Question #78b35

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.