Solving Electrolysis Questions: NaBr(aq) & Comparing Eo Values

In summary, the question is asking for the products formed at the anode and cathode when NaBr(aq) is electrolysed using inert electrodes. The answers vary, with one stating that Br2 gas is liberated at the anode and H2 gas at the cathode, while the other states H2 gas, H2O, and O2 at the anode. The correct answer is most likely the one stating Br2 gas, as it is provided by the school. The difference in the two answers lies in the equation used to compare Eo values, with one using O2 + 4H+ + 4e ::equil:: 2H2O and the other using O2 + 2H2
  • #1
Thevanquished
26
0
Here is the question:

Predict the products formed at the anode and at the cathode when NaBr(aq) is electrolysed using inert electrodes

Answer:

Somehow i got 2 different answers by different answer sheets

One of them stated that Br2 gas (anode) and H2 gas (cathode) would be liberated

However, the other one writes H2(cathode) , H2O and O2 (anode) are formed

There is no arguments for the products formed at the cathode as clearly it is H2 gas which is liberated there. However, the question lies in the products formed at the anode. Which one is correct?

Eo values if needed:
Br2 + 2e ::equil:: 2Br- Eo= +1.07V
Na+ + e ::equil:: Na Eo= -2.71V
O2 + 4H+ + 4e ::equil:: 2H2O Eo= +1.23V
O2 + 2H2O + 4e ::equil:: 4OH- Eo= +0.40V

2H2O + 2e ::equil:: H2 + 2OH- Eo= -0.83V

The ones in red are the equations in question. Which one should i use to compare Eo values with Br at the anode (oxidation)? And why?

The answer which states that bromine gas is liberated is most probably the right one (as it is provided by my school) but it used the O2 + 4H+ + 4e ::equil:: 2H2O equation to compare instead of O2 + 2H2O + 4e ::equil:: 4OH- why is this so?
 
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  • #2
Where would you get your "4OH-" from in an aqueous solution of NaBr? The equilibrium concentration of it from the self-ionisation of water is going to be very insignificant and negligible (hence E value of the half-cell would be very positive actually, since equilibrium position lies very far to the right)
 
  • #3
Aren't OH- ions produced in the cathode?
 

1. What is electrolysis and how does it work?

Electrolysis is a process in which an electric current is used to break down a substance into its component parts. In the case of NaBr(aq), the sodium bromide molecules are split into their individual sodium and bromine ions. This is achieved by passing an electric current through the solution, causing a chemical reaction to occur.

2. Why is it important to compare Eo values in electrolysis questions?

Eo (standard electrode potential) values are used to determine the feasibility and direction of an electrolytic reaction. By comparing the Eo values of different substances, we can determine which substance will be reduced (gain electrons) and which will be oxidized (lose electrons) during the electrolysis process.

3. What are the steps for solving electrolysis questions involving NaBr(aq)?

Step 1: Write out the balanced chemical equation for the electrolysis reaction.
Step 2: Identify the oxidizing agent and reducing agent based on their Eo values.
Step 3: Determine the number of moles of each substance involved in the reaction.
Step 4: Calculate the total charge required to complete the reaction.
Step 5: Use Faraday's constant (96485 C/mol) to convert the charge into moles of electrons.
Step 6: Use the mole ratio from the balanced equation to determine the number of moles of each product formed.

4. What is the significance of the Eo value in electrolysis questions?

The Eo value represents the tendency of a substance to gain or lose electrons. A higher Eo value indicates a stronger oxidizing agent, while a lower Eo value indicates a stronger reducing agent. In electrolysis questions, the Eo values are used to determine which substance will be reduced and oxidized.

5. How can the Eo values be used to predict the products of electrolysis reactions?

By comparing the Eo values of the reactants, we can predict which substance will be reduced and which will be oxidized during the electrolysis process. The reduced substance will form at the cathode (negative electrode), while the oxidized substance will form at the anode (positive electrode). Additionally, the Eo values can also be used to determine the overall voltage of the electrolytic cell.

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