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-)#
Start by assigning oxidation numbers to all the atoms that take part in the reaction.
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
The first thing you need to do is balance the chromium atoms
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.
Now balance the oxygen and hydrogen atoms by adding water molecules to the side that needs oxygen, and protons,
- Reduction half-reaction
Once again, balance the sulfur atoms and the number of electrons gained
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
Add these two half-reaction to get the balanced chemical equation
Finally, you'll get