Predict the products of electrolysis

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Discussion Overview

The discussion revolves around predicting the products of electrolysis for the compound CuBr2 in aqueous solution. Participants explore the possible reactions occurring at the cathode and anode, considering both reduction and oxidation processes.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why the reduction of Br2 is not included in the analysis, suggesting it may be due to Br- being an anion.
  • Another participant points out that CuBr2 dissociates into Cu2+ and 2Br-, implying that the reduction of Br- to Br2 should be considered at the cathode.
  • A participant references the Standard Reduction Table, noting the reduction potential for Br2 to Br- and indicating interest in the oxidation process instead.
  • There is a discussion about the significance of the standard reduction potential value of +1.07 for Br2 and its relevance to the problem.
  • A later reply suggests calculating the E values for various half-cell reactions to determine the most favorable redox reaction, emphasizing the need for both oxidation at the anode and reduction at the cathode.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the inclusion of Br2 reduction and its implications for the electrolysis products. Multiple competing views remain on the reactions that should be considered at the cathode.

Contextual Notes

There are limitations regarding the assumptions made about the reactions at the cathode and anode, as well as the dependence on the definitions of the half-cell reactions. The discussion does not resolve the mathematical steps needed to determine the most favorable reactions.

jumbogala
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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?)
 
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jumbogala said:
What I'm wondering is why we didn't include the reduction of Br2.

Because there is no Br2?
 
Wouldn't you get the dissociation of CuBr2 --> Cu2+ + 2Br-

And so need to take into account that 2Br2- + 2e- --> Br2? Why can't that be happening at the cathode?
 
What does a Standard Reduction Table tell you about
Br2 + 2e- -------> 2Br- ?
Note that you are interested in the oxidation, not the reduction.
 
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...
 
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.
 

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