# Question #4e8ae

May 6, 2015

Start by assigning oxidation numbers to all the atoms that take part in the reaction

$\stackrel{\textcolor{b l u e}{+ 1}}{{H}_{2}} \stackrel{\textcolor{b l u e}{- 2}}{{S}_{\left(s\right)}} + \stackrel{\textcolor{b l u e}{+ 5}}{N} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{3 \left(a q\right)}^{-}} \to \stackrel{\textcolor{b l u e}{0}}{{S}_{8 \left(s\right)}} + \stackrel{\textcolor{b l u e}{+ 2}}{N} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{\left(g\right)}}$

Notice that sulfur goes from an oxidation state of -2 on the reactants' side, to an oxidation state of zero on the products' side $\to$ sulfur is being oxidized.

Likewise, nitrogen's oxidation state changes from +5 to +2 $\to$ nitrogen is being reduced.

The oxidation and reduction half-reactions will look like this

• Oxidation half-reaction

${H}_{2} \stackrel{\textcolor{b l u e}{- 2}}{S} \to \stackrel{\textcolor{b l u e}{0}}{{S}_{8}} + 2 {e}^{-}$

Balance the sulfur atoms by multiplying the left side of the reaction by 8

$8 {H}_{2} \stackrel{\textcolor{b l u e}{- 2}}{S} \to \stackrel{\textcolor{b l u e}{0}}{{S}_{8}} + 16 {e}^{-}$

Balance the hydrogen atoms by adding 16 protons, ${H}^{+}$, to the products' side

$8 {H}_{2} \stackrel{\textcolor{b l u e}{- 2}}{S} \to \stackrel{\textcolor{b l u e}{0}}{{S}_{8}} + 16 {e}^{-} + 16 {H}^{+}$

• Reduction half-reaction

$\stackrel{\textcolor{b l u e}{+ 5}}{N} {O}_{3}^{-} + 3 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 2}}{N} O$

Balance the oxygens by adding 2 water molecules to the products' side

$\stackrel{\textcolor{b l u e}{+ 5}}{N} {O}_{3}^{-} + 3 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 2}}{N} O + 2 {H}_{2} O$

Now balance the hydrogens by adding 4 protons to the reactants' side

$4 {H}^{+} + \stackrel{\textcolor{b l u e}{+ 5}}{N} {O}_{3}^{-} + 3 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 2}}{N} O + 2 {H}_{2} O$

In a redox reaction, the number of electrons lost during oxidation and gained during reduction must be equal. At this point, you have $3 {e}^{-}$ gained in the reduction half-reaction, and $16 {e}^{-}$ lost during the oxidation half-reaction.

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 16 to balance the electrons out

$\left\{\begin{matrix}24 {H}_{2} \stackrel{\textcolor{b l u e}{- 2}}{S} \to 3 \stackrel{\textcolor{b l u e}{0}}{{S}_{8}} + 48 {e}^{-} + 48 {H}^{+} \\ 62 {H}^{+} + 16 \stackrel{\textcolor{b l u e}{+ 5}}{N} {O}_{3}^{-} + 48 {e}^{-} \to 16 \stackrel{\textcolor{b l u e}{+ 2}}{N} O + 32 {H}_{2} O\end{matrix}\right.$

Add the two half-reactions and cancel the species present on both sides of the equation to get

$24 {H}_{2} S + 16 {H}^{+} + \cancel{48 {e}^{-}} + 16 N {O}_{3}^{-} \to 3 {S}_{8} + \cancel{48 {e}^{-}} + 16 N O + 32 {H}_{2} O$

The balanced equation will thus be

$24 {H}_{2} S + 16 {H}^{+} + 16 N {O}_{3}^{-} \to 3 {S}_{8} + 16 N O + 32 {H}_{2} O$