Q.When acidulated water (dil. $H_2SO_4$ solution) is electrolysed, will the pH of the solution be affected? Justify your answer.

The key idea is that electrolysis of acidulated water decomposes water itself, not the acid. The $H_2SO_4$ merely provides ions to conduct electricity.

Reasoning:

1. At the cathode: $2H^+ + 2e^- \rightarrow H_2(g)$ — hydrogen ions are consumed.
2. At the anode: $2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$ — water is oxidised, producing hydrogen ions. …

Why the question matters

Many students instinctively think: "If we pass current through an acidic solution, the H⁺ ions get discharged at the cathode, so the solution should become less acidic — pH should rise." That sounds logical, but it misses a crucial point: in acidulated water, the acid (H₂SO₄) is not the reactant. It's only there to provide conductivity. The actual substance being electrolysed is water itself.

Let's see why.

Step-by-step reasoning

1. What does "acidulated water" mean?

Acidulated water is simply water made conducting by adding a few drops of dilute sulphuric acid. The acid dissociates completely:

These ions carry the current, but they are not consumed in the electrode reactions — they just shuttle charge.

2. What happens at the electrodes?

At the cathode (negative electrode), reduction occurs. Which species gets reduced? Compare the reduction potentials:

• $2H^+ + 2e^- \rightarrow H_2$ ($E^\circ = 0.00$ V)
• $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ ($E^\circ = -0.83$ V)

The H⁺ reduction is much easier (higher potential). So H⁺ from the acid is reduced to H₂ gas.

At the anode (positive electrode), oxidation occurs:

• $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ ($E^\circ = +1.23$ V)
• $SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-$ ($E^\circ = +2.01$ V)

Water oxidation is far easier, so water is oxidised to oxygen gas, releasing H⁺ ions.

3. The net effect on H⁺ concentration

Write the half-reactions with balanced electrons:

Notice: for every 4 H⁺ consumed at the cathode, exactly 4 H⁺ are produced at the anode. The H⁺ concentration remains unchanged.

4. What about the SO₄²⁻ ions?

They do nothing — they simply migrate to the anode but are not discharged (their oxidation potential is too high). So the acid concentration stays constant. …

Method: Conceptual Analysis Using Faraday’s Laws & Electrode Reactions

Step 1 – Identify the electrolyte and its ions

Acidulated water is dilute $H_2SO_4$ in water.

Ions present:

• From $H_2SO_4$: $H^+$, $SO_4^{2-}$
• From water: $H^+$, $OH^-$

Step 2 – Write the electrode reactions (using Faraday’s principle of selective discharge)

At cathode (reduction):

$2H^+ + 2e^- \rightarrow H_2(g)$

At anode (oxidation):

$2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$

Key observation:

At the cathode, $H^+$ is consumed.

At the anode, $H^+$ is produced.

Step 3 – Balance the electron transfer

Multiply cathode reaction by 2 to match electrons:

• Cathode: $4H^+ + 4e^- \rightarrow 2H_2$
• Anode: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$

Net reaction:

$2H_2O \rightarrow 2H_2 + O_2$

Step 4 – Interpret the effect on pH

• $H^+$ consumed at cathode = $4H^+$
• $H^+$ produced at anode = $4H^+$ …

Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Thinking pH changes because water is consumed

Many students see that water is being split into $H_2$ and $O_2$ and assume that since water is used up, the solution becomes more concentrated in acid, lowering the pH.

Why this is wrong:

In acidulated water (dilute $H_2SO_4$), the net reaction is:

Only water is consumed. The $H^+$ ions (from $H_2SO_4$) are not consumed — they act as charge carriers. At the cathode, $H^+$ is reduced:

But at the anode, water is oxidised:

For every $4H^+$ consumed at the cathode, $4H^+$ are produced at the anode. The $[H^+]$ remains constant.

How to avoid: Always write both half-reactions and check the net change in $H^+$ moles. If the number of $H^+$ consumed equals the number produced, pH is unchanged.

Mistake 2: Confusing this with electrolysis of a salt solution (e.g., NaCl)

Students sometimes apply the logic used for NaCl electrolysis (where pH changes due to $OH^-$ formation) to acidulated water.

Why this is wrong:

• In NaCl electrolysis, water is reduced at the cathode, producing $OH^-$ and $H_2$, so pH rises.
• In acidulated water, $H^+$ (from the acid) is reduced instead of water. No $OH^-$ is formed.

How to avoid: Identify the electrolyte. If the electrolyte is an acid (like $H_2SO_4$), the $H^+$ from the acid is reduced. If it's a salt of an active metal, water is reduced and pH changes.

Mistake 3: Forgetting that $H_2SO_4$ is a strong acid and fully dissociated

Some students think that as electrolysis proceeds, $SO_4^{2-}$ might get oxidised or react, altering the acid concentration.

Why this is wrong:

$SO_4^{2-}$ is not oxidised easily (it requires a very high potential). In dilute $H_2SO_4$, water is oxidised instead. The $SO_4^{2-}$ ions remain in solution, and the $H^+$ concentration stays constant. …