1. Given reaction

$2Fe^{3+}(aq) + 2I^-(aq) \to 2Fe^{2+}(aq) + I_2(s)$

We use

$\Delta_r G_m^\circ = -nFE_\text{cell}^\circ$

So first we need $E_\text{cell}^\circ$.

2. Find $E^\circ$ for the couple $Fe^{3+}/Fe^{2+}$

Given:

$E^\circ_{Fe^{2+}/Fe(s)} = -0.440\,V$ $E^\circ_{Fe^{3+}/Fe(s)} = -0.036\,V$

Write the corresponding Gibbs energy relations:

For $Fe^{2+} + 2e^- \to Fe(s)$

$\Delta G_1^\circ = -2F(-0.440)=0.880F$

For $Fe^{3+} + 3e^- \to Fe(s)$

$\Delta G_2^\circ = -3F(-0.036)=0.108F$

Now,

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

is obtained by:

$[Fe^{3+}+3e^-\to Fe] + [Fe\to Fe^{2+}+2e^-]$

Hence

$\Delta G^\circ = 0.108F - 0.880F = -0.772F$

Since this involves 1 electron,

$\Delta G^\circ = -FE^\circ$

So,

$-FE^\circ_{Fe^{3+}/Fe^{2+}} = -0.772F$

$E^\circ_{Fe^{3+}/Fe^{2+}} = 0.772\,V$

3. Identify cathode and anode

Reaction given:

• $Fe^{3+} \to Fe^{2+}$ is reduction
• $I^- \to I_2$ is oxidation

Given reduction potential:

$I_2 + 2e^- \to 2I^- ,\quad E^\circ = 0.539\,V$

Thus,

• Cathode: $Fe^{3+}/Fe^{2+}$, $E^\circ = 0.772\,V$
• Anode: $I_2/2I^-$, $E^\circ = 0.539\,V$

Therefore,

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

$E_\text{cell}^\circ = 0.772 - 0.539 = 0.233\,V$

4. Calculate $\Delta_r G_m^\circ$

In the balanced reaction, number of electrons transferred:

$n=2$

So,

$\Delta_r G_m^\circ = -nFE_\text{cell}^\circ$

$= -2 \times 96500 \times 0.233$

$= -44969\,J\,mol^{-1}$

$= -44.969\,kJ\,mol^{-1}$

Rounded to nearest integer:

$\Delta_r G_m^\circ \approx -45\,kJ\,mol^{-1}$

Since the question asks for the magnitude and shows

$\Delta_r G_m^\circ = -\underline{\hspace{1cm}}\text{ kJ}$

the required integer is:

$\boxed{45}$

5. Comparison with stored answer

Stored correct answer: $46$

My derived value is $45$ because:

$2 \times 96500 \times 0.233 = 44969\,J \approx 44.969\,kJ$

which rounds to $45$, not $46$.

So I disagree with the stored answer.