Question

Question #0c2a7

Answer 1

You could do it like this:

Explanation:

At the cathode, water molecules are reduced:

`sf(H_2O_((l))+erarrOH_((aq))^(-)+1/2H_(2(g)))`

Hydrogen gas is evolved and `sf(OH^-)` form around the electrode.

If the solution is sufficiently concentrated then chlorine is evolved at the anode:

`sf(2Cl_((aq))^(-)rarrCl_(2(g))+2e)`

The charge on 1 mole of electrons `sf(=9.65xx10^(4)" "C)`

`:.` `sf(9.65xx10^(4)" ""C")` will discharge 1 mole of `sf(OH^-)`

`:.` `sf(1/(360)" "C)` will discharge `sf(1/(9.65xx10^(4))xx(1)/(360)" ""mol")`

`sf(=2.88xx10^(-8)" ""mol")`

`:.` `sf([OH_((aq))^(-)]=(2.88xx10^(-8))/1=2.88xx10^(-8)"mol/l")`

Now this is a very low concentration so I would predict the pH to be very slightly above 7.

It is actually less than the concentration of `sf(OH^(-))` ions from the auto - ionisation of water which is `sf(10^(-7)"mol/l")` at `sf(25^@C)`.

I have explained how to calculate the pH more precisely in a similar answer here:

https://socratic.org/questions/what-is-the-ph-of-a-3-9-10-8-m-oh-solution#285796

In reality it is important that the products at each electrode are kept separate as they will react with each other.

This process forms the basis of the chlor - alkali industry where hydrogen, chlorine and sodium hydroxide are manufactured from the electrolysis of brine.