Question

Question #5e7c6

Answer 1

Here's what I got.

Explanation:

Start by writing the unbalanced chemical equation and assigning oxidation numbers to all the atoms that take part in the reaction

`stackrel(color(blue)(+7))("Cl")stackrel(color(blue)(-2))("O")""_ (4(aq))^(-) + stackrel(color(blue)(+4))("N")stackrel(color(blue)(-2))("O")_ (2(aq)) -> stackrel(color(blue)(-1))("Cl")""_ ((aq))^(-) + stackrel(color(blue)(+5))("N")stackrel(color(blue)(-2))("O")"" _(3(aq))^(-)`

Notice that the oxidation number of chlorine is going from `color(blue)(+7)` on the reactants' side to `color(blue)(-1)` on the products' side, which means that chlorine is being reduced.

On the other hand, the oxidation number of nitrogen is going from `color(blue)(+4)` on the reactants' side to `color(blue)(+5)` on the products' side, which means that nitrogen is being oxidized.

The reduction half-reaction looks like this--I'll skip the states from this point on to keep the equations simple

`stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-)`

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

You should end up with

`8"H"^(+) + stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-) + 4"H"_ 2"O"`

Notice that the charge is balanced because you have

`8 * (1+) + (1-) + 8 * (1-) = (1-)`

The oxidation half-reaction looks like this

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

Once again, balance the oxygen atoms by adding water and the hydrogen atoms by adding protons.

You should end up with

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

Once again, the charge is balanced because you have

`0 = (1-) + (1-) + 2 * (1+)`

Now, in any 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.

To balance the electrons out, multiply the oxidation half-reaction by `8`

`{(8"H"^(+) + stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-) + 4"H"_ 2"O"), (color(white)(aaaaaaa)"H"_ 2"O" + stackrel(color(blue)(+4))("N") "O"_ 2 -> stackrel(color(blue)(+5))("N") "O"_ 3^(-) + "e"^(-) + 2"H"^(+)color(white)(aaaa)| xx 8) :}`

and add the two half-reactions to get

`{(8"H"^(+) + stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-) + 4"H"_ 2"O"), (color(white)(aaaaa)8"H"_ 2"O" + 8stackrel(color(blue)(+4))("N") "O"_ 2 -> 8stackrel(color(blue)(+5))("N") "O"_ 3^(-) + 8"e"^(-) + 16"H"^(+)) :}`
`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)`

`8"H"^(+) + "ClO"_ 4^(-) + color(red)(cancel(color(black)(8"e"^(-)))) + 8"H"_ 2"O" + 8"NO"_ 2 -> "Cl"^(-) + 4"H"_ 2"O" + 8"NO"_ 3^(-) + color(red)(cancel(color(black)(8"e"^(-)))) + 16"H"^(+)`

This is equivalent to

`"ClO"_ (4(aq))^(-) + 4"H"_ 2"O"_ ((l)) + 8"NO"_ (2(aq)) -> "Cl"_ ((aq))^(-) + 8"NO"_ (3(aq))^(-) + 8"H"_ ((aq))^(+)`