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Friday, May 29, 2009

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- → I2 + 2e-

5 e- + 8 H+ + MnO4- → Mn2+ + 4 H2O

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- → I2 +2e-)

2(5e- + 8H+ + MnO4- → Mn2+ + 4H2O)

Balancing Redox 5 of 6 Reactions - Add the Half-Reactions


Now add the two half-reactions:

10 I- → 5 I2 + 10 e-

16 H+ + 2 MnO4- + 10 e- → 2 Mn2+ + 8 H2O

This yields the following final equation:

10 I- + 10 e- + 16 H+ + 2 MnO4- → 5 I2 + 2 Mn2+ + 10 e- + 8 H2O

Get the overall equation by canceling out the electrons and H2O, H+, and OH- that may appear on both sides of the equation:

10 I- + 16 H+ + 2 MnO4- → 5 I2 + 2 Mn2+ + 8 H2O

Balancing Redox 6 of 6 Reactions - Check Your Work


Check your numbers to make certain that the mass and charge are balanced. In this example, the atoms are now stoichiometrically balanced with a +4 net charge on each side of the reaction.



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