Q.Depict the galvanic cell in which the cell reaction is $Cu + 2Ag^+ \rightarrow 2Ag + Cu^{2+}$

The key idea is to represent the cell using the standard line notation, which separates the anode (oxidation) and cathode (reduction) half-cells with a salt bridge.

Step 1: Identify the half-reactions.

Oxidation (anode): $Cu \rightarrow Cu^{2+} + 2e^-$

Reduction (cathode): $Ag^+ + e^- \rightarrow Ag$

Step 2: Write the anode on the left and cathode on the right.

The solid electrode is placed at the extreme ends, and the ion in solution is written next to the salt bridge. …

The heart of this problem is translating a chemical reaction into a cell diagram and then writing the Nernst equation that governs its voltage. Let’s build this from the ground up.

A galvanic cell is a device that converts chemical energy into electrical energy through spontaneous redox reactions. The key is to separate the oxidation and reduction half-reactions into two compartments (half-cells), connected by a salt bridge to maintain charge balance. The cell representation is a shorthand: the anode (oxidation) is written on the left, the cathode (reduction) on the right, and a double vertical line ($||$) represents the salt bridge. A single vertical line ($|$) denotes a phase boundary (e.g., between a solid electrode and its ion solution).

For the given reaction:

We need to identify which species is oxidized and which is reduced. Copper metal ($\mathrm{Cu}$) loses electrons to become $\mathrm{Cu^{2+}}$ ions — that’s oxidation. Silver ions ($\mathrm{Ag^+}$) gain electrons to become silver metal ($\mathrm{Ag}$) — that’s reduction. So:

• Anode (oxidation): $\mathrm{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-}$
• Cathode (reduction): $\mathrm{2Ag^+(aq) + 2e^- \rightarrow 2Ag(s)}$

Now, let’s construct the cell representation step by step.

1. Write the anode half-cell. The anode is where oxidation occurs. Here, solid copper is in contact with a solution containing $\mathrm{Cu^{2+}}$ ions. The standard notation is: solid electrode | ion solution. So we write: $\mathrm{Cu(s) | Cu^{2+}(aq)}$.

2. Write the cathode half-cell. The cathode is where reduction occurs. Here, silver ions in solution deposit onto solid silver. The notation is: ion solution | solid electrode. So we write: $\mathrm{Ag^+(aq) | Ag(s)}$.

3. Connect them with the salt bridge. The salt bridge is represented by $||$. The anode is always on the left, cathode on the right. Thus, the complete cell representation is:

Often, the state symbols (s, aq) are omitted for brevity, but they are implied. So a common simplified form is: $\mathrm{Cu | Cu^{2+} || Ag^+ | Ag}$.

Now, for the Nernst equation. The Nernst equation gives the cell potential under non-standard conditions. For a general cell reaction:

The Nernst equation is:

where $Q$ is the reaction quotient, $n$ is the number of moles of electrons transferred, $R$ is the gas constant, $T$ is temperature in Kelvin, and $F$ is Faraday’s constant. At 298 K, using base-10 logarithms, this simplifies to:

For our reaction:

The reaction quotient $Q$ is: …

Method: Standard Cell Notation (IUPAC Convention)

This method converts a given redox reaction into the shorthand cell diagram that represents the galvanic cell.

Steps

1. Identify the two half-reactions

   • Oxidation (anode, left side): $\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$
   • Reduction (cathode, right side): $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ (Balance electrons: multiply reduction by 2 to match the 2 electrons lost by Cu)

2. Write the anode (oxidation) half-cell on the left

   • Solid electrode | ions in solution $\text{Cu} \mid \text{Cu}^{2+}$

3. Write the cathode (reduction) half-cell on the right

   • Ions in solution | solid electrode $\text{Ag}^+ \mid \text{Ag}$

4. Connect the two half-cells with a salt bridge

   • Represented by a double vertical line: $\parallel$

5. Combine into the full cell diagram …

Common Mistakes When Depicting This Galvanic Cell

The cell reaction is $Cu + 2Ag^+ \rightarrow 2Ag + Cu^{2+}$. Here are the errors students commonly make when writing its cell diagram.

Mistake 1: Reversing the anode and cathode order

What students do: Write $Ag \mid Ag^+ \parallel Cu^{2+} \mid Cu$ (cathode on the left, anode on the right).

Why it's wrong: By convention, the anode (oxidation) is always written on the left, and the cathode (reduction) on the right. Here, Cu loses electrons (is oxidised) to become $Cu^{2+}$, so Cu is the anode; $Ag^+$ gains electrons (is reduced) to become Ag, so Ag is the cathode.

How to avoid: Always identify which species is oxidised (loses electrons → anode, left) and which is reduced (gains electrons → cathode, right) before writing the diagram.

Correct representation: $Cu(s) \mid Cu^{2+}(aq) \parallel Ag^+(aq) \mid Ag(s)$

Mistake 2: Carrying the stoichiometric coefficient into the cell diagram

What students do: Write $Cu \mid Cu^{2+} \parallel 2Ag^+ \mid Ag$, carrying the "2" from the balanced equation into the diagram.

Why it's wrong: The cell diagram shows only the species present in each half-cell, not the stoichiometric coefficients from the balanced equation — those matter only when writing the Nernst equation's reaction quotient $Q$.

How to avoid: Write the diagram using bare species: $Cu \mid Cu^{2+} \parallel Ag^+ \mid Ag$. Save the coefficient for $Q = \frac{[Cu^{2+}]}{[Ag^+]^2}$ in the Nernst equation.

Mistake 3: Using a single line instead of a double line for the salt bridge …