# Question #6f441

May 24, 2015

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

$\stackrel{\textcolor{b l u e}{+ 4}}{S} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{2}} + \stackrel{\textcolor{b l u e}{+ 6}}{C {r}_{2}} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{7}^{2 -}} \to \stackrel{\textcolor{b l u e}{+ 6}}{S} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{4}^{2 -}} + \stackrel{\textcolor{b l u e}{+ 3}}{C {r}^{3 +}}$

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

At the same time, 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.

The oxidation and reduction half-reactions will look like this

• Oxidation half-reaction

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

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

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

• Reduction half-reaction

$\stackrel{\textcolor{b l u e}{+ 6}}{C {r}_{2}} {O}_{7}^{2 -} + 3 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 3}}{C {r}^{3 +}}$

Notice that you have two cxhromium atoms on the reactants' side, and only one on the products' side. Multiply the product by 2 to get

$\stackrel{\textcolor{b l u e}{+ 6}}{C {r}_{2}} {O}_{7}^{2 -} + 6 {e}^{-} \to 2 \stackrel{\textcolor{b l u e}{+ 3}}{C {r}^{3 +}}$

Notice that two chromium atoms lose a total of 6 electrons to go from an oxidation state of +6 to an aoxidation state of +3 per atom.

Once again, balance the oxygen and hydrogens atoms by adding water and protons.

$14 {H}^{+} + \stackrel{\textcolor{b l u e}{+ 6}}{C {r}_{2}} {O}_{7}^{2 -} + 6 {e}^{-} \to 2 \stackrel{\textcolor{b l u e}{+ 3}}{C {r}^{3 +}} + 7 {H}_{2} O$

In redox reactions, 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 to balance the number of transferred electrons.

$\left\{\begin{matrix}6 {H}_{2} O + 3 \stackrel{\textcolor{b l u e}{+ 4}}{S} {O}_{2} \to 3 \stackrel{\textcolor{b l u e}{+ 6}}{S} {O}_{4}^{2 -} + 6 {e}^{-} + 4 {H}^{+} \\ 14 {H}^{+} + \stackrel{\textcolor{b l u e}{+ 6}}{C {r}_{2}} {O}_{7}^{2 -} + 6 {e}^{-} \to 2 \stackrel{\textcolor{b l u e}{+ 3}}{C {r}^{3 +}} + 7 {H}_{2} O\end{matrix}\right.$

Add the two half-reactions together to get

$6 {H}_{2} O + 3 S {O}_{2} + 14 {H}^{+} + C {r}_{2} {O}_{7}^{2 -} + \cancel{6 {e}^{-}} \to 3 S {O}_{4}^{2 -} + 2 C {r}^{3 +} + 4 {H}^{+} + 7 {H}_{2} O + \cancel{6 {e}^{-}}$

The balanced net ionic equation will be

${H}_{2} {O}_{\left(l\right)} + 3 S {O}_{2 \left(g\right)} + C {r}_{2} {O}_{7 \left(a q\right)}^{2 -} \to 3 S {O}_{4 \left(a q\right)}^{2 -} + 2 C {r}_{\left(a q\right)}^{3 +} + 2 {H}_{\left(a q\right)}^{+}$