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

## $K C l {O}_{3} \to K C l + {O}_{2}$ $S b C {l}_{5} + K I \to S b C {l}_{3} + K C l + {I}_{2}$

May 19, 2017

$\text{KClO"_3"(s)" + Deltararr"KCl"+3/2"O"_2"(g)} \uparrow$

#### Explanation:

In the given equation, $\stackrel{+ V}{\text{Cl}}$ is reduced to $\stackrel{- I}{\text{Cl}}$; oxide is oxidized to dioxygen gas.

And for $S b C {l}_{5}$ we use the method of half equations:

${\text{SbCl"_5+"2e"^(-) rarr "SbCl"_3 + "2Cl}}^{-}$ $\left(i i\right)$

${\text{I"^(-) rarr 1/2"I}}_{2} + {e}^{-}$ $\left(i i i\right)$

And we takes $\left(i i\right) + 2 \times \left(i i i\right)$ to get............

${\text{2I"^(-) +"SbCl"_5rarr "I"_2 + "SbCl"_3 + "2Cl}}^{-}$

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

May 19, 2017

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

#### Explanation:

${\text{Sb""Cl"_5 + "K""I" → "Sb""Cl"_3 + "K""Cl" + "I}}_{2}$

Step 1. Identify the atoms that change oxidation number

$\stackrel{\textcolor{b l u e}{\text{+5")("Sb")stackrelcolor(blue)("-1")("Cl")_5 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("-1")("I") → stackrelcolor(blue)("+3")("Sb")stackrelcolor(blue)("-1")("Cl")_3 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("-1")("Cl") + stackrelcolor(blue)(0)("I}}}{_} 2$

The atoms that change oxidation number are:

$\text{Sb: +5 → +3; Change ="color(white)(ll) "-2 (reduction)}$
$\text{I:"color(white)(mll) "-1" → color(white)(ll)"0; Change = +1 (oxidation)}$

Step 2. Equalize the changes in oxidation number

We need 2 atoms of $\text{I}$ for every 1 atom of $\text{Sb}$.

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

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{1} {\text{Sb""Cl"_5 + color(red)(2)"K""I" → color(red)(1)"Sb""Cl"_3 + "K""Cl" + color(red)(1)"I}}_{2}$

Step 4. Balance $\text{Cl}$

We have fixed 5 $\text{Cl}$ atoms on the left and 3 on the right. We need 2 more$\text{Cl}$ atoms on the right. Put a 2 before $\text{KCl}$.

$\textcolor{red}{1} {\text{Sb""Cl"_5 + color(red)(2)"K""I" → color(red)(1)"Sb""Cl"_3 + color(blue)(2)"K""Cl" + color(red)(1)"I}}_{2}$

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

Step 5. Check that all atoms are balanced.

$\boldsymbol{\text{On the left"color(white)(l)bb "On the right}}$
$\textcolor{w h i t e}{m m} \text{1 Sb"color(white)(mmmmll) "1 Sb}$
$\textcolor{w h i t e}{m m} \text{5 Cl"color(white)(mmmmm) "5 Cl}$
$\textcolor{w h i t e}{m m} \text{2 K"color(white)(mmmmml) "2 K}$
$\textcolor{w h i t e}{m m} \text{2 I"color(white)(mmmmmll) "2 I}$

The balanced equation is

$\textcolor{b l u e}{{\text{Sb""Cl"_5 + "2K""I" → "Sb""Cl"_3 + "2K""Cl" + "I}}_{2}}$