(N/A) The ionization of $H_2SO_4 \rightarrow 2H^{+}{_{\text{(aq)}}} + SO_4^{2-}{_{\text{(aq)}}}$.
Near the anode,both $SO_4^{2-}$ ions and $H_2O$ molecules are present. Therefore,oxidation of either $SO_4^{2-}$ or $H_2O$ is possible.
$1$. Oxidation of $H_2O$: $2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} + 4e^- \quad E^{\circ} = 1.23 \ V$
$2$. Oxidation of $2SO_4^{2-}{_{\text{(aq)}}} \rightarrow S_2O_8^{2-}{_{\text{(aq)}}} + 2e^{-} \quad E^{\circ} = 1.96 \text{ V}$
Since the standard electrode potential $(E^{\circ})$ for the oxidation of water is lower than that of the sulfate ion,water is preferentially oxidized at the anode in dilute solutions,producing oxygen gas. However,in highly concentrated $H_2SO_4$ solutions,the oxidation of $SO_4^{2-}$ to form peroxodisulfate $(S_2O_8^{2-})$ becomes possible.