Question

Question #78b35

Answer 1

`"Br"^(-) + 3"H"_3"AsO"_4 -> "BrO"_3^(-) + 3"HAsO"_2 + 3"H"_2"O"`

Explanation:

Start by assigning oxidation numbers to the atoms that take part in the reaction

`stackrel(color(blue)(-1))("Br")""^(-) + stackrel(color(blue)(+1))("H")_ 3 stackrel(color(blue)(+5))("As") stackrel(color(blue)(-2))("O")_ 4 -> stackrel(color(blue)(+5))("Br") stackrel(color(blue)(-2))("O")_ 3""^(-) + stackrel(color(blue)(+1))("H") stackrel(color(blue)(+3))("As") stackrel(color(blue)(-2))("O")_ 2`

Now, notice that the oxidation number of bromine is going from `color(blue)(-1)` on the reactants' side to `color(blue)(+5)` on the products' side. This implies that bromine is being oxidized, i.e. its oxidation number is increasing.

Similarly, the oxidation number of astatine if going from `color(blue)(+5)` on the reactants' side to `color(blue)(+3)` on the products' side. This implies that astatine is being reduced, i.e. its oxidation number is decreasing.

The oxidation half-reaction looks like this

`stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-)`

Now, the bromine atoms are balanced. To balance the oxygen atoms, you can use water molecules on the side that needs oxygen. In this case, you need `3` oxygen atoms on the reactants' side, so add `3` water molecules to that side of the equation

`3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-)`

To balance the hydrogen atoms, add protons, `"H"^(+)`, to the side that needs hydrogen. In this case, you have `6` hydrogen atoms on the reactants' side, so add `6` protons to the products' side

`3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-) + 6"H"^(+)`

Notice that the charge is balanced because you have

`(1-) = (1-) + (6-) + (6+)`

The reduction half-reaction looks like this

`"H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_2`

Once again, the astatine atoms are balanced, so focus on the oxygen atoms first. You will have

`"H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_ 2 + 2"H"_ 2"O"`

You now need `2` protons on the reactants' side to balance the hydrogen atoms

`2"H"^(+) + "H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_ 2 + 2"H"_ 2"O"`

Once again, notice that the charge is balanced because you have

`(2+) + (2-) = 0`

Now, in every redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

This means that you must multiply the reduction half-reaction by `3`

#{(color(white)(aaaaaaaa)3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-) + 6"H"^(+)), (2"H"^(+) + "H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_ 2 + 2"H"_ 2"O" " " | xx 3) :}#

to get

`{(color(white)(aaaaaaaa)3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-) + 6"H"^(+)), (6"H"^(+) + 3"H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 6"e"^(-) -> 3"H" stackrel(color(blue)(+3))("As") "O"_ 2 + 6"H"_ 2"O") :}`
`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)`

`3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) + color(red)(cancel(color(black)(6"H"^(+)))) + 3"H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + color(red)(cancel(color(black)(6"e"^(-)))) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 3"H" stackrel(color(blue)(+3))("As") "O"_ 2 + color(red)(cancel(color(black)(6"e"^(-)))) + color(red)(cancel(color(black)(6"H"^(+)))) + 6"H"_2"O"`

This will be equivalent to

`color(darkgreen)(ul(color(black)("Br"^(-) + 3"H"_3"AsO"_4 -> "BrO"_3^(-) + 3"HAsO"_2 + 3"H"_2"O")))`

Therefore, you can say that water appears in the balanced chemical equation as a product with a coefficient of `3`.

The bromide anion, `"Br"^(-)` is being oxidized to the bromate anion, `"BrO"_3^(-)`, which implies that arsenic acid, `"H"_3"AsO"_4`, is acting as an oxidizing agent.

Consequently, you can say that arsenic acid is being reduced to arsonite, `"HAsO"_2`, which implies that the bromide anion is acting as a reducing agent.