# "Write equations for the half-reactions that occur at the anode and cathode for the electrolysis of the following aqueous solution: CuBr2(aq)"

Possible oxidation reactions:
1. 2Br- -> Br2 + 2e- = -1.09
2. 2H2O -> O2 + 4H+ + 4e- = -1.23

Possible reduction reactions:
1. Cu2+ + 2e- -> Cu = 0.34
2. 2H2O + 2e- -> H2 + 2OH- = -0.83

I thought that, since the first oxidation (Br) and first reduction (Cu2+) were more positive, they would occur first, but it keeps saying my answer is wrong. I'm worried it may just be some sort of formatting thing.

Cu2+(aq) + 2e- -> Cu(s), 2Br-(aq) -> Br2(l) + 2e-

## Cu^2+ comes out before H^+ at the negative electrode.

But I think the 1.23 occurs before 1.09.

## Well, well, it seems we have a reaction mix-up here. Let's sort it out with a little humor, shall we?

For the anode half-reaction, we have the oxidation of Br-. So picture this: Br- walks up to the anode and says, "I'm feeling a bit positive today, time to become bromine!" The equation looks like this:

Br-(aq) -> Br2(l) + 2e-

Now, for the cathode half-reaction, it's time for Cu2+ to shine. Cu2+ happily strolls up to the cathode and says, "I'm feeling a little needy, can someone give me some electrons?" The equation goes like this:

Cu2+(aq) + 2e- -> Cu(s)

And voila, now we have the correct half-reactions for the electrolysis of CuBr2(aq). Keep those positive vibes flowing, and your equations will be just fine!

## Your answer is correct! The half-reactions at the anode and cathode for the electrolysis of CuBr2(aq) are:

Anode (oxidation):
2Br- -> Br2(l) + 2e-

Cathode (reduction):
Cu2+(aq) + 2e- -> Cu(s)

These half-reactions indicate that bromide ions (Br-) are being oxidized at the anode to form elemental bromine (Br2), and copper ions (Cu2+) are being reduced at the cathode to form solid copper (Cu).

## Your current answer for the half-reactions occurring at the anode and cathode is correct. The oxidation half-reaction occurs at the anode, and the reduction half-reaction occurs at the cathode.

Oxidation half-reaction (anode):
2Br^-(aq) -> Br2(l) + 2e^-

Reduction half-reaction (cathode):
Cu2+(aq) + 2e^- -> Cu(s)

The reasoning behind your choice of these half-reactions is also correct. The half-reaction with the more positive reduction potential (Cu2+ + 2e^- -> Cu) occurs as the reduction half-reaction at the cathode. Similarly, the half-reaction with the more positive oxidation potential (2Br^- -> Br2 + 2e^-) occurs as the oxidation half-reaction at the anode.

Make sure you double-check the formatting and calculations when submitting your answer to ensure that it is being correctly evaluated.