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-)#

1 Answer
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(color(blue)(+3))(Cr^(3+)) -> stackrel(color(blue)(+6))(Cr_2) O_7^(2-) + 3e^(-)#

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

#2stackrel(color(blue)(+3))(Cr^(3+)) -> stackrel(color(blue)(+6))(Cr_2) O_7^(2-) + 3e^(-)#

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.

#2stackrel(color(blue)(+3))(Cr^(3+)) -> stackrel(color(blue)(+6))(Cr_2) O_7^(2-) + 6e^(-)#

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.

#7H_2O + 2stackrel(color(blue)(+3))(Cr^(3+)) -> stackrel(color(blue)(+6))(Cr_2) O_7^(2-) + 6e^(-) + 14H^(+)#

  • Reduction half-reaction

#stackrel(color(blue)(+7))(S_2) O_8^(2-) + 1e^(-) -> stackrel(color(blue)(+6))(S) O_4^(2-)#

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

#stackrel(color(blue)(+7))(S_2) O_8^(2-) + 2e^(-) -> 2stackrel(color(blue)(+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

#{ (7H_2O + 2stackrel(color(blue)(+3))(Cr^(3+)) -> stackrel(color(blue)(+6))(Cr_2) O_7^(2-) + 6e^(-) + 14H^(+)), (3stackrel(color(blue)(+7))(S_2) O_8^(2-) + 6e^(-) -> 6stackrel(color(blue)(+6))(S) O_4^(2-)) :}#

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

#7H_2O + 3S_2O_8^(2-) + 2Cr^(3+) + cancel(6e^(-)) -> 6SO_4^(2-) + Cr_2O_7^(2-) + cancel(6e^(-)) + 14H^(+)#

Finally, you'll get

#7H_2O + 3S_2O_8^(2-) + 2Cr^(3+) -> 6SO_4^(2-) + Cr_2O_7^(2-) + 14H^(+)#