# Question 306cf

Jul 20, 2017

Left side:

$\text{K}$: $1 +$

$\text{Cl}$: $7 +$

$\text{O}$: $2 -$

Right side:

$\text{K}$: $1 +$

$\text{Cl}$: $1 -$

$\text{O}$: $0$

#### Explanation:

We're asked to assign an oxidation state to each element on both sides of the equation.

All pure elements have an oxidation state of $0$, so the oxygen on the right side has an oxidation state of $0$.

An important fact worth knowing is that oxygen will almost always have an oxidation state of $2 -$ when in a compound. (The most common exception to this are superoxides, such as ${\text{KO}}_{2}$).

Thus, oxygen on the left side has an oxidation state of $2 -$.

Group $1$ metals such as $\text{Na}$ and $\text{K}$ will almost always have a $1 +$ oxidation state, so the potassium on both sides has an oxidation state of $1 +$.

Now, we only have $\text{Cl}$ left.

All neutral compounds have a net oxidation state of $0$, so we use a little simple math to find the oxidation state of chlorine on both sides:

Left side:

${\text{KClO}}_{4}$: overbrace(1+)^"K" + overbrace(x)^"Cl" + overbrace(4(2-))^(4color(white)(l)"O atoms") = 0

x = color(red)(+7

The oxidation state of $\text{Cl}$ on the left side is thus color(red)(7+.

Right side:

$\text{KCl}$: ${\overbrace{1 +}}^{\text{K" + overbrace(x)^"Cl}} = 0$

x = color(blue)(-1

The oxidation state of $\text{Cl}$ on the right side is thus color(blue)(1-#.