For the reaction, 2Fe3+(aq) + 2I-(aq) - 2Fe2+(aq) + I2(s). The magnitude of the standard molar Gibb

Q: Why can't we just add or subtract the electrode potentials directly?

Electrode potentials are intensive properties, meaning they don't depend on the amount of substance. You can't just add them algebraically. However, Gibbs free energy is an extensive property, so we convert potentials to $\Delta G^\circ$, add them, and then convert back!

Q: How do we know $n=2$ for the main reaction?

Look at the balanced equation: $2\text{Fe}^{3+}$ reduces to $2\text{Fe}^{2+}$, which requires 2 electrons. Simultaneously, $2\text{I}^-$ oxidizes to $\text{I}_2$, releasing 2 electrons. Thus, the total number of electrons transferred is 2.

Q: Why is the formula for $E^\circ_{\text{cell}}$ written as Cathode - Anode?

By convention, standard potentials are always given as reduction potentials. The cell potential is the reduction potential of the substance actually being reduced (cathode) minus the reduction potential of the substance being oxidized (anode).

Q: What does the negative sign in $\Delta G^\circ$ indicate?

A negative $\Delta G^\circ$ indicates that the reaction is thermodynamically spontaneous in the forward direction under standard conditions.

Q: Why did we take the magnitude at the end?

The question specifically asked for the 'magnitude' of the standard molar Gibbs free energy change. Magnitude refers to the absolute value, so we drop the negative sign.

Question

For the reaction,
$2	ext{Fe}^{3+}(aq) + 2	ext{I}^-(aq) \longrightarrow 2	ext{Fe}^{2+}(aq) + 	ext{I}_2(s)$
The magnitude of the standard molar Gibbs free energy change, $\Delta_r G^\circ_m = \dots\dots\dots$ kJ (Round off to the nearest integer).
$\left[ \begin{array}{l} E^\circ_{	ext{Fe}^{2+}/	ext{Fe}(s)} = -0.440 	ext{ V}; \ E^\circ_{	ext{I}_2/2	ext{I}^-} = 0.539 	ext{ V}; \ E^\circ_{	ext{Fe}^{3+}/	ext{Fe}(s)} = -0.036 	ext{ V} \ F = 96500 	ext{ C} \end{array} ight]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are setting up an electrochemical cell in the lab. On one side, you have Iron(III) ions ready to grab electrons, and on the other side, Iodide ions ready to give them up.

The overall reaction is given by:
$2	ext{Fe}^{3+}(aq) + 2	ext{I}^-(aq) \longrightarrow 2	ext{Fe}^{2+}(aq) + 	ext{I}_2(s)$

Our ultimate goal is to find the magnitude of the standard Gibbs free energy change, $\Delta_r G^\circ$, for this exact reaction.

To do this, we need our master thermodynamic equation:
$\Delta_r G^\circ = -n F E^\circ_{	ext{cell}}$

Here, $n$ is the number of electrons transferred, $F$ is Faraday's constant, and $E^\circ_{	ext{cell}}$ is the standard cell potential.

So, our first mission is clear: we must find $E^\circ_{	ext{cell}}$.

### The Missing Piece: Latimer Diagram

The standard cell potential is the difference between the reduction potentials of the cathode and the anode:
$E^\circ_{	ext{cell}} = E^\circ_{	ext{Fe}^{3+}/	ext{Fe}^{2+}} - E^\circ_{	ext{I}_2/	ext{I}^-}$

We are given the potential for the Iodine couple ($0.539 	ext{ V}$), but there is a catch! We don't have the standard reduction potential for the $	ext{Fe}^{3+}$ to $	ext{Fe}^{2+}$ transition.

Instead, we are given the potentials for $	ext{Fe}^{3+}$ to solid $	ext{Fe}$, and $	ext{Fe}^{2+}$ to solid $	ext{Fe}$.

**Never add or subtract electrode potentials directly!** They are intensive properties.

However, Gibbs free energy is an extensive property. We can use a thermodynamic cycle, often visualized as a Latimer diagram, to find our missing potential.

The direct jump from $	ext{Fe}^{3+}$ to $	ext{Fe}$ is thermodynamically equivalent to going from $	ext{Fe}^{3+}$ to $	ext{Fe}^{2+}$, and then from $	ext{Fe}^{2+}$ to $	ext{Fe}$.
$\Delta G^\circ_{	ext{Fe}^{3+} 	o 	ext{Fe}} = \Delta G^\circ_{	ext{Fe}^{3+} 	o 	ext{Fe}^{2+}} + \Delta G^\circ_{	ext{Fe}^{2+} 	o 	ext{Fe}}$

Substituting the formula $\Delta G^\circ = -nFE^\circ$ into our cycle:
$-3 F E^\circ_{	ext{Fe}^{3+}/	ext{Fe}} = -1 F E^\circ_{	ext{Fe}^{3+}/	ext{Fe}^{2+}} - 2 F E^\circ_{	ext{Fe}^{2+}/	ext{Fe}}$

Notice how Faraday's constant, $F$, and the negative signs cancel out beautifully.

Now, we plug in the given potential values:
$3(-0.036) = 1(E^\circ_{	ext{Fe}^{3+}/	ext{Fe}^{2+}}) + 2(-0.440)$

Rearranging the terms to solve for our unknown:
$E^\circ_{	ext{Fe}^{3+}/	ext{Fe}^{2+}} = -0.108 + 0.880 = 0.772 	ext{ V}$

Awesome! We found our missing piece.

### Calculating the Cell Potential

Now we can finally calculate the standard cell potential for our main reaction.

We simply subtract the anode's potential from the cathode's potential:
$E^\circ_{	ext{cell}} = 0.772 - 0.539 = 0.233 	ext{ V}$

A positive cell potential means our reaction is spontaneous in the forward direction.

### The Final Thermodynamic Calculation

Let's bring back our master equation.

Looking at the balanced reaction, $2	ext{Fe}^{3+}$ reduces to $2	ext{Fe}^{2+}$, which requires 2 electrons. Thus, $n = 2$.
$\Delta_r G^\circ = -2 	imes 96500 	imes 0.233 	ext{ J}$

Multiplying these values out gives:
$\Delta_r G^\circ = -44969 	ext{ J}$

Converting this to kilojoules, we get $-44.969 	ext{ kJ}$.

Rounding to the nearest integer, it becomes $-45 	ext{ kJ}$.

The question specifically asks for the **magnitude** of the standard molar Gibbs free energy change. Therefore, we drop the negative sign.

The final answer is $45$. What a beautiful application of thermodynamics!

Enter Numerical Value:

Visualized Solution

$2 Fe^{3 +} + 2 I^{-} ⟶ 2 Fe^{2 +} + I_{2}$

$Δ_{r} G^{\circ} = - n F E_{cell}^{\circ}$

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$

$Δ G_{Fe^{3 +} \to Fe}^{\circ} = Δ G_{Fe^{3 +} \to Fe^{2 +}}^{\circ} + Δ G_{Fe^{2 +} \to Fe}^{\circ}$

$- 3 F E_{Fe^{3 +} / Fe}^{\circ} = - 1 F E_{Fe^{3 +} / Fe^{2 +}}^{\circ} - 2 F E_{Fe^{2 +} / Fe}^{\circ}$

$3 (- 0.036) = 1 (E_{Fe^{3 +} / Fe^{2 +}}^{\circ}) + 2 (- 0.440)$

$E_{Fe^{3 +} / Fe^{2 +}}^{\circ} = 0.772 V$

$E_{cell}^{\circ} = 0.233 V$

$Δ_{r} G^{\circ} = - 2 \times 96500 \times 0.233 J$

$Magnitude = 45$

$What if the reaction was reversed?$

The Sigma Insight: Electrochemical Cells

Analyzing the Setup

The Missing Piece

Calculating the Cell Potential

The Final Thermodynamic Calculation

Similar Questions

The Gibbs energy change (in J) for the given reaction at $[C u^{2 +}] = [S n^{2 +}] = 1 M$ and $298 K$ is $C u (s) + S n^{2 +} (a q) ⟶ C u^{2 +} (a q) + S n (s)$ ; $(E_{S n^{2 +} / S n}^{\circ} = - 0.16 V, E_{C u^{2 +} / C u}^{\circ} = 0.34 V$ , Take, $F = 96500 C mol^{- 1})$

The standard Gibbs energy for the given cell reaction in $kJ mol^{- 1}$ at $298 K$ is $Zn (s) + Cu^{2 +} (a q) ⟶ Zn^{2 +} (a q) + Cu (s), E^{\circ} = 2 V$ at $298 K$ (Faraday's constant, $F = 96000 C mol^{- 1}$ )

The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $2 5^{\circ} C$ . The equilibrium constant of the reaction is ( $F = 96500 C mol^{- 1}$ )

Consider the following $E^{\circ}$ values :\n $E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V$ \n $E_{S n^{2 +} / S n}^{\circ} = - 0.14 V$ \nUnder standard conditions, the potential for the reaction\n $S n (s) + 2 F e^{3 +} (a q) \to 2 F e^{2 +} (a q) + S n^{2 +} (a q)$ is

Given the equilibrium constant ( $K_{c}$ ) of the reaction : $C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s)$ is $10 \times 1 0^{15}$ , calculate the $E_{ce l l}^{\circ}$ of this reaction at 298 K. $[2.303 \frac{R T}{F} at 298 K = 0.059 V]$

For the given cell; $C u (s) ∣ C u^{2 +} (C_{1} M) ∣∣ C u^{2 +} (C_{2} M) ∣ C u (s)$ change in Gibbs energy ( $Δ G$ ) is negative, if

Given, $E_{F e^{3 +} / F e}^{\circ} = - 0.036 V$ , $E_{F e^{2 +} / F e}^{\circ} = - 0.439 V$ . The value of standard electrode potential for the charge, $F e^{3 +} (a q) + e^{-} ⟶ F e^{2 +} (a q)$ will be

For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 V$ at $2 5^{\circ} C$ . The equilibrium constant of the reaction at $2 5^{\circ} C$ will be

Given, $E_{C r^{3 +} / C r}^{\circ} = - 0.72 V$ , $E_{F e^{2 +} / F e}^{\circ} = - 0.42 V$ . The potential for the cell $C r ∣ C r^{3 +} (0.1 M) ∣∣ F e^{2 +} (0.01 M) ∣ F e$ is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2Fe3++2I−⟶2Fe2++I2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δr​G∘=−nFEcell∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ecell∘​=Ecathode∘​−Eanode∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔGFe3+→Fe∘​=ΔGFe3+→Fe2+∘​+ΔGFe2+→Fe∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −3FEFe3+/Fe∘​=−1FEFe3+/Fe2+∘​−2FEFe2+/Fe∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3(−0.036)=1(EFe3+/Fe2+∘​)+2(−0.440)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EFe3+/Fe2+∘​=0.772 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ecell∘​=0.233 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δr​G∘=−2×96500×0.233 J

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnitude=45

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if the reaction was reversed?