# Question f0628

Jan 4, 2017

Warning! Long answer! Here's how to balance your equation by the half-reaction method.

#### Explanation:

You can find the technique for balancing redox equations in this Socratic answer, so I will just go through the steps.

Step 1: Write the two half-reactions.

${\text{PbS" → "PbSO}}_{4}$
${\text{O"_3 → "O}}_{2}$

Step 2: Balance all atoms other than $\text{H}$ and $\text{O}$.

Done.

Step 3: Balance $\text{O}$.

${\text{PbS + 4H"_2"O" → "PbSO}}_{4}$
$\text{O"_3 → "O"_2 + "H"_2"O}$

Step 4: Balance $\text{H}$.

$\text{PbS + 4H"_2"O" → "PbSO"_4 + "8H"^"+}$
$\text{O"_3 + "2H"^"+" → "O"_2 + "H"_2"O}$

Step 5: Balance charge.

$\text{PbS + 4H"_2"O" → "PbSO"_4 + "8H"^"+" + "8e"^"-}$
$\text{O"_3 + "2H"^"+" + "2e"^"-" → "O"_2 + "H"_2"O}$

Step 6: Equalize electrons transferred.

1 × ["PbS + 4H"_2"O" → "PbSO"_4 + "8H"^"+" + "8e"^"-"]
4 × ["O"_3 + "2H"^"+" + "2e"^"-" → "O"_2 + "H"_2"O"]

Step 7: Add the two half-reactions.

"PbS" + color(red)(cancel(color(black)("4H"_2"O"))) → "PbSO"_4 + color(red)(cancel(color(black)("8H"^"+"))) + color(red)(cancel(color(black)("8e"^"-")))
4"O"_3 + color(red)(cancel(color(black)("8H"^"+"))) + color(red)(cancel(color(black)("8e"^"-"))) → "4O"_2 + color(red)(cancel(color(black)("4H"_2"O")))
stackrel(——————————————————)("PbS + 4O"_3 → "PbSO"_4 + "4O"_2)#

Step 8: Check mass balance.

$m a t h b f \left(\text{Atom"color(white)(m)"On the left"color(white)(m)"On the right}\right)$
$\textcolor{w h i t e}{m l} \text{Pb} \textcolor{w h i t e}{m m m m l l} 1 \textcolor{w h i t e}{m m m m m m m} 1$
$\textcolor{w h i t e}{m l} \text{S} \textcolor{w h i t e}{m m m m m l} 1 \textcolor{w h i t e}{m m m m m m m} 1$
$\textcolor{w h i t e}{m l} \text{O} \textcolor{w h i t e}{m m m m l l} 12 \textcolor{w h i t e}{m m m m m m l} 12$

Step 9: Check charge balance.

$m a t h b f \left(\text{On the left"color(white)(m)"On the right}\right)$
$\textcolor{w h i t e}{m m m} 0 \textcolor{w h i t e}{m m m m m m l l} 0$

Everything balances, so the balanced equation is

${\text{PbS + 4O"_3 → "PbSO"_4 + "4O}}_{2}$

Note: The trick is in the ozone half-reaction.

At first it appears as if ${\text{O"_3 → "O}}_{2}$ involves no change in oxidation number.

However, the half-reaction is really $\text{O"_3 → "H"_2"O}$, and you don’t use $\text{H"_2"O}$ in Step 1 because you use it later in Step 3 for balancing $\text{O}$.