How to I write half reactions and balance this oxidation-reduction reaction? S_2O_(8(aq))^(2-) + Cr_((aq))^(3+) -> SO_(4(aq))^(2-) + Cr_2O_(7(aq))^(2-)

May 22, 2015

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

stackrel(color(blue)(+7))(S_2) stackrel(color(blue)(-2))(O_8^(2-)) + stackrel(color(blue)(+3))(Cr^(3+)) -> stackrel(color(blue)(+6))(S) stackrel(color(blue)(-2))(O_4^(2-)) + stackrel(color(blue)(+6))(Cr_2) stackrel(color(blue)(-2))(O_7^(2-)

Notice that the oxidation state of sulfur goes from +7 on the rectants' side, to +6 on the products' side, which means that sulfur is being reduced.

At the same time, cromium's oxidtion state is going from +3 on the products' side, to +6 on the reactants' side, which menas that it is being oxidized.

The oxidation and reduction half-reactions are going to look like this

• Oxidation half-reaction

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

The first thing you need to do is balance the chromium atoms

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

Notice that you go from two chromium atoms that have an oxidation state of +3, to two chromium atoms that have an oxidation state of +6, which means that a total of 6 electrons are being lost.

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

Now 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.

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

• Reduction half-reaction

$\stackrel{\textcolor{b l u e}{+ 7}}{{S}_{2}} {O}_{8}^{2 -} + 1 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 6}}{S} {O}_{4}^{2 -}$

Once again, balance the sulfur atoms and the number of electrons gained

$\stackrel{\textcolor{b l u e}{+ 7}}{{S}_{2}} {O}_{8}^{2 -} + 2 {e}^{-} \to 2 \stackrel{\textcolor{b l u e}{+ 6}}{S} {O}_{4}^{2 -}$

You don't need to balance any oxygen or hydrogen atoms this time.

THe number of electrons gained in the reduction half-reaction must be equal to the number of electrons lost in the oxidation half-reaction, so multiply the reduction half-reaction by 3 to get

$\left\{\begin{matrix}7 {H}_{2} O + 2 \stackrel{\textcolor{b l u e}{+ 3}}{C {r}^{3 +}} \to \stackrel{\textcolor{b l u e}{+ 6}}{C {r}_{2}} {O}_{7}^{2 -} + 6 {e}^{-} + 14 {H}^{+} \\ 3 \stackrel{\textcolor{b l u e}{+ 7}}{{S}_{2}} {O}_{8}^{2 -} + 6 {e}^{-} \to 6 \stackrel{\textcolor{b l u e}{+ 6}}{S} {O}_{4}^{2 -}\end{matrix}\right.$

Add these two half-reaction to get the balanced chemical equation

$7 {H}_{2} O + 3 {S}_{2} {O}_{8}^{2 -} + 2 C {r}^{3 +} + \cancel{6 {e}^{-}} \to 6 S {O}_{4}^{2 -} + C {r}_{2} {O}_{7}^{2 -} + \cancel{6 {e}^{-}} + 14 {H}^{+}$

Finally, you'll get

$7 {H}_{2} O + 3 {S}_{2} {O}_{8}^{2 -} + 2 C {r}^{3 +} \to 6 S {O}_{4}^{2 -} + C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+}$