How to Balance Redox Reactions - Balancing Redox Reactions
Balancing Redox 4 of 6 Reactions - Balance the Charge Next, balance the charges in each half-reaction so that the reduction half-reaction consumes the same number of electrons as the oxidation half-reaction supplies. This is accomplished by adding electrons to the reactions: 2 I - → I 2 + 2e - 5 e - + 8 H + + MnO 4 - → Mn 2+ + 4 H 2 O Now multiple the oxidations numbers so that the two half-reactions will have the same number of electrons and can cancel each other out: 5(2I - → I 2 +2e - ) 2(5e - + 8H + + MnO 4 - → Mn 2+ + 4H 2 O) Balancing Redox 5 of 6 Reactions - Add the Half-Reactions Now add the two half-reactions: 10 I - → 5 I 2 + 10 e - 16 H + + 2 MnO 4 - + 10 e - → 2 Mn 2+ + 8 H 2 O This yields the following final equation: 10 I - + 10 e - + 16 H + + 2 MnO 4 - → 5 I 2 + 2 Mn 2+ + 10 e - + 8 H 2 O Get the overall equation by canceling out the electrons and H 2 O, H + , and OH - that may appear on both sides of the equation: 10 I - + 16 H + ...