Question #0c2a7

1 Answer
Jul 13, 2016

Answer:

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.