# Question 0c2a7

Jul 13, 2016

You could do it like this:

#### Explanation:

At the cathode, water molecules are reduced:

$\textsf{{H}_{2} {O}_{\left(l\right)} + e \rightarrow O {H}_{\left(a q\right)}^{-} + \frac{1}{2} {H}_{2 \left(g\right)}}$

Hydrogen gas is evolved and $\textsf{O {H}^{-}}$ form around the electrode.

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

$\textsf{2 C {l}_{\left(a q\right)}^{-} \rightarrow C {l}_{2 \left(g\right)} + 2 e}$

The charge on 1 mole of electrons $\textsf{= 9.65 \times {10}^{4} \text{ } C}$

$\therefore$ $\textsf{9.65 \times {10}^{4} \text{ ""C}}$ will discharge 1 mole of $\textsf{O {H}^{-}}$

$\therefore$$\textsf{\frac{1}{360} \text{ } C}$ will discharge $\textsf{\frac{1}{9.65 \times {10}^{4}} \times \frac{1}{360} \text{ ""mol}}$

$\textsf{= 2.88 \times {10}^{- 8} \text{ ""mol}}$

$\therefore$$\textsf{\left[O {H}_{\left(a q\right)}^{-}\right] = \frac{2.88 \times {10}^{- 8}}{1} = 2.88 \times {10}^{- 8} \text{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 $\textsf{O {H}^{-}}$ ions from the auto - ionisation of water which is $\textsf{{10}^{- 7} \text{mol/l}}$ at $\textsf{{25}^{\circ} C}$.

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-solution285796

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.