Question

Question #2ce32

Answer 1

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

Explanation:

Mercury is below hydrogen in the activity series, so `"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

`"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.

`"Hg" → "Hg"^"2+"`
`"NO"_3^"-" → "NO"`

Step 2: Balance all atoms other than `"H"` and `"O"`.

Done.

Step 3: Balance `"O"`.

`"Hg" → "Hg"^"2+"`
`"NO"_3^"-" → "NO" + 2"H"_2"O"`

Step 4: Balance `"H"`.

`"Hg" → "Hg"^"2+"`
`"NO"_3^"-" + "4H"^"+" → "NO" + 2"H"_2"O"`

Step 5: Balance charge.

`"Hg" → "Hg"^"2+" + "2e"^"-"`
`"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.

`"3Hg" → "3Hg"^"2+" + "6e"^"-"`
`"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

`"3Hg + 2NO"_3^"-" + "8H"^"+" → "3Hg"^"2+" + "2NO" + 4"H"_2"O"`
`color(white)(mm)+ "6NO"_3^"-" color(white)(mmmm)+"6NO"_3^"-"`
`stackrel(————————————————————)("3Hg + 8HNO"_3 → "3Hg"("NO"_3)_2 + "2NO" + 4"H"_2"O")`

∴ The balanced equation is

`"3Hg + 8HNO"_3 → "3Hg"("NO"_3)_2 + "2NO" + 4"H"_2"O"`