In the electrolysis of dilute sulphuric acid, I'm given these two half-equations:
anode: 4OH-(aq) -> 2H2O(l) + O2(g) + 4e-
cathode: 4H+(aq) + 4e- -> 2H2(g)
Why does the concentration of hydrogen ions increase?
Isn't the concentration of hydrogen ions decreasing here, rather than increasing as it is being discharged at the cathode?
N.B: (This question originates from a past AQA GCSE Question: see question 2 part (d) here.)
Replace the anode equation with your new equation. Then find the overall equation for the reaction. You are correct that the net number of $\ce{H+}$ is not changing as a result of this process. However, something is being consumed. What is it? How does that change the concentrations of the other species present?
The sulphuric acid does not take part in any reactions, and is only present to ensure that water conducts electricity.
Only the water self-ionises to form hydroxide (OH- and hydrogen (H+) ions, and its concentration decreases as it decomposes to form O2 and H2.
Therefore, the concentration of hydrogen ions from sulphuric acid increases during the electrolysis of dilute sulfuric acid.
There does seem to be 'resistance' to be accurate on the underlying mechanics of electrochemical reactions, likely due to complexity and the presence of radicals and associated short-lived intermediates (see, for example, this discussion on the electrolysis of water presented on this forum).
In the case of dilute H2SO4 with H+ ions, expect the reaction:
$\ce{ H+ + e- -> .H }$
And, the kinetically possible self-reaction of the hydrogen atom radical:
$\ce{ .H + .H -> H2 (g) }$
So, the H+ ion is indeed being consumed and the dissociation of H2SO4 increases to re-balance the equilibrium:
$\ce{ H2SO4 <=> H+ + HSO4- }$
As support for my discussion I also note the electrolysis of dilute H2SO4 in organic reduction reactions (see this ebook).
To reiterate, electrolysis half-cell reactions do not relate to mechanics, more informative as to a possible reaction's likelihood only.
