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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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