How do you balance the following redox equation in acidic solution: P_4 + HOCl -> H_3PO_4 + Cl^-?

Jan 3, 2017

Warning! Long answer! The balanced equation is

$\text{P"_4 + 6"H"_2"O" + "10HOCl" → "4H"_3"PO"_4+ 10"H"^"+" + "10Cl"^"-}$

Explanation:

You can find the general technique for balancing redox equations in acid solution in this Socratic answer.

We see that ${\text{P}}_{4}$ is oxidized to ${\text{H"_3"PO}}_{4}$ and $\text{HOCl}$ is reduced to $\text{Cl"^"-}$.

Step 1: Write the two half-reactions.

${\text{P"_4 → "H"_3"PO}}_{4}$
$\text{HOCl" → "Cl"^"-}$

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

${\text{P"_4 → "4H"_3"PO}}_{4}$
$\text{HOCl" → "Cl"^"-}$

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

${\text{P"_4 + "16H"_2"O" → "4H"_3"PO}}_{4}$
$\text{HOCl" → "Cl"^"-" + "H"_2"O}$

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

$\text{P"_4 + "16H"_2"O" → "4H"_3"PO"_4+ "20H"^"+}$
$\text{HOCl" +"H"^"+" → "Cl"^"-" + "H"_2"O}$

Step 5: Balance charge.

$\text{P"_4 + "16H"_2"O" → "4H"_3"PO"_4+ "20H"^"+" + "20e"^"-}$
$\text{HOCl" +"H"^"+" + "2e"^"-" → "Cl"^"-" + "H"_2"O}$

Step 6: Equalize electrons transferred.

1 ×["P"_4 + "16H"_2"O" → "4H"_3"PO"_4+ "20H"^"+" + "20e"^"-"]
10 × ["HOCl" +"H"^"+" + "2e"^"-" → "Cl"^"-" + "H"_2"O"]

Step 7: Add the two half-reactions.

"P"_4 + stackrelcolor(blue)(6)(color(red)(cancel(color(black)(16))))"H"_2"O" → "4H"_3"PO"_4+ stackrelcolor(blue)(10)(color(red)(cancel(color(black)(20))))"H"^"+" + color(red)(cancel(color(black)(20"e"^"-")))
"10HOCl" + color(red)(cancel(color(black)(10"H"^"+"))) + color(red)(cancel(color(black)("20e"^"-"))) → "10Cl"^"-" + color(red)(cancel(color(black)(10"H"_2"O")))
stackrel(————————————————————)("P"_4 + 6"H"_2"O" + "10HOCl" → "4H"_3"PO"_4+ 10"H"^"+" + "10Cl"^"-")

Step 8: Check mass balance.

$m a t h b f \left(\text{Atom"color(white)(m)"On the left"color(white)(m)"On the right}\right)$
$\textcolor{w h i t e}{m l} \text{P} \textcolor{w h i t e}{m m m m m l} 4 \textcolor{w h i t e}{m m m m m m m} 4$
$\textcolor{w h i t e}{m l} \text{H} \textcolor{w h i t e}{m m m m l l} 22 \textcolor{w h i t e}{m m m m m m l} 22$
$\textcolor{w h i t e}{m l} \text{O} \textcolor{w h i t e}{m m m m l l} 16 \textcolor{w h i t e}{m m m m m m l} 16$
$\textcolor{w h i t e}{m l} \text{Cl} \textcolor{w h i t e}{m m m m l} 10 \textcolor{w h i t e}{m m m m m m l} 10$

Step 9: Check charge balance.

$m a t h b f \left(\text{On the left"color(white)(m)"On the right}\right)$
$\textcolor{w h i t e}{m m m} 0 \textcolor{w h i t e}{m m m l l} 10 + \left(\text{-10}\right) = 0$

∴ The balanced equation is

$\text{P"_4 + 6"H"_2"O" + "10HOCl" → "4H"_3"PO"_4+ 10"H"^"+" + "10Cl"^"-}$