In aqueous electrolysis, water can compete with solute ions at both electrodes. At cathode: if $E_{reduction}$($M^{n+}$) > $E_{reduction}$(H2O) = -0.83 V, the metal deposits. If $E_{reduction}$($M^{n+}$) < -0.83 V (active metals like Na, K, Al), H2O is reduced to H2. At anode: if $E_{oxidation}$(anion) > $E_{oxidation}$(H2O) = -1.23 V... actually, the species with less negative (more positive) oxidation potential is oxidised preferentially. In practice, overpotential (extra voltage needed due to kinetics) complicates predictions. For dilute NaCl: O2 at anode, H2 at cathode. For concentrated NaCl: Cl2 at anode (overpotential of O2 is high), H2 at cathode. Na+ is never reduced in aqueous solution.