Electrolysis of dilute aqueous solutions

[Tryt]

Homework Statement

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.

Homework 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

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?

 

[nate808]

Your argument is correct in terms of the standard reduction potentials and the effect of dilution on the equilibrium. However, in practical electrolysis, other factors such as concentration, temperature, and pH can also play a role in determining the products formed at the electrodes. In this case, the concentration of magnesium bromide and the pH of the solution may also influence the products formed.

Additionally, the electrolysis process involves the passage of current through the solution, which can cause changes in the concentrations of the ions and thus affect the equilibrium. This can lead to a different product being formed at the anode compared to what is predicted based on the standard reduction potentials.

In summary, while your argument is valid in terms of the theoretical principles, it is important to consider other factors that may affect the products formed in practical electrolysis.