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

#"KMnO"_4 +"H"_2"SO"_4 + "KI" → "K"_2"SO"_4 + "MnSO"_4 + "H"_2"O" + "I"_2#

1 Answer
Jun 13, 2017

Answer:

The oxidizing agent is #"KMnO"_4# and the reducing agent is #"KI"#.

Explanation:

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

#stackrelcolor(blue)("+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 #"Mn"# has decreased from +7 in #"KMnO"_4# to +2 in #"MnSO"_4#.

The decrease in oxidation number shows that the #"KMnO"_4# has been reduced, so #"KMnO"_4# is the oxidizing agent.

Also, the oxidation number of #"I"# has increased from -1 in #"KI"# to 0 in #"I"_2#.

The increase in oxidation number shows that the #"KI"# has been oxidized, so #"KI"# is the reducing agent.