What Is the Correct Equation for the Electrolysis of NaOH and Water?

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SUMMARY

The correct equation for the electrolysis of sodium hydroxide (NaOH) and water is represented as 2NaOH + 2H2O -> 2H2 + O2 + 2NaOH. In this reaction, NaOH acts as a spectator ion, enhancing the conductivity of the solution without participating in the electrochemical reaction. The amount of gases produced during electrolysis is determined by Faraday's law, rather than the stoichiometric amount of NaOH present. Therefore, when given a specific mass of NaOH, such as 40g, it is not possible to calculate the mass of water (H2O) involved in the reaction due to insufficient information.

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Homework Statement


Write the water electrolise of NaOH.

Homework Equations


NaOH+H2O->H2+O2+NaOH

The Attempt at a Solution


2NaOH+2H2o->2H2+O2+2NaOH
I thought to write 2H2O->2H2+O2
If I am given the mass of NaOH (which in one problem I am), then how am I supposed to find the mas of H2O for example?
Is it right to write this way
2NaOH+2H2o->2H2+O2+2NaOH
or
NaOH+2H2o->2H2+O2+NaOH?
 
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NaOH is just a spectator, it doesn't react. What it does it increases the solution conductivity.

Amount of gases produced doesn't depend on the amount of NaOH (at least not stoichiometrically), instead it is given by Faraday's law.
 
Borek said:
NaOH is just a spectator, it doesn't react. What it does it increases the solution conductivity.

Amount of gases produced doesn't depend on the amount of NaOH (at least not stoichiometrically), instead it is given by Faraday's law.
I am given the mass of NaOH 40g. I have to find the mass of substances added in cathode and anode. How can I find it?
 
You can't, there is not enough information.
 

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