# Question 53cf1

Aug 16, 2017

WARNING! Long answer! The balanced equation is

$\text{I"_2 + "10HNO"_3 → "2HIO"_3 + "10NO"_2 + 4"H"_2"O}$

#### Explanation:

You can find the general technique for balancing redox equations in acid solution here.

We see that ${\text{I}}_{2}$ is oxidized to ${\text{HIO}}_{3}$ and ${\text{HNO}}_{3}$ is reduced to ${\text{NO}}_{2}$.

Step 1: Write the two half-reactions.

${\text{I"_2 → "HIO}}_{3}$
${\text{HNO"_3 → "NO}}_{2}$

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

${\text{I"_2 → "2HIO}}_{3}$
${\text{HNO"_3 → "NO}}_{2}$

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

${\text{I"_2 + 6"H"_2"O" → "2HIO}}_{3}$
$\text{HNO"_3 → "NO"_2 + "H"_2"O}$

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

$\text{I"_2 + 6"H"_2"O" → "2HIO"_3 + 10"H"^"+}$
$\text{HNO"_3 + "H"^"+"→ "NO"_2 + "H"_2"O}$

Step 5: Balance charge.

$\text{I"_2 + 6"H"_2"O" → "2HIO"_3 + 10"H"^"+" + 10"e"^"-}$
$\text{HNO"_3 + "H"^"+" + "e"^"-"→ "NO"_2 + "H"_2"O}$

Step 6: Equalize electrons transferred.

1 × ["I"_2 + 6"H"_2"O" → "2HIO"_3 + 10"H"^"+" + 10"e"^"-"]
10 × ["HNO"_3 + "H"^"+" + "e"^"-"→ "NO"_2 + "H"_2"O"]

Step 7: Add the two half-reactions.

"I"_2 + color(red)(cancel(color(black)(6"H"_2"O"))) → "2HIO"_3 + color(red)(cancel(color(black)(10"H"^"+"))) + color(red)(cancel(color(black)(10"e"^"-")))#
$\underline{10 \text{HNO"_3 + color(red)(cancel(color(black)(10"H"^"+"))) + color(red)(cancel(color(black)(10"e"^"-")))→ 10"NO"_2 + stackrelcolor(blue)(4)(color(red)(cancel(color(black)(10))))"H"_2"O}}$
$\text{I"_2 + "10HNO"_3 → "2HIO"_3 + "10NO"_2 + 4"H"_2"O}$

Step 8: Check mass balance.

$\underline{m a t h b f \left(\text{Atom"color(white)(m)"On the left"color(white)(m)"On the right}\right)}$
$\textcolor{w h i t e}{m l} \text{I} \textcolor{w h i t e}{m m m m m l} 2 \textcolor{w h i t e}{m m m m m m m} 2$
$\textcolor{w h i t e}{m l} \text{H} \textcolor{w h i t e}{m m m m l} 10 \textcolor{w h i t e}{m m m m m m l} 10$
$\textcolor{w h i t e}{m l} \text{N} \textcolor{w h i t e}{m m m m l} 10 \textcolor{w h i t e}{m m m m m m l} 10$
$\textcolor{w h i t e}{m l} \text{O} \textcolor{w h i t e}{m m m m l} 30 \textcolor{w h i t e}{m m m m m m l} 30$

Step 9: Check charge balance.

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

∴ The balanced equation is

$\text{I"_2 + "10HNO"_3 → "2HIO"_3 + "10NO"_2 + 4"H"_2"O}$