Q.In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?

Concept: Electrolysis of aqueous NaCl — at the anode, oxidation occurs. Standard electrode potentials alone favour water oxidation, but in practice, overpotential reverses this.

Reasoning:

1. At the anode, we compare the oxidation of water ($E^\circ = 1.23\ \text{V}$) and oxidation of chloride ions ($E^\circ = 1.36\ \text{V}$).
2. Purely by standard potentials, water (lower potential) should oxidise first. …

1. Identify what happens at the anode.

   The anode is the electrode where oxidation takes place — loss of electrons. So we need to look at the half‑reactions that are written as oxidations (electrons on the right). Options (i) and (iii) are reductions (electrons on the left), so they cannot occur at the anode. That leaves (ii) and (iv).

2. Understand the competition.

   In aqueous NaCl, the solution contains Na+, Cl-, H+, OH-, and water molecules. At the anode, two species can be oxidised:

   • Water: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$
   • Chloride ions: $Cl^-(aq) \rightarrow \frac12 Cl_2(g) + e^-$

3. Compare the standard potentials.

   • Water oxidation: $E^\circ = +1.23\ \text{V}$
   • Chloride oxidation: $E^\circ = +1.36\ \text{V}$

   Purely thermodynamically (standard conditions, no kinetic barriers), the reaction needing the lower potential is favoured — so water oxidation looks like it should win.

4. But real electrodes have overpotential — the deciding factor. …

Method: Standard Electrode Potential Comparison + Overpotential for Anode Reactions

In electrolysis, the anode is where oxidation occurs (loss of electrons). Comparing standard oxidation potentials alone is not sufficient — the reaction that actually occurs also depends on overpotential (a kinetic factor).

Steps

1. Identify the species present at the anode

   In aqueous NaCl:

   • $Cl^-$ (from NaCl)
   • $H_2O$ (solvent)

2. Write possible oxidation half-reactions

   • (ii) $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$; $E^\circ = +1.23\ V$
   • (iv) $Cl^-(aq) \rightarrow \frac{1}{2}Cl_2(g) + e^-$; $E^\circ = +1.36\ V$

   (Options A and C are reduction reactions — they occur at the cathode, not anode.)

3. Compare standard oxidation potentials

   Water oxidation ($+1.23\ V$) has the lower standard potential, so purely thermodynamically it should be favoured.

4. Apply the overpotential correction …

Common Mistakes Students Make on This Question

Mistake 1: Confusing Anode and Cathode Reactions

• The error: Students often pick reduction reactions (like A or C) for the anode, forgetting that oxidation always occurs at the anode.
• How to avoid: Memorise the mnemonic "An Ox – Red Cat" (Anode = Oxidation, Cathode = Reduction). Before answering, ask yourself: "Is this a loss of electrons?" Only reactions where electrons appear on the right side (products) are oxidation.

Mistake 2: Not Checking Which Species Actually Gets Oxidised

• The error: Students see option D ($Cl^- \rightarrow Cl_2$) and assume it's correct without comparing the overpotential and concentration effects in aqueous NaCl.
• How to avoid: In aqueous solutions, water itself can oxidise (option B, $E^\circ = 1.23\ V$). Even though $Cl^-$ oxidation has $E^\circ = 1.36\ V$, the overpotential for oxygen evolution is high (~0.4–0.6 V) on common electrodes like platinum. This means the actual voltage needed for water oxidation is higher than 1.23 V, making chloride oxidation kinetically favoured in concentrated NaCl. Always consider:
  • Standard potentials
  • Overpotential (kinetic barrier)
  • Concentration (Le Chatelier's principle)

Mistake 3: Misreading the Sign Convention for $E^\circ$

• The error: Students see $E^\circ_{cell}$ labelled on each half-reaction and treat them as if they are reduction potentials. But option B and D are written as oxidation half-reactions — the given $E^\circ$ values are actually the oxidation potentials (reverse of standard reduction potentials).
• How to avoid: Always convert to a standard reduction potential table in your mind. For oxidation reactions, the true tendency to occur is opposite to the sign shown. For example:
  • $Cl^- \rightarrow \frac{1}{2}Cl_2 + e^-$ has $E^\circ_{ox} = 1.36\ V$ → reduction potential for $Cl_2 + e^- \rightarrow Cl^-$ is +1.36 V.
  • $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ has $E^\circ_{ox} = 1.23\ V$ → reduction potential for $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$ is +1.23 V.
  • Higher reduction potential = easier to reduce = harder to oxidise. So water ($E^\circ_{red} = 1.23\ V$) is harder to oxidise than chloride ($E^\circ_{red} = 1.36\ V$) based on thermodynamics alone — but overpotential reverses this.

Mistake 4: Ignoring the "Aqueous" Condition

• The error: Students pick option A ($Na^+ \rightarrow Na$) because they think of molten NaCl electrolysis. In aqueous solution, water is present and $Na^+$ reduction ($E^\circ = -2.71\ V$) is not possible because water reduction to $H_2$ ($E^\circ = -0.83\ V$ at pH 7) occurs first.
• How to avoid: For aqueous solutions, always check if water can react more easily than the ion. Use the electrochemical series and remember:
  • Cations of highly active metals (Group 1, 2) are not reduced in water — $H_2O$ gets reduced instead.
  • Anions like $Cl^-$ may or may not be oxidised depending on concentration and electrode material. …