Question

How to write half reaction for H2O2(aq) -> O2(g)

Answer 1

Follow the usual procedure:

(1) Assign oxidation states. In this case, the H is +1, the O in peroxide is -1 and in O2 is 0. Since you've only got something being oxidized here, you're only going to get an oxidation half-reaction.

(2) Write the half-reaction skeleton, with just the species being oxidized or reduced, and the electrons in the right place:\[{\rm H}_2{\rm O}_2(aq) \rightarrow {\rm O}_2(g) + 2 e^-\]Notice I put two electrons being released, because two oxygen atoms are being oxidized from -1 to 0.

(3) Balance elements other than O and H. Already done.

(4) Balance oxygen by adding water to the appropriate side. Don't need to.

(5) Balance hydrogen by adding H+:\[{\rm H}_2{\rm O}_2(aq) \rightarrow 2 {\rm H}^+(aq) + {\rm O}_2(g) + 2 e^-\](6) If your reaction is in acid solution, you're done. If it's in basic solution, add OH- to both sides to cancel the H+:\[{\rm H}_2{\rm O}_2(aq) + 2 {\rm OH}^-(aq) \rightarrow 2 {\rm H}_2{\rm O}(l) + {\rm O}_2(g) + 2 e^-\]