Question

Balance this redox reaction in acidic media? `"As"_2"O"_3(s) + "NO"_3^(-)(aq) -> "H"_3"AsO"_4(aq) + "N"_2"O"_3(aq)`

Answer 1

`"As"_2"O"_text(3(s]) + 2"NO"_text(3(aq])^(-) + 2"H"_text((aq])^(+) + 2"H"_2"O"_text((l]) -> 2"H"_3"AsO"_text(4(aq]) + "N"_2"O"_text(3(aq])`

Explanation:

Start by assigning oxidation numbers to the atoms that are taking part in the reaction.

For simplicity, I won't add the state symbols for the chemical species involved in the reaction.

`stackrel(color(blue)(+3))("As")_2stackrel(color(blue)(-2))("O")_3 + stackrel(color(blue)(+5))("N")stackrel(color(blue)(-2))("O"_3^(-)) -> stackrel(color(blue)(+1))("H")_3stackrel(color(blue)(+5))("As")stackrel(color(blue)(-2))("O")_4 + stackrel(color(blue)(+3))("N")_2stackrel(color(blue)(-2))("O")_3`

Notice that the oxidation number of arsenic, `"As"`, changes from `color(blue)(+3)` on the reactants' side, to `color(blue)(+5)` on the products' side. This tells you that arsenic is being oxidized, since its oxidation number increases.

On the other hand, the oxidation number of nitrogen, `"N"`, changes from `color(blue)(+5)` on the reactants' side, to `color(blue)(+3)` on the products' side. This tells you that nitrogen is being reduced, since its oxidation number decreases.

Write and balance the oxidation and reduction half-reactions.

• the oxidation-half reaction

`stackrel(color(blue)(+3))("As")_2"O"_3 -> "H"_3stackrel(color(blue)(+5))("As")"O"_4`

Balance the atoms of arsenic first

`stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4`

Here each atom of arsenic loses two electrons, which means that two atoms of arsenic will lose a total of four electrons.

`stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-)`

Since you're in acidic solution, you can balance the oxygen atoms by adding water to the side of the reaction that needs oxygen atoms, and the hydrogen atoms by adding protons, `"H"^(+)`, to the side that needs hydrogen atoms.

In this case, you need five extra atoms of oxygen on the reactants' side, so you'll have

`5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-)`

Similarly, the half-reaction needs four extra atoms of hydrogen on the products' side, so add four protons in there, too.

`5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-) + 4"H"^(+)`

• the reduction half-reaction

`stackrel(color(blue)(+5))("N")"O"_3^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3`

Balance the nitrogen atoms first

`2stackrel(color(blue)(+5))("N")"O"_3^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3`

Here each atom of nitrogen gains two electrons, so two atoms of nitrogen will gain a total of four electrons

`2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3`

Balance the oxygen by adding three molecules of water on the products' side

`2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O"`

Finally, balance the hydrogen by adding six protons on the reactants' side

`6"H"^(+) + 2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O"`

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

Since `4"e"^(-)` are both gained and lost in the respective half-reactions, you can go ahead and add the two half-reactions to get the balanced overall chemical equation

`{(5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-) + 4"H"^(+)), (6"H"^(+) + 2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O") :}`
`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)`

`color(red)(cancel(color(black)(3"H"_2"O"))) + 2"H"_2"O" + "As"_2"O"_3 + color(red)(cancel(color(black)(4"H"^(+)))) + 2"H"^(+) + 2"NO"_3^(-) + color(red)(cancel(color(black)(4"e"^(-)))) -> 2"H"_3"AsO"_4 + color(red)(cancel(color(black)(4"e"^(-)))) + color(red)(cancel(color(black)(4"H"^(+)))) + "N"_2"O"_3 + color(red)(cancel(color(black)(3"H"_2"O")))`

This will be equivalent to

`"As"_2"O"_3 + 2"NO"_3^(-) + 2"H"^(+) + 2"H"_2"O" -> 2"H"_3"AsO"_4 + "N"_2"O"_3`

The balanced chemical equation - with state symbols included - will thus be

`color(green)(|bar(ul(color(white)(a/a)color(black)("As"_2"O"_text(3(s]) + 2"NO"_text(3(aq])^(-) + 2"H"_text((aq])^(+) + 2"H"_2"O"_text((l]) -> 2"H"_3"AsO"_text(4(aq]) + "N"_2"O"_text(3(aq]))|)))`