How to do this question? Apr 8, 2018

See below:

Explanation:

The keyword they use here is concentrated Sodium Chloride solution.

Normally in the case of electrolysis of $N a C l$ dissolved in a dilute aqueous solution, it is the same as "the electrolysis of water"- you will oxidize and reduce water at the anode and cathode respectively. This is because of their electrode potentials, Which are found in this booklet in table 24

At cathode we have the following possibilities:
$N {a}^{+} \left(a q\right) + {e}^{-} \to N a \left(s\right)$ / $E = - 2.71 V$
$2 {H}_{2} O \left(l\right) + 2 {e}^{-} \to {H}_{2} \left(g\right) + 2 O {H}^{-}$ / $E = - 0.83 V$ So water is favoured at the anode.

At the anode we have the following possiblities:
$2 C {l}^{-} \left(a q\right) \to C {l}_{2} \left(g\right) + 2 {e}^{-}$/ $E = - 1.36 V$
$2 {H}_{2} O \left(l\right) \to 4 {H}^{+} \left(a q\right) + {O}_{2} \left(g\right) + 4 {e}^{-}$/ $E = - 1.23 V$
Again, water is favoured as its potential is lower.

However, with a concentrated saline solution, the concentration of $C {l}^{-}$ matters. My prof told me that if the concentration by mass of $N a C l$ is more than 25% of the solution, the oxidation of $C {l}^{-}$ to $C {l}_{2}$ becomes favoured, despite having a higher electrode potential. So beware if it states that it has a high concentration of $C {l}^{-}$!

So the two half equations in the electrolysis of concentrated salt solutions should be:
Anode:
$2 C {l}^{-} \left(a q\right) \to C {l}_{2} \left(g\right) + 2 {e}^{-}$/ $E = - 1.36 V$

Cathode:
$2 {H}_{2} O \left(l\right) + 2 {e}^{-} \to {H}_{2} \left(g\right) + 2 O {H}^{-}$ / $E = - 0.83 V$

Overall:
$2 N a C l \left(a q\right) + 2 {H}_{2} O \left(l\right) \to {H}_{2} \left(g\right) + C {l}_{2} \left(g\right) + 2 N {a}^{+} \left(a q\right) + 2 O {H}^{-} \left(a q\right)$
(Sodium ions just dissolve into the solution and are only spectators)
Results:
We form gas at both electrodes in a 1:1 molar ratio. $C {l}_{2}$ gives a strong smell, coming from the anode.
$p H$ will increase as we form Hydroxide ions.