Q.Consider a cell given below:
$Cu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, Pt$
Write the reactions that occur at anode and cathode.

The key idea is that the cell notation convention directly tells you which electrode is the anode and which is the cathode: by convention, the anode (oxidation) is written on the left and the cathode (reduction) is written on the right.

Step 1: Identify the half-reactions.

Copper metal ($Cu$) is in contact with $Cu^{2+}$ ions — it can oxidise to $Cu^{2+}$ by losing electrons. On the other side, $Cl_2$ gas on platinum is in contact with $Cl^-$ ions — $Cl_2$ can reduce to $Cl^-$ by gaining electrons.

Step 2: Determine anode and cathode. …

Let's understand what this cell notation actually means before jumping into the reactions.

The Language of Cell Notation

The notation $Cu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, Pt$ follows a standard convention. The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double vertical line $\parallel$ represents the salt bridge that connects the two half-cells.

By convention, the anode (where oxidation happens) is written on the left, and the cathode (where reduction happens) is written on the right. This is a critical rule to remember.

Step-by-Step Breakdown

1. Identify the two half-cells

The left half-cell is $Cu \mid Cu^{2+}$. This means a copper metal electrode is in contact with a solution containing Cu²⁺ ions.

The right half-cell is $Cl^- \mid Cl_2, Pt$. Here, a platinum electrode (inert, written last) is in contact with a solution containing Cl⁻ ions and chlorine gas (Cl₂). Platinum is used because it doesn't participate chemically — it just conducts electrons.

2. Determine the reaction at the anode (left side)

At the anode, oxidation occurs — the species loses electrons. Looking at the left half-cell, copper metal (Cu) can lose two electrons to become Cu²⁺ ions:

This is oxidation because the oxidation state of copper increases from 0 to +2.

3. Determine the reaction at the cathode (right side)

At the cathode, reduction occurs — the species gains electrons. Looking at the right half-cell, chlorine gas (Cl₂) can gain two electrons to become two chloride ions (Cl⁻):

This is reduction because the oxidation state of chlorine decreases from 0 to -1.

4. Verify the overall cell reaction (optional but helpful) …

Method: Electrode Identification & Half-Reaction Writing

This method uses the cell diagram convention to identify which electrode is anode (oxidation) and which is cathode (reduction), then writes the balanced half-reactions.

Steps

Step 1: Identify the electrodes from the cell diagram

The cell diagram is:

$Cu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, Pt$

• Left side (before $\parallel$): Anode (oxidation occurs here)
• Right side (after $\parallel$): Cathode (reduction occurs here)

So:

• Anode: $Cu \mid Cu^{2+}$
• Cathode: $Cl^- \mid Cl_2, Pt$

Step 2: Write the oxidation half-reaction (at anode)

At the anode, the solid copper metal loses electrons to form copper ions:

$Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$

Step 3: Write the reduction half-reaction (at cathode)

At the cathode, chlorine gas is produced from chloride ions gaining electrons:

$Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq)$

Step 4: Verify electron balance …

Here are the most common mistakes students make with this specific electrochemical cell setup, along with how to avoid each.

Mistake 1: Misidentifying the Anode and Cathode

The Error:

Students often assume the left side is always the anode and the right side is always the cathode. In this cell, they might write the oxidation of $Cu$ at the left electrode (which is correct) but then incorrectly write the reduction of $Cl_2$ at the right electrode (which is also correct, but for the wrong reason).

Why it happens:

They memorize "anode on left, cathode on right" without understanding the underlying chemistry. The cell notation $Cu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, Pt$ tells us the anode is on the left and the cathode is on the right only if the cell is spontaneous. Here, it is.

How to avoid it:

Always determine the direction of electron flow based on the standard reduction potentials ($E^\circ$).

• Step 1: Write the two half-reactions.

  • Left half-cell: $Cu^{2+} + 2e^- \rightarrow Cu$ (Reduction potential $E^\circ = +0.34$ V)
  • Right half-cell: $Cl_2 + 2e^- \rightarrow 2Cl^-$ (Reduction potential $E^\circ = +1.36$ V)

• Step 2: The half-cell with the higher reduction potential (more positive) will undergo reduction (gain electrons). Here, $Cl_2/Cl^-$ has $+1.36$ V > $+0.34$ V, so $Cl_2$ is reduced at the cathode.

• Step 3: The other half-cell (with lower $E^\circ$) will undergo oxidation (lose electrons). Here, $Cu$ is oxidized at the anode.

Result: Anode = Left (Cu electrode), Cathode = Right (Pt electrode).

Mistake 2: Writing the Wrong Half-Reaction at the Anode

The Error:

Students write the reduction of $Cu^{2+}$ at the anode (e.g., $Cu^{2+} + 2e^- \rightarrow Cu$) instead of the oxidation of $Cu$.

Why it happens:

They confuse the species present. The anode is where oxidation occurs (loss of electrons). The cell notation shows $Cu$ (solid) in contact with $Cu^{2+}$ (aqueous). The only species that can be oxidized is the solid $Cu$ metal.

How to avoid it:

Remember the mnemonic: "An Ox, Red Cat" (Anode = Oxidation, Cathode = Reduction).

• At the anode, look for a species that can lose electrons (increase in oxidation state).

  • $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$ (Oxidation: $Cu$ goes from 0 to +2)

• At the cathode, look for a species that can gain electrons (decrease in oxidation state).

  • $Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq)$ (Reduction: $Cl$ goes from 0 to -1)

Correct Anode Reaction:

Mistake 3: Forgetting the Inert Electrode (Pt) in the Cathode Reaction

The Error:

Students write the cathode reaction as $Pt + Cl_2 \rightarrow ...$ or simply $Cl_2 + 2e^- \rightarrow 2Cl^-$ but then forget to mention that Pt is just an inert conductor.

Why it happens:

They see $Pt$ in the cell notation and think it participates chemically. In reality, $Pt$ is inert (does not react). It only provides a surface for the $Cl_2$ gas to interact with the $Cl^-$ solution.

How to avoid it: …