Q.How will the pH of brine (aq. $NaCl$ solution) be affected when it is electrolysed?

The key idea is that electrolysis of aqueous NaCl (brine) produces NaOH at the cathode, making the solution basic.

Reasoning:

1. In aqueous NaCl, the possible cathode reactions are reduction of $Na^+$ (very negative $E^\circ$) or reduction of water: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$. Water is preferentially reduced.
2. This generates $OH^-$ ions, increasing the concentration of hydroxide in the solution. …

The key to understanding this lies in Faraday’s laws of electrolysis and the relative ease of reduction/oxidation of the species present. Brine is an aqueous solution of sodium chloride — it contains $Na^+$, $Cl^-$, $H_2O$ (which gives $H^+$ and $OH^-$ in tiny amounts), and the water molecules themselves. During electrolysis, two competing reactions happen at each electrode.

At the cathode (negative electrode), reduction occurs. Two species can be reduced: $Na^+$ ions and water molecules. The standard reduction potentials tell us which is easier:

• $Na^+ + e^- \rightarrow Na$ has $E^\circ = -2.71\ \text{V}$
• $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ has $E^\circ = -0.83\ \text{V}$ (in neutral water)

The less negative (higher) potential is thermodynamically favoured. Water reduction is far easier than sodium ion reduction. So at the cathode, water is reduced to hydrogen gas and hydroxide ions:

This produces $OH^-$ ions, which immediately increase the concentration of hydroxide in the solution near the cathode — making it basic.

At the anode (positive electrode), oxidation occurs. The possible oxidations are:

• $2Cl^- \rightarrow Cl_2 + 2e^-$ has $E^\circ = +1.36\ \text{V}$
• $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ has $E^\circ = +1.23\ \text{V}$

Thermodynamically, water oxidation (to oxygen) has a lower (less positive) potential and should be easier. However, in practice, the overpotential for oxygen evolution on common electrode materials (like graphite or titanium) is high, while chlorine evolution has a low overpotential. This kinetic factor makes chlorine the dominant product at the anode in concentrated brine:

Chlorine gas bubbles off, and the $Cl^-$ ions are depleted locally. No $H^+$ is produced here (unlike water oxidation), so the anode reaction does not acidify the solution.

Now, look at the overall cell reaction. Combine the two half-reactions:

• Cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$
• Anode: $2Cl^- \rightarrow Cl_2 + 2e^-$ …

Concept: Electrolysis of Aqueous Sodium Chloride (Brine)

The relevant concept is the electrolysis of brine — an aqueous solution of NaCl. During electrolysis, both water and the dissolved ions compete at the electrodes. The key idea is that water is more easily reduced than Na⁺ ions, and water is more easily oxidised than Cl⁻ ions under certain conditions.

Method: Competitive Discharge Theory

This theory explains which ions get discharged at the electrodes based on their standard electrode potentials and concentration effects.

Steps:

1. Identify all ions present in brine

   • Cations: $\text{Na}^+$ and $\text{H}^+$ (from water)
   • Anions: $\text{Cl}^-$ and $\text{OH}^-$ (from water)

2. Determine which cation gets reduced at the cathode

   • Compare reduction potentials:
     • $\text{Na}^+ + e^- \rightarrow \text{Na}$; $E^\circ = -2.71\ \text{V}$
     • $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$; $E^\circ = -0.83\ \text{V}$
   • Water is reduced (less negative potential), so $\text{H}_2$ gas is produced at the cathode.
   • Result: $\text{OH}^-$ ions accumulate in the solution near the cathode.

3. Determine which anion gets oxidised at the anode

   • Compare oxidation tendencies:
     • $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$; $E^\circ = -1.36\ \text{V}$
     • $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$; $E^\circ = -1.23\ \text{V}$
   • Water has a less negative oxidation potential, so oxygen should form.
   • BUT — at high $\text{Cl}^-$ concentration (brine), overpotential for oxygen is high. So chlorine is actually discharged preferentially. …

Here are the common mistakes students make when analyzing the pH change during the electrolysis of brine (aqueous $NaCl$), along with how to avoid each.

Mistake 1: Forgetting that water competes with $Na^+$ and $Cl^-$ at the electrodes

Many students assume that since $NaCl$ is present, $Na$ metal will plate out at the cathode and $Cl_2$ gas will form at the anode. This leads to the wrong conclusion that the pH remains neutral.

Why it’s wrong:

In aqueous solution, water itself provides $H^+$ and $OH^-$ ions. The reduction potential of $H_2O$ (to produce $H_2$ gas) is much higher (less negative) than that of $Na^+$. So water is reduced at the cathode, not sodium.

• Cathode reaction: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ This produces $OH^-$ ions, making the solution basic.

How to avoid:

Always check the electrochemical series or standard reduction potentials. For Group 1 and 2 metal ions in water, water is reduced instead of the metal ion. Memorize: In brine electrolysis, $H_2O$ is reduced, not $Na^+$.

Mistake 2: Thinking the anode reaction is $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$

Some students apply the same logic as the cathode and assume water is oxidised at the anode, producing $H^+$ and $O_2$.

Why it’s wrong:

In concentrated $NaCl$ solution (brine), the concentration of $Cl^-$ is high. The oxidation potential of $Cl^-$ to $Cl_2$ is lower (easier) than that of water to $O_2$. So chloride ions are oxidised instead of water.

• Anode reaction: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$ This does not produce $H^+$ directly.

How to avoid:

Remember the overpotential of oxygen: in concentrated chloride solutions, $Cl^-$ oxidation is kinetically and thermodynamically favoured. The rule of thumb: At the anode, if halide ions are present in high concentration, they get oxidised, not water.

Mistake 3: Claiming the pH remains neutral because $H^+$ and $OH^-$ are produced in equal amounts

A student might think: “At the cathode, $OH^-$ is made; at the anode, $H^+$ is made (from water oxidation), so they cancel out.”

Why it’s wrong:

As explained above, the anode does not produce $H^+$ in brine electrolysis. Only the cathode produces $OH^-$. There is no compensating acid production.

• Net effect: $OH^-$ accumulates in the solution.
• Result: pH increases (becomes basic, >7).

How to avoid:

Write the overall cell reaction:

The product $NaOH$ is a strong base. So the solution becomes alkaline. Always check the products — if $NaOH$ is formed, pH must rise.

Mistake 4: Confusing “brine” with dilute $NaCl$ solution

Some students treat all $NaCl$ solutions the same. In dilute $NaCl$, the anode reaction can be water oxidation (producing $O_2$ and $H^+$), which would keep pH nearly neutral.

Why it’s wrong:

The question specifically says brine — a concentrated $NaCl$ solution. In concentrated solution, $Cl^-$ oxidation dominates.

How to avoid: …