The iodine clock reaction involves 2 equations:
1. H2O2 (aq) + 2I-(aq)+ 2H+(aq) → 2H2O(l) + I2 (aq)
2. I2 (aq) + 2S2O32-(aq) → 2I-(aq) + S4O62-(aq)
Adding a solution of MoO4 catalyses the first reaction by providing an alternative pathway with a lower activation energy - 43 kJ/mol as compared to 52 kJ/mol in the uncatalysed reaction. What specifically is this 'alternative pathway' and in full detail, how does the MoO2 speed up the reaction rate of the first reaction?