"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.
This is my current answer:
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:
2Br- -> Br2(l) + 2e-
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.