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My teacher told me that electrolysis of dil. $\ce{H2SO4}$ (concentration below $50\%$) using inert electrodes results in gradual increase of the concentration of $\ce{H2SO4}$.

I can't seem to figure out why. I know that the sulfate ions will remain in the solution, but aren't the $\ce{H+}$ ions of $\ce{H2SO4}$ reduced at the cathode as well?

Another doubt I have is that I've seen different equations for the oxidation at the anode for this electrolysis in different books. Some of them show that $\ce{OH-}$ will be reduced, and some show that $\ce{H2O}$ will be reduced! I'm really confused about all this.

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In a dilute solution of sulfuric acid, there are the following species present: $\ce {H_2O}$, $\ce {H^+}$, $\ce {OH^-}$, $ \ce {HSO_4^-}$, $\ce {SO_4^2-}$. Of course, the water molecules are present in the highest concentration, much higher than the other species, since it is a dilute solution. For simplicity, we would also represent the proton and hydroxide ion as unsolvated. However, do take note that they are in fact heavily solvated in aqueous solution.

In some textbooks, it may be said that for the electrolysis of dilute $ \ce H_{2}\ce S\ce O_{4}$, the oxidation half-equation is written as $\ce {2H^+ + 2e^- -> H_2}$ and the reduction half-equation is written as $\ce {4OH^- -> 2H_2O + O_2 + 4e^-}$. However, the concentrations of these ions in solution are rather small compared to the concentration of the water molecule in solution. Albeit these reactions may be taking place, they should not be the main reactions and are unlikely to be representative of what is actually happening in solution.

In fact, since water molecules are the dominant species in solution, it is more likely that they undergo the redox reactions. The following half-equations represent the reactions happening in the solution more accurately:

Reduction: $\ce {2H_2O + 2e^- -> H_2 + 2OH^-}$

Oxidation: $\ce {2H_2O -> 4H^+ + O_2 + 4e^-}$

Since water molecules are being used up in the electrolysis process, the concentration of the remaining ions increase as the solution is electrolysed. Thus, the concentration of the sulfuric acid solution should increase.

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$\ce{OH-}$ can't be reduced. It may only be oxidized by giving an electron at the anode, while $\ce{H+}$ is reduced instead of $\ce{OH-}$. Secondly, the reducing property of $\ce{H2O}$ is greater than that of $\ce{H+}$ ion. Moreover, it is greater in number than other ions like $\ce{H+}$ ions etc. So it will be preferentially discharged although it is already discharged but preferentially reduced at the cathode.

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