Cl2 •O no oxygen atoms = no change 2Cl- → Cl2 •H no hydrogen atoms = no change 2Cl- → Cl2 •E charge on left = -1 x 2 = -2 charge on right = 0 difference in charge is 2 = 2 e- 2Cl- → Cl2 + 2 e- •S include states 2Cl- (aq) → Cl2 (aq) + 2 e- |