Mercury is below hydrogen in the activity series, so "Hg"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"3Hg + 8HNO3→3Hg(NO3)2+2NO+4H2O
You can find the general technique for balancing redox equations in acid solution here.
Step 1: Write the two half-reactions.
"Hg" → "Hg"^"2+"Hg→Hg2+
"NO"_3^"-" → "NO"NO-3→NO
Step 2: Balance all atoms other than "H"H and "O"O.
Done.
Step 3: Balance "O"O.
"Hg" → "Hg"^"2+"Hg→Hg2+
"NO"_3^"-" → "NO" + 2"H"_2"O"NO-3→NO+2H2O
Step 4: Balance "H"H.
"Hg" → "Hg"^"2+"Hg→Hg2+
"NO"_3^"-" + "4H"^"+" → "NO" + 2"H"_2"O"NO-3+4H+→NO+2H2O
Step 5: Balance charge.
"Hg" → "Hg"^"2+" + "2e"^"-"Hg→Hg2++2e-
"NO"_3^"-" + "4H"^"+" + "3e"^"-"→ "NO" + 2"H"_2"O"NO-3+4H++3e-→NO+2H2O
Step 6: Equalize electrons transferred.
3×["Hg" → "Hg"^"2+" + "2e"^"-"]3×[Hg→Hg2++2e-]
2×["NO"_3^"-" + "4H"^"+" + "3e"^"-"→ "NO" + 2"H"_2"O"]2×[NO-3+4H++3e-→NO+2H2O]
Step 7: Add the two half-reactions.
"3Hg" → "3Hg"^"2+" + "6e"^"-"3Hg→3Hg2++6e-
"2NO"_3^"-" + "8H"^"+" + "6e"^"-"→ "2NO" + "4H"_2"O"2NO-3+8H++6e-→2NO+4H2O
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"
