You have got the sign wrong (or alternatively, you have got the log expression upside down, which amounts to the same thing). This is something I have difficulty remembering, so I always think in terms of Le Chatelier's principle. What will be the effect of increasing [H2O2]? It will push the equilibrium to the right; that is, it will make the oxidation (of something else by H2O2) more favourable, so it will increase the electrode potential.