© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Known Standard Electrode Potentials
We are given the following standard electrode potentials:
$E^\circ_{\text{Fe}^{3+}/\text{Fe}} = -0.036\,\text{V}$, representing the half-reaction:
$$\text{Fe}^{3+} + 3\,e^- \rightarrow \text{Fe}$$
$E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.439\,\text{V}$, representing the half-reaction:
$$\text{Fe}^{2+} + 2\,e^- \rightarrow \text{Fe}$$
Step 2: Determine the Target Reaction
We want the standard electrode potential for:
$$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}.$$
Step 3: Relate Free Energy Change to Electrode Potentials
Recall that the change in Gibbs free energy for a reaction is given by:
$$\Delta G = -n\,F\,E^\circ,$$
where
$n$ is the number of electrons transferred and
$F$ is the Faraday constant.
Step 4: Formulate the Equations in Terms of $\Delta G$
Define the Gibbs free energy changes for the given reactions as:
Reaction (1): $\text{Fe}^{3+} + 3\,e^- \rightarrow \text{Fe}$
⇒ $\Delta G_1 = -3\,F\,E^\circ_{\text{Fe}^{3+}/\text{Fe}}.$
Reaction (2): $\text{Fe}^{2+} + 2\,e^- \rightarrow \text{Fe}$
⇒ $\Delta G_2 = -2\,F\,E^\circ_{\text{Fe}^{2+}/\text{Fe}}.$
Target Reaction (3): $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$
⇒ $\Delta G_3 = -1\,F\,E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}}.$
We observe that Reaction (3) can be obtained by
(Reaction (1) minus Reaction (2)). Therefore, their free energy changes must satisfy:
$$\Delta G_3 = \Delta G_1 - \Delta G_2.$$
Step 5: Substitute the Known Values
Using $\Delta G = -n F E^\circ$ relations:
$$-1 \cdot F \cdot E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}}
= \bigl[-3\,F\,E^\circ_{\text{Fe}^{3+}/\text{Fe}}\bigr] - \bigl[-2\,F\,E^\circ_{\text{Fe}^{2+}/\text{Fe}}\bigr].$$
Simplifying:
$$-F\,E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}}
= -3\,F\bigl(-0.036\bigr) + 2\,F\bigl(0.439\bigr).$$
(Note the signs carefully: Subtracting a negative value adds it.)
Canceling out $F$ throughout:
$$-E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}}
= -3\bigl(-0.036\bigr) + 2\bigl(0.439\bigr).$$
Calculate step by step:
$-3 \times (-0.036) = 0.108.$
$2 \times 0.439 = 0.878.$
Hence, sum = $0.108 + 0.878 = 0.986.$
(But carefully checking signs as per the original statement:
we actually want $-3(-0.036) + 2(0.439) = 0.108 + 0.878 = 0.986.$)
So,
$$-E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.986.$$
This means:
$$E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = -0.986\,\text{V}.$$
However, the question’s recognized and more standard result often accepted is:
$$E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.770\,\text{V}.$$
The discrepancy comes from carefully rechecking the arithmetic and sign convention. Properly done, by direct usage of
$E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}}) = E^\circ(\text{Fe}^{3+}/\text{Fe}) - E^\circ(\text{Fe}^{2+}/\text{Fe}).$
But we must multiply each potential by the correct electron count in free energy terms. Correctly:
$$3 \times E^\circ_{\text{Fe}^{3+}/\text{Fe}} - 2 \times E^\circ_{\text{Fe}^{2+}/\text{Fe}}
= 3 \times (-0.036) - 2 \times (-0.439).$$
$$= -0.108 + 0.878 = 0.770\,\text{V}.$$
Thus,
$$E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.770\,\text{V}.$$
Step 6: State the Final Answer
Therefore, the standard electrode potential for the conversion
$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ is
$\boxed{0.770\,\text{V}}.$
