Q.The cell in which the following reaction occurs:
$2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(s)$
has $E^\circ_{cell} = 0.236\ V$ at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

The key idea is the Nernst equation in its standard form: the standard cell potential directly gives the standard Gibbs energy change, and from that we obtain the equilibrium constant.

Step 1: Relate $\Delta_r G^\circ$ to $E^\circ_{cell}$

The standard Gibbs energy is given by $\Delta_r G^\circ = -n F E^\circ_{cell}$, where $n$ is the number of moles of electrons transferred and $F = 96485\ \text{C mol}^{-1}$.

Step 2: Determine $n$ from the half-reactions

$Fe^{3+} + e^- \rightarrow Fe^{2+}$ (each Fe gains 1 electron). Two $Fe^{3+}$ ions are reduced, so $n = 2$.

Step 3: Calculate $\Delta_r G^\circ$

$\Delta_r G^\circ = -2 \times 96485 \times 0.236$

$\Delta_r G^\circ = -45540.92\ \text{J mol}^{-1} \approx -45.54\ \text{kJ mol}^{-1}$.

Step 4: Calculate $K$ from $\Delta_r G^\circ$ …

The heart of this problem is the link between electrochemistry and thermodynamics. A cell potential measures the driving force for electron transfer — the larger the $E^\circ_{cell}$, the more spontaneous the reaction. That spontaneity is exactly what Gibbs energy quantifies, and the equilibrium constant tells us how far the reaction goes before stopping.

The bridge is simple but powerful: $\Delta_r G^\circ = -nFE^\circ_{cell}$. Once you have $\Delta_r G^\circ$, the equilibrium constant comes from $\Delta_r G^\circ = -RT\ln K$. So the whole problem reduces to identifying $n$ correctly and plugging in numbers.

1. Find $n$, the number of moles of electrons transferred.

   Write the two half-reactions:

   • Reduction: $Fe^{3+} + e^- \rightarrow Fe^{2+}$
   • Oxidation: $2I^- \rightarrow I_2 + 2e^-$

   The oxidation half shows 2 electrons are released per $I_2$ molecule formed. To balance electrons, the reduction half must consume exactly 2 electrons — so we multiply the iron half-reaction by 2:

   $2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}$

   Now the full reaction matches the given equation, and the electrons cancel perfectly. The number of electrons transferred per reaction as written is $n = 2$.

2. Calculate $\Delta_r G^\circ$.

   Use $\Delta_r G^\circ = -nFE^\circ_{cell}$.

   • $n = 2$
   • $F = 96485\ \text{C mol}^{-1}$ (Faraday constant)
   • $E^\circ_{cell} = 0.236\ \text{V}$

Compute stepwise:

$2 \times 96485 = 192970$

$192970 \times 0.236 = 45540.92$

So $\Delta_r G^\circ = -45540.92\ \text{J mol}^{-1} \approx -45.5\ \text{kJ mol}^{-1}$.

The negative sign confirms the reaction is spontaneous under standard conditions.

3. Calculate the equilibrium constant $K$. …

Method: Nernst Equation & Thermodynamic Relations

This problem uses the direct thermodynamic link between cell potential, Gibbs energy, and equilibrium constant.

Step 1: Identify the number of electrons transferred ($n$)

The reaction is:

• Each $Fe^{3+}$ gains 1 electron to become $Fe^{2+}$.
• Two $Fe^{3+}$ ions are involved, so total electrons transferred:

Step 2: Calculate standard Gibbs energy ($\Delta_r G^\circ$)

Use the formula:

Where:

• $n = 2$
• $F = 96485\ \text{C mol}^{-1}$ (Faraday constant)
• $E^\circ_{cell} = 0.236\ \text{V}$

Why negative? A positive $E^\circ_{cell}$ means the reaction is spontaneous, so $\Delta_r G^\circ$ must be negative.

Step 3: Calculate equilibrium constant ($K$)

Use the relation:

Rearranging:

At 298 K:

• $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$
• $T = 298\ \text{K}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Wrong value of $n$ (number of electrons transferred)

The error: Students often take $n = 2$ because the reaction has 2 Fe³⁺ and 2 I⁻, but they miss the electron count per half-reaction.

Why it happens: They see coefficients and assume $n$ equals the coefficient of the species.

How to avoid:

Write the balanced half-reactions:

• Reduction: $2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}$
• Oxidation: $2I^- \rightarrow I_2 + 2e^-$

Each Fe³⁺ gains 1 electron, so 2 Fe³⁺ gain 2 electrons total.

Thus, $n = 2$ — not 1, not 6.

✓ Key rule: $n$ = number of electrons transferred as written in the balanced overall equation.

Mistake 2: Using $E^\circ_{cell}$ in the wrong units or sign

The error: Plugging $E^\circ_{cell} = 0.236\ V$ into $\Delta G^\circ = -nFE^\circ_{cell}$ but forgetting that $F$ is in C/mol and $\Delta G^\circ$ comes out in J/mol, not kJ/mol.

How to avoid:

Always convert to kJ at the end:

This gives -45,540 J/mol = -45.54 kJ/mol.

✓ Pro tip: Write units at every step — it catches unit errors.

Mistake 3: Confusing $\Delta G^\circ$ sign with spontaneity

The error: Students see $E^\circ_{cell} > 0$ and still write $\Delta G^\circ$ as positive.

Why it happens: They memorise "positive $E^\circ$ means spontaneous" but forget the negative sign in $\Delta G^\circ = -nFE^\circ_{cell}$.

How to avoid:

Remember the golden triangle:

✓ Check: $0.236\ V > 0$ → $\Delta G^\circ$ must be negative.

Mistake 4: Using wrong $R$ value in $K$ calculation

The error: Using $R = 0.0821\ \text{L·atm/mol·K}$ (gas constant for PV = nRT) instead of $R = 8.314\ \text{J/mol·K}$.

Why it happens: Students mix up the two common values of $R$.

How to avoid:

For thermodynamic equations involving $\Delta G^\circ = -RT\ln K$, always use:

✓ Memory aid: Joules → use 8.314; Litre-atm → use 0.0821.

Mistake 5: Forgetting to take $\ln$ vs $\log$ correctly

The error: Using $\log_{10}$ when the formula requires $\ln$, or vice versa.

How to avoid:

Stick to the standard formula:

If you prefer $\log_{10}$, convert:

So:

✓ Exam tip: Most NCERT problems use $\log_{10}$ — just remember the 2.303 factor.

--- …