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Answer 1 · 2016-01-25T19:31:38

For #7+# to be equal to #2+# we need to add #5-# which means adding five electrons #5e^-#.

This is the reduction half-equation in a redox reaction involving the oxidizing agent #KMnO_4#.

The equation is then:

#MnO_4^(-)(aq)+8H^(+)(aq)+ 5e^(-) -> Mn^(2+)(aq) + 4H_2O(l)#

The 5 electrons here are added to balance the charges to be equal to the left and right sides.

To the left side we have:

#1- " from " MnO_4^(-) " and "8+ " from " H^+=7+#

To the right side we have:

#2+ " from " Mn^(2+) " and "0 " from " H_2O=2+#

For #7+# to be equal to #2+# we need to add #5-# which means adding five electrons #5e^-#.

Here is a video that explains the balancing of a redox reaction in acidic medium.

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Balancing Redox Reactions | Acidic Medium.