Question #306cf

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Question #306cf

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Answer 1 · 2017-07-20T02:24:04

Left side:

$K$ $"K"$ : $1 +$ $1+$

$Cl$ $"Cl"$ : $7 +$ $7+$

$O$ $"O"$ : $2 -$ $2-$

Right side:

$K$ $"K"$ : $1 +$ $1+$

$Cl$ $"Cl"$ : $1 -$ $1-$

$O$ $"O"$ : $0$ $0$

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 $0$ $0$ , so the oxygen on the right side has an oxidation state of $0$ $0$ .

An important fact worth knowing is that oxygen will almost always have an oxidation state of $2 -$ $2-$ when in a compound. (The most common exception to this are superoxides, such as ${KO}_{2}$ $"KO"_2$ ).

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

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

Now, we only have $Cl$ $"Cl"$ left.

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

Left side:

${KClO}_{4}$ $"KClO"_4$ : $overbrace(1+)^"K" + overbrace(x)^"Cl" + overbrace(4(2-))^(4color(white)(l)"O atoms") = 0$

$x = color(red)(+7$

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 $color(blue)(1-$ .

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Question #306cf

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