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.

Question

Accepted Answer

> [!TLDR]
> The standard Gibbs energy is calculated directly from $E^\circ_{cell}$ using $\Delta_r G^\circ = -nFE^\circ_{cell}$, and the equilibrium constant follows from $\Delta_r G^\circ = -RT\ln K$. For this reaction, $n=2$, giving $\Delta_r G^\circ = -45.5\ 	ext{kJ mol}^{-1}$ and $K \approx 9.6 	imes 10^7$.

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.

> [!WARNING]
> The most common mistake here is getting $n$ wrong. Students often count the total electrons transferred in the balanced equation without checking the half-reactions carefully. For this reaction, $n=2$, not 1 and not 6 — verify it yourself below.

1. **Find $n$, the number of moles of electrons transferred.**  
   Write the two half-reactions:
   - Reduction: $Fe^{3+} + e^- ightarrow Fe^{2+}$  
   - Oxidation: $2I^- ightarrow I_2 + 2e^-$

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:  
   $2Fe^{3+} + 2e^- ightarrow 2Fe^{2+}$

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 $\Delta_r G^\circ = -nFE^\circ_{cell}$.  
   - $n = 2$  
   - $F = 96485\ 	ext{C mol}^{-1}$ (Faraday constant)  
   - $E^\circ_{cell} = 0.236\ 	e

Concept	Physical Meaning
$E^\circ_{\text{cell}}$	Voltage when all species are at 1 M (standard state)
$-\frac{RT}{nF} \ln Q$	Correction factor — adjusts voltage for real concentrations
$Q$	Tells you how far the reaction is from equilibrium

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.

Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

The Intuition First

The Precise Statement

Cell Representation: How We Write It

Applying the Nernst Equation to a Cell Representation

A Worked Example

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

2. The Chemical Side: $\Delta G$ Depends on Concentration

3. Combining Both Sides: The Derivation

4. The "Why" in Plain Terms

5. The Common Form (log base 10)

Method: Nernst Equation & Thermodynamic Relations

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

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

Step 3: Calculate equilibrium constant ( $K$ )

Common Mistakes & How to Avoid Them

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

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

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

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

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

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