How do I balance this reaction using half-redox equations? $$\ce{2SO2 + O2->2SO3}$$

My work is:

Oxidation: $\ce{SO2 + H2O -> SO3 + 2e- + 2H+}$

Reduction: $\ce{H2O + 2e- + O2 -> SO3 + 2H+}$

Total: $\ce{SO2 + 2H2O + O2 -> 2SO3 + 4H+}$

BUT it's false. What did I do wrong?

  • $\begingroup$ Your reduction formula is wrong. For one thing you have no sulfur containing species on the reactant side and the reactant side has a charge of -2 while the product side shows a charge of +2, this is a violation of charge neutrality, both sides must have the same overall charge. $\endgroup$ – Philipp Nov 10 '14 at 15:32

WELL I think you put something erroneous in your reduction formula.

You will never get SO3 out of just O2 and water. Instead of SO3

The reduction formla will be 4e- + O2 + 4H+ --> 2 H2O

Multiply oxidation formula by 2 and add it to new reduction formula and you got your answer


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