1. The problem statement, all variables and given/known data If dilute, aqueous magnesium bromide is electrolyzed, one product is formed at each electrode. Identify the product, state the electrode it occurs at (anode or cathode) and state a reason for your choice of product. 2. Relevant equations Mg2+(aq) + 2e- --> Mg(s) E°red = -2.37 V 0.5 Br2(l) + e- --> Br-(aq) E°red = +1.07 V H2O(l) + e- --> 0.5 H2(g) + OH-(aq) E°red = -0.83 V 0.5 O2(g) + 2H+(aq) + 2e- --> H2O(l) E°red = +1.23 V 3. The attempt at a solution At the cathode hydrogen gas and OH- ions would be formed since the reduction potential of water is less negative than the reduction potential of magnesium. This makes it easier to reduce water at the cathode. At the anode, oxidation occurs so the equations and potentials for the anions will have to be flipped. Br-(aq) --> 0.5 Br2(l) + e- E°ox = -1.07 V H2O(l) --> 0.5O2(g) + 2H+(aq) + 2e- E°ox = -1.23 V It would appear that it is easier to oxidize bromine ions since the oxidation potential is less negative. However, the solution is dilute. In the equilibrium equation 0.5 Br2(l) + e- <--> Br-(aq) The equilibrium would shift to the right and the number of electrons would be reduced. This would make the standard reduction potential of bromine more positive and hence the standard oxidation potential would be more negative. Since the oxidation potentials of bromine and water are close together, the bromine solution being dilute would cause water to be easier to oxidize instead of bromine (similar to how concentrated chlorine is oxidized over water in the electrolysis of brine). Thus oxygen gas and H+ ions are formed at the anode. However, my teacher said that bromine would still be oxidized regardless. Can anyone please tell me what is wrong with my argument?