Question

Question #59ff2

Answer 1

Bromine is being oxidized and manganese is being reduced.

Explanation:

In this redox reaction, hydrobromic acid, `"HBr"`, will reduce manganese(IV) oxide, `"MnO"_2`, to manganese(II) cations, `"Mn"^(2+)`, while being oxidized to molecular bromine, `"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(color(blue)(+1))("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(l))`

The oxidation number of bromine changes from `color(blue)(-1)` on the reactants' side, to `color(blue)(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 `color(blue)(+4)` on the reactants' side, to `color(blue)(+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, `"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"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

`{(color(white)(aaaaaaaaaaaa)2"Br"^(-) -> "Br"_2 + 2"e"^(-)), (4"H"^(+) + "MnO"_2 + 2"e"^(-) -> "Mn"^(2+) + 2"H"_2"O") :}`
`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaa)`
`4"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"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

`color(green)(|bar(ul(color(white)(a/a)color(black)(4"HBr"_ ((aq)) + "MnO"_ (2(aq)) -> "MnBr"_ (2(aq)) + 2"H"_ 2"O"_ ((l)) + "Br"_ (2(l)))color(white)(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.