Given `E_(Fe^(3+)//Fe)^(@)=-0.036V,E_(Fe^(2+)//Fe)^(@)=-0.439V`. The value of `E_(Fe^(3+)//Fe^(2+))^(@)` will be

Options

A. `-0.072V`
B. 0.385V
C. 0.770V
D. `-0.270V`

Answer

To find the standard electrode potential \( E^\circ_{Fe^{3+}/Fe^{2+}} \), we will use the given standard electrode potentials for the half-reactions involving iron ions. The given values are:

- \( E^\circ_{Fe^{3+}/Fe} = -0.036 \, V \)
- \( E^\circ_{Fe^{2+}/Fe} = -0.439 \, V \)

### Step 1: Write the half-reactions

1. The reduction half-reaction for \( Fe^{3+} \) to \( Fe \):
\[ Fe^{3+} + 3e^- \rightarrow Fe \quad (E^\circ = -0.036 \, V) \]

2. The reduction half-reaction for \( Fe^{2+} \) to \( Fe \):
\[ Fe^{2+} + 2e^- \rightarrow Fe \quad (E^\circ = -0.439 \, V) \]

### Step 2: Reverse the second half-reaction

To find the potential for the reaction \( Fe^{3+} + e^- \rightarrow Fe^{2+} \), we need to reverse the second half-reaction (from \( Fe^{2+} \) to \( Fe \)). When we reverse a reaction, we change the sign of the electrode potential:

\[ Fe \rightarrow Fe^{2+} + 2e^- \quad (E^\circ = +0.439 \, V) \]

### Step 3: Combine the half-reactions

Now we will combine the two half-reactions. We need to adjust the number of electrons transferred so that they cancel out. The first half-reaction involves 3 electrons, while the reversed second half-reaction involves 2 electrons. To balance the electrons, we can multiply the first half-reaction by 2 and the second half-reaction by 3:

1. Multiply the first half-reaction by 2:
\[ 2Fe^{3+} + 6e^- \rightarrow 2Fe \quad (E^\circ = -0.036 \, V) \]

2. Multiply the reversed second half-reaction by 3:
\[ 3Fe \rightarrow 3Fe^{2+} + 6e^- \quad (E^\circ = +0.439 \, V) \]

### Step 4: Add the half-reactions

Now we can add the two half-reactions:

\[ 2Fe^{3+} + 6e^- + 3Fe \rightarrow 2Fe + 3Fe^{2+} + 6e^- \]

The \( 6e^- \) cancels out, and we are left with:

\[ 2Fe^{3+} + 3Fe \rightarrow 2Fe + 3Fe^{2+} \]

### Step 5: Calculate the overall standard electrode potential

To find the standard electrode potential for the reaction \( Fe^{3+} + e^- \rightarrow Fe^{2+} \), we need to subtract the potentials:

\[ E^\circ_{Fe^{3+}/Fe^{2+}} = E^\circ_{Fe^{3+}/Fe} - E^\circ_{Fe^{2+}/Fe} \]

Substituting the values:

\[ E^\circ_{Fe^{3+}/Fe^{2+}} = (-0.036) - (-0.439) = -0.036 + 0.439 = 0.403 \, V \]

### Final Answer

Thus, the value of \( E^\circ_{Fe^{3+}/Fe^{2+}} \) is approximately \( 0.403 \, V \).

---