Question

Question #306cf

Answer 1

Left side:

`"K"`: `1+`

`"Cl"`: `7+`

`"O"`: `2-`

Right side:

`"K"`: `1+`

`"Cl"`: `1-`

`"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 `"KO"_2`).

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

Group `1` metals such as `"Na"` and `"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 `"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:

`"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 `"Cl"` on the left side is thus `color(red)(7+`.

Right side:

`"KCl"`: `overbrace(1+)^"K" + overbrace(x)^"Cl" = 0`

`x = color(blue)(-1`

The oxidation state of `"Cl"` on the right side is thus `color(blue)(1-`.