Question

How would you balance the following reactions using the oxidation number change method?

`KClO_3 -> KCl + O_2`
`SbCl_5 + KI -> SbCl_3 + KCl + I_2`

Answer 1

`"KClO"_3"(s)" + Deltararr"KCl"+3/2"O"_2"(g)"uarr`

Explanation:

In the given equation, `stackrel(+V)"Cl"` is reduced to `stackrel(-I)"Cl"`; oxide is oxidized to dioxygen gas.

And for `SbCl_5` we use the method of half equations:

`"SbCl"_5+"2e"^(-) rarr "SbCl"_3 + "2Cl"^(-)` `(ii)`

`"I"^(-) rarr 1/2"I"_2 + e^(-)` `(iii)`

And we takes `(ii)+2xx(iii)` to get............

`"2I"^(-) +"SbCl"_5rarr "I"_2 + "SbCl"_3 + "2Cl"^(-)`

Charge is balanced and mass is balanced, so the equation is reasonable.

Answer 2

Here's how to balance the second equation by the oxidation number method.

Explanation:

We start with the unbalanced equation:

`"Sb""Cl"_5 + "K""I" → "Sb""Cl"_3 + "K""Cl" + "I"_2`

Step 1. Identify the atoms that change oxidation number

`stackrelcolor(blue)("+5")("Sb")stackrelcolor(blue)("-1")("Cl")_5 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("-1")("I") → stackrelcolor(blue)("+3")("Sb")stackrelcolor(blue)("-1")("Cl")_3 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("-1")("Cl") + stackrelcolor(blue)(0)("I")_2`

The atoms that change oxidation number are:

`"Sb: +5 → +3; Change ="color(white)(ll) "-2 (reduction)"`
`"I:"color(white)(mll) "-1" → color(white)(ll)"0; Change = +1 (oxidation)"`

Step 2. Equalize the changes in oxidation number

We need 2 atoms of `"I"` for every 1 atom of `"Sb"`.

This gives us total changes of +2 and -2.

Step 3. Insert coefficients to get these numbers

`color(red)(1)"Sb""Cl"_5 + color(red)(2)"K""I" → color(red)(1)"Sb""Cl"_3 + "K""Cl" + color(red)(1)"I"_2`

Step 4. Balance `"Cl"`

We have fixed 5 `"Cl"` atoms on the left and 3 on the right. We need 2 more`"Cl"` atoms on the right. Put a 2 before `"KCl"`.

`color(red)(1)"Sb""Cl"_5 + color(red)(2)"K""I" → color(red)(1)"Sb""Cl"_3 + color(blue)(2)"K""Cl" + color(red)(1)"I"_2`

Every formula now has a coefficient. The equation should be balanced.

Step 5. Check that all atoms are balanced.

`bb"On the left"color(white)(l)bb "On the right"`
`color(white)(mm)"1 Sb"color(white)(mmmmll) "1 Sb"`
`color(white)(mm)"5 Cl"color(white)(mmmmm) "5 Cl"`
`color(white)(mm)"2 K"color(white)(mmmmml) "2 K"`
`color(white)(mm)"2 I"color(white)(mmmmmll) "2 I"`

The balanced equation is

`color(blue)("Sb""Cl"_5 + "2K""I" → "Sb""Cl"_3 + "2K""Cl" + "I"_2)`