# Question 91e2a

Jan 10, 2015

$2 \text{KMnO"_4 + 16"HCl" -> 2"KCl" + 2"MnCl"_2 + 5"Cl"_2 + 8"H"_2"O}$

#### Explanation:

Equations of this sort can be easily balanced by using oxidation numbers for every atom involved in the reaction.

stackrel(color(blue)(+1))("K")stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O")_4 + stackrel(color(blue)(+1))("H")stackrel(color(blue)(-1))("Cl") -> stackrel(color(blue)(+1))("K")stackrel(color(blue)(-1))("Cl") + stackrel(color(blue)(+2))("Mn")stackrel(color(blue)(-1))("Cl")_2 + stackrel(color(blue)(+1))("H")_2stackrel(color(blue)(-1))("O") + stackrel(color(blue)(0))"Cl"_2

The next step is to look for atoms that have undergone a change in their oxidation numbers (ON), which correlates to either a reduction, or an oxidation of that particular element.

You can see that manganese, $\text{Mn}$, went from an ON of $\textcolor{b l u e}{+ 7}$ on the reactants' side to an ON of $\textcolor{b l u e}{+ 2}$ on the products' side, which means that it has been reduced, i.e. it gained electrons.

The reduction half-reaction looks like this

$\stackrel{\textcolor{b l u e}{+ 7}}{{\text{Mn")"O"_4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn}}^{2 +}}$

Since you're in acidic solution, you can balance the oxygen and hydrogen atoms by adding water molecules to the side that needs oxygen, and protons, ${\text{H}}^{+}$, to the side that needs hydrogen.

In this case, you need four oxygen atoms on the products' side, so add four water molecules on that side

stackrel(color(blue)(+7))("Mn")"O"_4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn"^(2+)) + 4"H"_2"O"

To balance the hydrogen atoms, add eight protons on the reactants' side

$8 \text{H"^(+) + stackrel(color(blue)(+7))("Mn")"O"_4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn"^(2+)) + 4"H"_2"O}$

Chlorine, $\text{Cl}$, on the other hand, went from an ON of $\textcolor{b l u e}{- 1}$ on the reactants' side to an ON of $\textcolor{b l u e}{0}$ on the products' side, which means it has been oxidized, i.e. it lost electrons.

The oxidation half-reaction looks like this

2stackrel(color(blue)(-1))("Cl"^(-)) -> stackrel(color(blue)(0))"Cl"_2 + 2"e"^(-)

Since one chlorine atom will lose one electron, it follows that two chlorine atoms will lose a total of two electrons.

During an oxidation-reduction reaction, the total number of electrons lost in the oxidation half-reaction must be equal to the total number of electrons gained in the reduction half-reaction.

This means that you must multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $2$, for a total of $10$ electrons transferred.

{ ([2stackrel(color(blue)(-1))("Cl"^(-)) -> stackrel(color(blue)(0))"Cl"_2 + 2"e"^(-)] xx 5), ([8"H"^(+) + stackrel(color(blue)(+7))("Mn")"O"_4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn"^(2+)) + 4"H"_2"O"] xx 2) :}

Add the resulting half-equations to get

{ (10stackrel(color(blue)(-1))("Cl"^(-)) -> 5stackrel(color(blue)(0))"Cl"_2 + 10"e"^(-)), (16"H"^(+) + 2stackrel(color(blue)(+7))("Mn")"O"_4^(-) + 10"e"^(-) -> 2stackrel(color(blue)(+2))("Mn"^(2+)) + 8"H"_2"O") :}#
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}$
$10 \text{Cl"^(-) + 16"H"^(+) + 2"MnO"_4^(-) + color(red)(cancel(color(black)(10"e"^(-)))) -> 5"Cl"_2 + color(red)(cancel(color(black)(10"e"^(-)))) + 2"Mn"^(2+) + 8"H"_2"O}$

Now, since hydrochloric acid, $\text{HCl}$, is a strong acid, it dissociates completely in aqueous solution. This means that the chloride anions, ${\text{Cl}}^{-}$, and the protons, ${\text{H}}^{+}$, present on the reactants' side are both coming from the hydrochloric acid.

Since you need $16$ protons present in solution, you can say that you will also need $16$ molecules of hydrochloric acid. This means that you have

$2 \text{MnO"_4^(-) + 16"HCl" ->2"Mn"^(2+) + 5"Cl"_2 + 8"H"_2"O}$

Add the potassium cations and the rest of the chloride anions back to the equation to get

$2 \text{KMnO"_4 + 16"HCl" -> "KCl" + 2"MnCl"_2 + 5"Cl"_2 + 8"H"_2"O}$

Now all you have to do is multiply the potassium chloride, $\text{KCl}$, by $2$ to get the balanced chemical equation

$2 \text{KMnO"_4 + 16"HCl" -> 2"KCl" + 2"MnCl"_2 + 5"Cl"_2 + 8"H"_2"O}$