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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.)

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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?

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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.

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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.

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