Brine is electrolysed by using inert electrodes. The reaction at the anode is:

1. Cl-(aq) → $$\frac{1}{2}$$Cl2(g) + e-$$E_{cell}^{o}$$ = 1.36 V

2. 2H2O(l) → O2(g) + 4H+ +  4 e-;  $$E_{cell}^{o}$$ = 1.23 V

3. Na+(aq) + e- → Na(s); $$E_{cell}^{o}$$ = 2.71 V

4. H+(aq) + e-$$\frac{1}{2}$$H2(g); $$E_{cell}^{o}$$ = 0.00 V

HINT: Cl2 produces at anode.
Explanation:
STEP 1:
Brine is electrolysed by using inert electrodes. The possible reactions occurring at anode are :
${\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to \frac{1}{2}{\mathrm{Cl}}_{2}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}$;                                                       ${\mathrm{E}}_{\mathrm{Cell}}^{⊝}$=1.36V
;                                            ${\mathrm{E}}_{\mathrm{Cell}}^{⊝}$= 1.23 V

STEP 2:
The reaction at anode with lower value of E° is preferred and therefore water should get oxidised in preference to ${\mathrm{Cl}}^{-}$(aq). However, ${\mathrm{Cl}}_{2}$ is produced instead of ${\mathrm{O}}_{2}$.
This unexpected result is explained on the basis of the fact that water needs greater voltage for oxidation to ${\mathrm{O}}_{2}$ (as it is kinetically slow process) than that needed for oxidation of ${\mathrm{Cl}}^{-}$ions to ${\mathrm{Cl}}_{2}$