# Question 5e7c6

Jun 9, 2017

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 $\textcolor{b l u e}{+ 7}$ on the reactants' side to $\textcolor{b l u e}{- 1}$ on the products' side, which means that chlorine is being reduced.

On the other hand, the oxidation number of nitrogen is going from $\textcolor{b l u e}{+ 4}$ on the reactants' side to $\textcolor{b l u e}{+ 5}$ 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, ${\text{H}}^{+}$, to the side that needs hydrogen.

You should end up with

$8 \text{H"^(+) + stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-) + 4"H"_ 2"O}$

Notice that the charge is balanced because you have

$8 \cdot \left(1 +\right) + \left(1 -\right) + 8 \cdot \left(1 -\right) = \left(1 -\right)$

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

${\text{H"_ 2"O" + stackrel(color(blue)(+4))("N") "O"_ 2 -> stackrel(color(blue)(+5))("N") "O"_ 3^(-) + "e"^(-) + 2"H}}^{+}$

Once again, the charge is balanced because you have

$0 = \left(1 -\right) + \left(1 -\right) + 2 \cdot \left(1 +\right)$

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 $8$

$\left\{\begin{matrix}8 {\text{H"^(+) + stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-) + 4"H"_ 2"O" \\ color(white)(aaaaaaa)"H"_ 2"O" + stackrel(color(blue)(+4))("N") "O"_ 2 -> stackrel(color(blue)(+5))("N") "O"_ 3^(-) + "e"^(-) + 2"H}}^{+} \textcolor{w h i t e}{a a a a} | \times 8\end{matrix}\right.$

and add the two half-reactions to get

$\left\{\begin{matrix}8 {\text{H"^(+) + stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-) + 4"H"_ 2"O" \\ color(white)(aaaaa)8"H"_ 2"O" + 8stackrel(color(blue)(+4))("N") "O"_ 2 -> 8stackrel(color(blue)(+5))("N") "O"_ 3^(-) + 8"e"^(-) + 16"H}}^{+}\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

$8 {\text{H"^(+) + "ClO"_ 4^(-) + color(red)(cancel(color(black)(8"e"^(-)))) + 8"H"_ 2"O" + 8"NO"_ 2 -> "Cl"^(-) + 4"H"_ 2"O" + 8"NO"_ 3^(-) + color(red)(cancel(color(black)(8"e"^(-)))) + 16"H}}^{+}$

This is equivalent to

${\text{ClO"_ (4(aq))^(-) + 4"H"_ 2"O"_ ((l)) + 8"NO"_ (2(aq)) -> "Cl"_ ((aq))^(-) + 8"NO"_ (3(aq))^(-) + 8"H}}_{\left(a q\right)}^{+}$