# Question #9acfd

Jan 25, 2016

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

#### Explanation:

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

The equation is then:

$M n {O}_{4}^{-} \left(a q\right) + 8 {H}^{+} \left(a q\right) + 5 {e}^{-} \to M {n}^{2 +} \left(a q\right) + 4 {H}_{2} O \left(l\right)$

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:

$1 - \text{ from " MnO_4^(-) " and "8+ " from } {H}^{+} = 7 +$

To the right side we have:

$2 + \text{ from " Mn^(2+) " and "0 " from } {H}_{2} O = 2 +$

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

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

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Balancing Redox Reactions | Acidic Medium.