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Answer 1 · 2016-01-25T19:31:38

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

This is the reduction half-equation in a redox reaction involving the oxidizing agent $K M n O_{4}$ $KMnO_4$ .

The equation is then:

$M n O_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (l)$ $MnO_4^(-)(aq)+8H^(+)(aq)+ 5e^(-) -> Mn^(2+)(aq) + 4H_2O(l)$

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:

$1 - from M n O_{4}^{-} and 8 + from H^{+} = 7 +$ $1- " from " MnO_4^(-) " and "8+ " from " H^+=7+$

To the right side we have:

$2 + from M n^{2 +} and 0 from H_{2} O = 2 +$ $2+ " from " Mn^(2+) " and "0 " from " H_2O=2+$

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

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

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Balancing Redox Reactions | Acidic Medium.