# Question 59ff2

May 15, 2016

Bromine is being oxidized and manganese is being reduced.

#### Explanation:

In this redox reaction, hydrobromic acid, $\text{HBr}$, will reduce manganese(IV) oxide, ${\text{MnO}}_{2}$, to manganese(II) cations, ${\text{Mn}}^{2 +}$, while being oxidized to molecular bromine, ${\text{Br}}_{2}$, in the process.

You can see what's going on here by assigning oxidation numbers to the atoms that take part in the reaction

${\stackrel{\textcolor{b l u e}{+ 1}}{\text{H")stackrel(color(blue)(-1))("Br")_ ((aq)) + stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O") _ (2(aq)) -> stackrel(color(blue)(+2))("Mn")stackrel(color(blue)(-1))("Br")_ ((aq)) + stackrel(color(blue)(+1))("H")_ 2stackrel(color(blue)(-2))("O")_ ((l)) + stackrel(color(blue)(0))("Br}}}_{2 \left(l\right)}$

The oxidation number of bromine changes from $\textcolor{b l u e}{- 1}$ on the reactants' side, to $\textcolor{b l u e}{0}$ on the products' side. Bromine's oxidation number is increasing, which means that it is being oxidized.

On the other hand, manganese's oxidation number goes from $\textcolor{b l u e}{+ 4}$ on the reactants' side, to $\textcolor{b l u e}{+ 2}$ on the products' side. Manganese's oxidation number is decreasing, which means that it is being reduced.

The oxidation half-reaction will look like this

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

Here every atom of bromine loses one electron, which means that two atoms will lose a total of two electrons.

The reduction half-reaction looks like this

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

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

Add two water molecules on the products' side to get two atoms of oxygen

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

Add four protons on the reactants' side to get four atoms of hydrogen

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

Notice that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction, which means that you can go ahead and add the two half-reactions to get

$\left\{\begin{matrix}\textcolor{w h i t e}{a a a a a a a a a a a a} 2 \text{Br"^(-) -> "Br"_2 + 2"e"^(-) \\ 4"H"^(+) + "MnO"_2 + 2"e"^(-) -> "Mn"^(2+) + 2"H"_2"O}\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a}}$
$4 {\text{H"^(+) + 2"Br"^(-) + color(red)(cancel(color(black)(2"e"^(-)))) -> "Mn"^(2+) + 2"H"_2"O" + color(red)(cancel(color(black)(2"e"^(-)))) + "Br}}_{2}$

which is equivalent to

$4 {\text{H"^(+) + 2"Br"^(-) + "MnO"_2 -> "Mn"^(2+) + 2"H"_2"O" + "Br}}_{2}$

The balanced chemical equation that describes this redox reaction will thus be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{4 {\text{HBr"_ ((aq)) + "MnO"_ (2(aq)) -> "MnBr"_ (2(aq)) + 2"H"_ 2"O"_ ((l)) + "Br}}_{2 \left(l\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, in this reaction, hydrobromic acid acts as a reducing agent and manganese(IV) oxide acts as an oxidizing agent. In other words, bromine is being oxidized and manganese is being reduced.