# Question 2ce32

May 18, 2017

The reduction-oxidation reaction gives mercury(II) nitrate and nitrogen monoxide.

#### Explanation:

Mercury is below hydrogen in the activity series, so $\text{Hg}$ does not displace hydrogen from nitric acid.

However, the nitrate ion is a strong oxidizing agent.

In acid solution it oxidizes mercury to the mercury(II) ion and becomes reduced to nitrogen monoxide.

The balanced equation is

$\text{3Hg + 8HNO"_3 → "3Hg"("NO"_3)_2 + "2NO" + 4"H"_2"O}$

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

Step 1: Write the two half-reactions.

$\text{Hg" → "Hg"^"2+}$
$\text{NO"_3^"-" → "NO}$

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

Done.

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

$\text{Hg" → "Hg"^"2+}$
$\text{NO"_3^"-" → "NO" + 2"H"_2"O}$

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

$\text{Hg" → "Hg"^"2+}$
$\text{NO"_3^"-" + "4H"^"+" → "NO" + 2"H"_2"O}$

Step 5: Balance charge.

$\text{Hg" → "Hg"^"2+" + "2e"^"-}$
$\text{NO"_3^"-" + "4H"^"+" + "3e"^"-"→ "NO" + 2"H"_2"O}$

Step 6: Equalize electrons transferred.

3×["Hg" → "Hg"^"2+" + "2e"^"-"]
2×["NO"_3^"-" + "4H"^"+" + "3e"^"-"→ "NO" + 2"H"_2"O"]

Step 7: Add the two half-reactions.

$\text{3Hg" → "3Hg"^"2+" + "6e"^"-}$
$\text{2NO"_3^"-" + "8H"^"+" + "6e"^"-"→ "2NO" + "4H"_2"O}$
stackrel(————————————————————)("3Hg + 2NO"_3^"-" + "8H"^"+" → "3Hg"^"2+" + "2NO" + 4"H"_2"O")

Step 8. Add the spectator ions

$\text{3Hg + 2NO"_3^"-" + "8H"^"+" → "3Hg"^"2+" + "2NO" + 4"H"_2"O}$
$\textcolor{w h i t e}{m m} + \text{6NO"_3^"-" color(white)(mmmm)+"6NO"_3^"-}$
stackrel(————————————————————)("3Hg + 8HNO"_3 → "3Hg"("NO"_3)_2 + "2NO" + 4"H"_2"O")#

∴ The balanced equation is

$\text{3Hg + 8HNO"_3 → "3Hg"("NO"_3)_2 + "2NO" + 4"H"_2"O}$