Question

Reduction

$MnO_{4(aq)}^{-}$ + $8H_{(aq)}^{+}$ + $5e^-$ $\longrightarrow$ $Mn_{(aq)}^{+2}$ + $4H_2O$

Answer 1

The given equation represents the reduction of the permanganate ion ($MnO_4^-$) to the manganese(II) ion ($Mn^{2+}$) in an acidic medium. In this process, manganese undergoes a decrease in its oxidation state from +7 to +2 by gaining 5 electrons.

Answer 2

The provided equation represents a balanced reduction half-reaction of the permanganate ion ($MnO_4^-$) in an acidic medium.

Explanation of the Solution:

1. Nature of Reaction: This is a reduction reaction because the permanganate ion gains 5 electrons ($5e^-$) as shown on the reactant side.
2. Oxidation State Change: The oxidation state of manganese (Mn) in $MnO_4^-$ is +7 (calculated as $x + 4(-2) = -1 \implies x = +7$). In the product, $Mn^{2+}$, the oxidation state of manganese is +2. The decrease in oxidation state from +7 to +2 confirms that manganese is reduced.
3. Electron Transfer: The change in oxidation state ($+7 \rightarrow +2$) corresponds to the gain of 5 electrons, which is correctly represented in the equation.
4. Role in Acidic Medium: The presence of $H^+$ ions indicates that the reaction occurs in an acidic environment. In acidic solutions, $MnO_4^-$ is a very strong oxidizing agent and is typically reduced to $Mn^{2+}$.