The keyword they use here is concentrated Sodium Chloride solution.
Normally in the case of electrolysis of NaCl 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:
Na^+(aq)+ e^(-) -> Na(s) / E=-2.71 V
2H_2O(l) + 2e^(-) -> H_2 (g) + 2OH^- / E=-0.83V So water is favoured at the anode.
At the anode we have the following possiblities:
2Cl^(-)(aq)->Cl_2(g) + 2e^-/ E=-1.36V
2H_2O(l) ->4H^(+)(aq) + O_2(g)+ 4e^-/ E=-1.23V
Again, water is favoured as its potential is lower.
However, with a concentrated saline solution, the concentration of Cl^- matters. My prof told me that if the concentration by mass of NaCl is more than 25% of the solution, the oxidation of Cl^- to Cl_2 becomes favoured, despite having a higher electrode potential. So beware if it states that it has a high concentration of Cl^-!
So the two half equations in the electrolysis of concentrated salt solutions should be:
Anode:
2Cl^(-)(aq)->Cl_2(g) + 2e^-/ E=-1.36V
Cathode:
2H_2O(l) + 2e^(-) -> H_2 (g) + 2OH^- / E=-0.83V
Overall:
2NaCl (aq) + 2H_2O (l) -> H_2(g) + Cl_2 (g) + 2 Na^(+)(aq) + 2OH^(-) (aq)
(Sodium ions just dissolve into the solution and are only spectators)
Results:
We form gas at both electrodes in a 1:1 molar ratio. Cl_2 gives a strong smell, coming from the anode.
pH will increase as we form Hydroxide ions.
