# Question #4b8db

May 23, 2015

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

$\stackrel{\textcolor{b l u e}{+ 1}}{{K}_{2}} \stackrel{\textcolor{b l u e}{+ 6}}{C r} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{4}} + \stackrel{\textcolor{b l u e}{+ 1}}{N {a}_{2}} \stackrel{\textcolor{b l u e}{+ 4}}{S} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{3}} + \stackrel{\textcolor{b l u e}{+ 1}}{H} \stackrel{\textcolor{b l u e}{- 1}}{C l} \to \stackrel{\textcolor{b l u e}{+ 1}}{K} \stackrel{\textcolor{b l u e}{- 1}}{C l} + \stackrel{\textcolor{b l u e}{+ 1}}{N {a}_{2}} \stackrel{\textcolor{b l u e}{+ 6}}{S} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{4}} + \stackrel{\textcolor{b l u e}{+ 3}}{C r} \stackrel{\textcolor{b l u e}{- 1}}{C {l}_{3}} + \stackrel{\textcolor{b l u e}{+ 1}}{{H}_{2}} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{2}}$

Notice that the oxidation state of chromium goes from +6 on the reactants' side, to +3 on the products' side, which means that it is being reduced.

At the same time, the oxidation number of sulfur goes from +4 on the reactants' side, to +6 on the products' side, which means that it is being oxidized.

Since the reaction takes place in acidic solution, you write the net ionic half-reactions like this

• Reduction half-reaction

$\stackrel{\textcolor{b l u e}{+ 6}}{C r} {O}_{4}^{2 -} + 3 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 3}}{C {r}^{3 +}}$

Balance the oxygen and hydrogen atoms by adding water molecules to the side of the equation that needs oxygen, and protons, ${H}^{+}$, to the side of the equation that needs hydrogens.

$8 {H}^{+} + \stackrel{\textcolor{b l u e}{+ 6}}{C r} {O}_{4}^{2 -} + 3 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 3}}{C {r}^{3 +}} + 4 {H}_{2} O$

• Oxidation half-reaction

$\stackrel{\textcolor{b l u e}{+ 4}}{S} {O}_{3}^{2 -} \to \stackrel{\textcolor{b l u e}{+ 6}}{S} {O}_{4}^{2 -} + 2 {e}^{-}$

Once again, use water molecules and protons to balance the oxygen and hydrogen atoms.

${H}_{2} O + \stackrel{\textcolor{b l u e}{+ 4}}{S} {O}_{3}^{2 -} \to \stackrel{\textcolor{b l u e}{+ 6}}{S} {O}_{4}^{2 -} + 2 {e}^{-} + 2 {H}^{+}$

In any redox reaction, the number of electrons gained in the reduction half-reaction must be equal to the number of electrons lost in the oxidation half-reaction.

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to get

$\left\{\begin{matrix}16 {H}^{+} + 2 \stackrel{\textcolor{b l u e}{+ 6}}{C r} {O}_{4}^{2 -} + 6 {e}^{-} \to 2 \stackrel{\textcolor{b l u e}{+ 3}}{C {r}^{3 +}} + 8 {H}_{2} O \\ 3 {H}_{2} O + 3 \stackrel{\textcolor{b l u e}{+ 4}}{S} {O}_{3}^{2 -} \to 3 \stackrel{\textcolor{b l u e}{+ 6}}{S} {O}_{4}^{2 -} + 6 {e}^{-} + 6 {H}^{+}\end{matrix}\right.$

Add the two half-reactions to get

$16 {H}^{+} + 2 C r {O}_{4}^{2 -} + 3 {H}_{2} O + 3 S {O}_{3}^{2 -} + \cancel{6 {e}^{-}} \to 2 C {r}^{3 +} + 3 S {O}_{4}^{2 -} + \cancel{6 {e}^{-}} + 6 {H}^{+} + 8 {H}_{2} O$

The balanced chemical equation will look like this

$10 {H}^{+} + 2 C r {O}_{4}^{2 -} + 3 S {O}_{3}^{2 -} \to 2 C {r}^{3 +} + 3 S {O}_{4}^{2 -} + 5 {H}_{2} O$

Add the spectator ions to get

$2 {K}_{2} C r {O}_{4} + 3 N {a}_{2} S {O}_{3} + 10 H C l \to K C l + 3 N {a}_{2} S {O}_{4} + 2 C r C {l}_{3} + 5 {H}_{2} O$

Finally, balance the potassium and chlorine atoms by multiplying $K C l$ by 4

$2 {K}_{2} C r {O}_{4} + 3 N {a}_{2} S {O}_{3} + 10 H C l \to 4 K C l + 3 N {a}_{2} S {O}_{4} + 2 C r C {l}_{3} + 5 {H}_{2} O$