Predict the products of electrolysis

[jumbogala]

Homework Statement

What products are expected in the electrolysis of CuBr₂(aq)?

Homework Equations

The Attempt at a Solution

This is actually a solved example from my textbook that I have a question about. It says to first consider the possible reactions at the cathode.

Since reduction happens at the cathode, it says the possible reductions are:
Cu²⁺(aq) + 2e_- --> Cu(s) E = +0.34
2H₂O(l) + 2e^- --> H₂(g) + 2OH^- E= -0.83

The first one happens because its E is the most positive.

What I'm wondering is why we didn't include the reduction of Br₂. Is it because Br^- is an anion that we don't consider it?

And, how did they know to use the reaction for water that I gave and not 2H₂0(l) --> O₂(g) + 4H⁺ + 4e^-? (Because it's oxidation, which can't happen at the cathode, right?)

 

[Borek]

What I'm wondering is why we didn't include the reduction of Br₂.

Because there is no Br₂?

 

[jumbogala]

Wouldn't you get the dissociation of CuBr₂ --> Cu²⁺ + 2Br^-

And so need to take into account that 2Br^2- + 2e^- --> Br₂? Why can't that be happening at the cathode?

 

[symbolipoint]

What does a Standard Reduction Table tell you about
Br₂ + 2e^- -------> 2Br^- ?
Note that you are interested in the oxidation, not the reduction.

 

[jumbogala]

It has a value of +1.07.

But I don't understand what that means for this problem =(

There's something I'm missing here...

 

[symbolipoint]

You would determine or calculate the E for any pair of halfcell reduction-oxidation pairs. You know what the possible reduction half cells could be to choose from; you know the possible oxidation half cells could be to choose from. If needed, choose to examine all possible combinations of redox pairs and determine the whole cell E's of each combination. The most positive value might correspond to the most favorable whole cell redox reaction. Just remind yourself that you need an oxidation to occur at the anode and you need a reduction to occur at the cathode. Fit the halfcell values accordingly.