Electrolysis Problem -- Half reactions and gasses produced

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Discussion Overview

The discussion revolves around the electrolysis of water in the presence of Na2SO4, focusing on the half-reactions occurring at the anode and cathode, as well as the role of Na2SO4 in the process. Participants explore the products of electrolysis, the behavior of ions, and the implications of ionic mobility.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the half-reactions for electrolysis, indicating that water is the primary species undergoing reduction and oxidation, while questioning the role of Na2SO4.
  • Another participant inquires about the products at the anode and cathode, suggesting uncertainty about the behavior of Na2SO4 during electrolysis.
  • A different participant speculates that SO3 gas might be produced at the cathode, expressing doubt about the oxidation of Na2SO4.
  • Discussion includes the dissociation of Na2SO4 in solution and the relative ionic mobilities of H+, Na+, OH-, and SO42-, with one participant referencing a "pecking order" of anions based on their mobility.
  • A hypothetical scenario is posed regarding the production of metallic Na on the electrode and its potential consequences.

Areas of Agreement / Disagreement

Participants express varying levels of uncertainty regarding the role of Na2SO4 in the electrolysis process, with no consensus on whether it is reduced or oxidized. Multiple competing views on the products of electrolysis and the behavior of ions remain unresolved.

Contextual Notes

Participants note limitations in their understanding of the actual processes involved in electrolysis and the decisive properties of the ions, indicating that further clarification may be needed.

Who May Find This Useful

This discussion may be useful for individuals interested in electrochemistry, particularly those exploring the effects of electrolytes in electrolysis processes.

i_love_science
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Homework Statement
Water is electrolyzed by passing a current of 0.500 amps through 1.00 L of 1.00 M Na2SO4 solution for 120 minutes. One gas is collected at the anode and a different gas is collected at the cathode.
a) Write the balanced equation for the cathode half-reaction.
b) Give the formula for the gas produced at the anode.
Relevant Equations
half reactions
The solution for a) is 2H2O+2e- → H2 +2OH-, and the solution for b) is O2, because the half reaction at the anode is
2H2O→O2 +4H+ + 4e-.
I don't understand why the electrolysis does not involve Na2SO4 being reduced or oxidized, and only considers water. Thanks.
 
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i_love_science said:
Na2SO4 being reduced or oxidized
What would be the result at anode ?
And at cathode ?
 
BvU said:
What would be the result at anode ?
And at cathode ?
For the cathode, SO3 gas might be produced, though I'm really not sure. I don't think Na2SO4 will be oxidized.
Do you imply that since Na2SO4 is not likely to be reduced or oxidized, and only water is reduced or oxidized?
 
Dissolved Na2SO4 is fully dissociated.

Ionic mobility of H+ is 36.23 10-4 cm2/(s.V) , Na+ has 5.19
Na has an ionization energy of 496 kJ/mol, H has 1312, so who wins ?

Ionic mobility of OH- is 20.64 10-4 cm2/(s.V) , SO42- has 8.29
SO42- ions are rather big and slow
(this guy, at 3'50" mentions a pecking order: halide anions, OH-, all other anions)

If I sound a bit careful, it's because I'm not entirely sure of the actual processes and the decisive properties.

@Borek ?

(numbers are from Peter Atkins: Physical Chemistry)
##\ ##
 
Last edited:
Imagine somehow metallic Na is produced on the electrode - what would happen?
 

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