Give the electrolytic reaction of a concentrated aqueous $NaCl$ solution using inert electrodes.

(N/A) Inert electrodes are electrodes that do not participate in the chemical reaction.
In an aqueous $NaCl$ solution,the ionization occurs as: $NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$.
At the cathode,the competition is between the reduction of $Na^{+}$ and the reduction of $H_{2}O$ (or $H^{+}$):
$Na^{+}_{(aq)} + e^{-} \rightarrow Na_{(s)} \quad E^{\ominus} = -2.71 \ V$
$2H_{2}O_{(l)} + 2e^{-} \rightarrow H_{2(g)} + 2OH^{-}_{(aq)} \quad E^{\ominus} = -0.83 \ V$
Since the reduction potential of water is higher,$H_{2}$ gas is produced at the cathode.
At the anode,the competition is between the oxidation of $Cl^{-}$ and the oxidation of $H_{2}O$:
$2Cl^{-}_{(aq)} \rightarrow Cl_{2(g)} + 2e^{-} \quad E^{\ominus} = 1.36 \ V$
$2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} + 4e^{-} \quad E^{\ominus} = 1.23 \ V$
Although the oxidation potential of water is lower,due to the overpotential of oxygen,$Cl^{-}$ is preferentially oxidized to $Cl_{2}$ gas.
Overall reaction:
$2NaCl_{(aq)} + 2H_{2}O_{(l)} \rightarrow Cl_{2(g)} + H_{2(g)} + 2NaOH_{(aq)}$