What are the oxidizing and reducing agent in the reaction below?

${\text{KMnO"_4 +"H"_2"SO"_4 + "KI" → "K"_2"SO"_4 + "MnSO"_4 + "H"_2"O" + "I}}_{2}$

Jun 13, 2017

The oxidizing agent is ${\text{KMnO}}_{4}$ and the reducing agent is $\text{KI}$.

Explanation:

We start by identifying the oxidation number of every atom in the equation.

$\stackrel{\textcolor{b l u e}{\text{+1")("K")stackrelcolor(blue)("+7")("Mn")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("-1")("I") → stackrelcolor(blue)("+1")("K")_2stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+2")("Mn")stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O")+stackrelcolor(blue)(0)("I}}}{_} 2$

We see that the oxidation number of $\text{Mn}$ has decreased from +7 in ${\text{KMnO}}_{4}$ to +2 in ${\text{MnSO}}_{4}$.

The decrease in oxidation number shows that the ${\text{KMnO}}_{4}$ has been reduced, so ${\text{KMnO}}_{4}$ is the oxidizing agent.

Also, the oxidation number of $\text{I}$ has increased from -1 in $\text{KI}$ to 0 in ${\text{I}}_{2}$.

The increase in oxidation number shows that the $\text{KI}$ has been oxidized, so $\text{KI}$ is the reducing agent.