Q.Represent the cell in which the following reaction takes place:
$Mg(s) + 2Ag^+(0.0001\ M) \rightarrow Mg^{2+}(0.130\ M) + 2Ag(s)$
Calculate its $E_{cell}$ if $E^\circ_{cell} = 3.17\ V$.

The key idea is the Nernst equation, which relates cell potential to concentration for a spontaneous redox reaction.

Step 1 – Write the cell representation

Anode (oxidation): Mg(s) | Mg²⁺(aq)

Cathode (reduction): Ag⁺(aq) | Ag(s)

The cell is:

Step 2 – Determine $n$ and the reaction quotient $Q$

Two electrons are transferred: $n = 2$.

Step 3 – Apply the Nernst equation

$E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q$ …

Cell representation

Oxidation (anode) is written on the left, reduction (cathode) on the right:

Here $Mg$ is oxidised to $Mg^{2+}$ and $Ag^+$ is reduced to $Ag$; two electrons are transferred, so $n = 2$.

Nernst equation

Step 1 — Reaction quotient. …

Method: Nernst Equation for Electrochemical Cell Representation and EMF Calculation

Step 1: Identify the Cell Representation (Cell Notation)

The reaction is:

Rules for cell notation:

• Anode (oxidation) is written on the left
• Cathode (reduction) is written on the right
• Single vertical line | represents phase boundary
• Double vertical line || represents salt bridge

Oxidation half-reaction (Anode):

Reduction half-reaction (Cathode):

Cell representation:

Step 2: Write the Nernst Equation

For the general cell reaction:

The Nernst equation at 298 K is:

Where:

• $n$ = number of electrons transferred
• $E^\circ_{cell}$ = standard cell potential

Step 3: Identify Values

• $n = 2$ (2 electrons are transferred)
• $E^\circ_{cell} = 3.17\ V$
• Reaction quotient $Q$:

Step 4: Calculate $Q$

Step 5: Apply Nernst Equation …

🧠 The Core Concept First

A cell representation is a shorthand notation of an electrochemical cell. The convention is:

• Anode (oxidation) on the left
• Cathode (reduction) on the right
• Single vertical line | for phase boundary
• Double vertical line || for salt bridge
• Concentrations written in parentheses

The Nernst equation for the cell reaction at 298 K is:

where $Q$ is the reaction quotient and $n$ is the number of electrons transferred.

✗ Common Mistake #1: Wrong order of electrodes in cell representation

What students do:

Write $Ag | Ag^+ || Mg^{2+} | Mg$ (reversing anode and cathode).

Why it’s wrong:

Oxidation happens at the anode. Here, Mg is oxidised to Mg²⁺, so Mg is the anode (left side). Ag⁺ is reduced to Ag, so Ag is the cathode (right side).

✓ How to avoid:

Always identify which species is oxidised (loses electrons) and which is reduced (gains electrons). Then write:

Correct representation:

✗ Common Mistake #2: Using wrong value of $n$ in Nernst equation

What students do:

Take $n = 1$ or $n = 3$ (guessing from coefficients).

Why it’s wrong:

$n$ is the number of electrons transferred in the balanced redox reaction. Here:

• Mg → Mg²⁺ loses 2 electrons
• 2Ag⁺ + 2e⁻ → 2Ag gains 2 electrons

So $n = 2$.

✓ How to avoid:

Write the half-reactions and balance electrons. Count electrons lost = electrons gained.

✗ Common Mistake #3: Writing $Q$ incorrectly

What students do:

Write $Q = \frac{[Mg^{2+}]}{[Ag^+]}$ or $Q = \frac{[Mg^{2+}]}{[Ag^+]^2}$ but forget to raise to stoichiometric coefficients.

Why it’s wrong:

For the reaction:

$Q = \frac{[Mg^{2+}]}{[Ag^+]^2}$ (solids are omitted).

✓ How to avoid:

Write the reaction quotient using products over reactants, each raised to its stoichiometric coefficient. Ignore solids and pure liquids.

✗ Common Mistake #4: Forgetting the minus sign or log in Nernst equation

What students do:

Use $E_{cell} = E^\circ_{cell} + \frac{0.0591}{n} \log Q$ or forget the log entirely.

Why it’s wrong:

The correct form for a cell at 298 K is:

✓ How to avoid:

Memorise the formula with the minus sign. Remember: as the reaction proceeds, $E_{cell}$ decreases from $E^\circ_{cell}$.

✓ Step-by-Step Solution (for reference) …