Given, standard electrode potentials
`Fe^(2+) +2e^(-) rarr Fe, E^(@)=-0.440 V`
`Fe^(3+)+3e^(-) rarr Fe, E^(@)=- 0.036 V`
The standarde potential `(E^(@))` for
`Fe^(2+)+e^(-) rarr Fe^(2+)`, is

Options

A. `+0.772 V`
B. `-0.772 V`
C. `+0.417 V`
D. `-0.417 V`

Explanation

To find the standard electrode potential \( E^\circ \) for the reaction

\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \]

we will use the given standard electrode potentials for the half-reactions:

1. \(\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}, \quad E^\circ = -0.440 \, \text{V}\)
2. \(\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}, \quad E^\circ = -0.036 \, \text{V}\)

### Step 1: Write the two half-reactions

We have the following half-reactions:

1. \(\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad (E^\circ_1 = -0.440 \, \text{V})\)
2. \(\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \quad (E^\circ_2 = -0.036 \, \text{V})\)

### Step 2: Calculate the Gibbs free energy changes

The Gibbs free energy change for a reaction can be calculated using the formula:

\[ \Delta G^\circ = -nFE^\circ \]

where \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant, and \( E^\circ \) is the standard electrode potential.

For the first reaction (1 electron transfer):

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

For the second reaction (3 electrons transfer):

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

### Step 3: Determine the Gibbs free energy change for the desired reaction

To find the Gibbs free energy change for the reaction

\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \]

we can use the relation:

\[ \Delta G_3^\circ = \Delta G_2^\circ - \Delta G_1^\circ \]

Substituting the values we calculated:

\[ \Delta G_3^\circ = 0.108F - 0.880F = -0.772F \]

### Step 4: Calculate the standard electrode potential for the desired reaction

Now we can find \( E^\circ_3 \) using:

\[ \Delta G_3^\circ = -F E^\circ_3 \]

Thus,

\[ -0.772F = -FE^\circ_3 \]

Cancelling \( F \) from both sides gives:

\[ E^\circ_3 = 0.772 \, \text{V} \]

### Final Answer

The standard electrode potential \( E^\circ \) for the reaction

\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \]

is

\[ E^\circ = 0.772 \, \text{V} \]