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)
By Anne Marie Helmenstine, Ph.D., About.com
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)
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