Question #2ce32

1 Answer
May 18, 2017

Answer:

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"#