Question

What is the value of $E^\circ(\text{Fe}^{3+}/\text{Fe}^0)$?
(The standard reduction potential values are $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = 0.77\text{ V}$, and $E^\circ(\text{Fe}^{2+}/\text{Fe}^0) = -0.44\text{ V}$)}

• A. $-0.04\text{ V}$
• B. $0.33\text{ V}$
• C. $0.11\text{ V}$
• D. $-0.11\text{ V}$

Hint:

Never add $E^\circ$ values directly!
Always use the formula:
\[ E^\circ_{\text{net}} = \frac{n_1 E^\circ_1 + n_2 E^\circ_2}{n_1 + n_2} \]
Plugging in:
\[ E^\circ_{\text{net}} = \frac{1(0.77) + 2(-0.44)}{3} = \frac{0.77 - 0.88}{3} = -0.037\text{ V} \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Standard reduction potentials (\( \text{E}^\circ \)) are intensive properties and cannot be added directly when combining half-reactions. Instead, we convert them into their corresponding standard Gibbs free energy changes (\( \Delta \text{G}^\circ \)), which are extensive and additive.
Step 2: Key Formula or Approach:
The relationship between Gibbs free energy and cell potential is:
\[ \Delta \text{G}^\circ = -n\text{FE}^\circ \]
where $n$ is the number of electrons transferred in the half-reaction, and $\text{F}$ is Faraday's constant.
Step 3: Detailed Explanation:
Let us list the given half-reactions along with their electron counts ($n$) and potentials:
1) \( \text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+} \) \quad (\( n_1 = 1, \text{ E}_1^\circ = 0.77 \text{ V} \))
\[ \Delta \text{G}_1^\circ = -1 \cdot \text{F} \cdot (0.77) = -0.77\text{F} \]
2) \( \text{Fe}^{2+} + 2\text{e}^- \rightarrow \text{Fe}^0 \) \quad (\( n_2 = 2, \text{ E}_2^\circ = -0.44 \text{ V} \))
\[ \Delta \text{G}_2^\circ = -2 \cdot \text{F} \cdot (-0.44) = 0.88\text{F} \]
We want to find the potential for the target half-reaction:
3) \( \text{Fe}^{3+} + 3\text{e}^- \rightarrow \text{Fe}^0 \) \quad (\( n_3 = 3, \text{ E}_3^\circ = ? \))
\[ \Delta \text{G}_3^\circ = -3 \cdot \text{F} \cdot \text{E}_3^\circ \]
Adding equation (1) and equation (2) gives the target equation (3):
\[ \Delta \text{G}_3^\circ = \Delta \text{G}_1^\circ + \Delta \text{G}_2^\circ \]
Substitute the values:
\[ -3\text{FE}_3^\circ = -0.77\text{F} + 0.88\text{F} \]
Divide both sides by $-\text{F}$:
\[ 3\text{E}_3^\circ = 0.77 - 0.88 \]
\[ 3\text{E}_3^\circ = -0.11 \]
\[ \text{E}_3^\circ = \frac{-0.11}{3} \approx -0.0366 \text{ V} \]
Rounding to two decimal places gives \( -0.04 \text{ V} \).
Step 4: Final Answer:
The value of \( \text{E}^\circ(\text{Fe}^{3+}/\text{Fe}^0) \) is \( -0.04 \text{ V} \).