Electrolysis of NaCl: Na+ + Cl- to NaOH + Cl2

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Discussion Overview

The discussion centers on the electrolysis of sodium chloride (NaCl) in different states: molten and aqueous. Participants explore why the products differ between these two processes, specifically focusing on the formation of sodium and chlorine from molten NaCl versus sodium hydroxide and chlorine from NaCl in water.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that the difference in products is due to the relative ease of reducing H+ compared to Na+.
  • Reduction potentials are presented, indicating that 2H+ + 2e- yields H2 with an Eo of 0, while 2Cl- yields Cl2 with an Eo of -1.358 V.
  • Another participant reiterates the point about H+ being more readily reduced than Na+ and emphasizes that the chlorine reaction remains unchanged.

Areas of Agreement / Disagreement

Participants generally agree on the role of H+ reduction in the electrolysis of NaCl solution, but the discussion does not reach a consensus on the implications or further details of the reactions involved.

Contextual Notes

There are unresolved aspects regarding the comparison of reduction potentials and the implications for the electrolysis products, as well as the specific conditions under which these reactions occur.

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Homework Statement


Why does electrolysis of molten NaCl yield sodium and chlorine, while electrolysis of a solution of NaCl in water yields sodium hydroxide and chlorine.

Homework Equations



The Attempt at a Solution


Is it because H+ is more readily reduced than Na+?

2H+ + 2e --> H2 Eo= 0
2Cl- --> Cl2 + 2e Eo=-1.358

Thus, 2H+ + 2Cl- --> Cl2 H2 Eorxn=-1.358 V

And so 2H+ --> H2, leaving OH- behind in place of Cl-
 
Last edited:
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Any ideas? You see, that template that you should use is not a joke, but a tool to force you follow forum rules.

--
methods
 
Is it because H+ is more readily reduced than Na+?

2H+ + 2e --> H2 Eo= 0
2Cl- --> Cl2 + 2e Eo=-1.358

Thus, 2H+ + 2Cl- --> Cl2 H2 Eorxn=-1.358 V

And so 2H+ --> H2, leaving OH- behind in place of Cl-
 
Last edited:
plexus0208 said:
Is it because H+ is more readily reduced than Na+?

Yes. Note that further discussion is irrelevant, as you should compare reduction potentials of H+ and Na+, chlorine reaction doesn't change.

--
methods
 

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