If `E_(Fe^(2+)|Fe)^0=-0.44V,E_((Fe^3+)||Fe^(2+))^0=0.77`. Calculate `E_(Fe^(3+)|Fe)^0`

Answer

To solve the problem, we need to calculate the standard electrode potential \( E^\circ_{(Fe^{3+}|Fe)} \) using the given standard electrode potentials:

1. \( E^\circ_{(Fe^{2+}|Fe)} = -0.44 \, V \)
2. \( E^\circ_{(Fe^{3+}|Fe^{2+})} = 0.77 \, V \)

### Step 1: Write the half-reactions
We have the following half-reactions based on the given potentials:

- For \( Fe^{2+} + 2e^- \rightarrow Fe \) (reduction of \( Fe^{2+} \) to \( Fe \)):
\[ E^\circ_1 = -0.44 \, V \]

- For \( Fe^{3+} + e^- \rightarrow Fe^{2+} \) (reduction of \( Fe^{3+} \) to \( Fe^{2+} \)):
\[ E^\circ_2 = 0.77 \, V \]

### Step 2: Combine the half-reactions
To find \( E^\circ_{(Fe^{3+}|Fe)} \), we need to combine the two half-reactions. The overall reaction can be written as:
\[ Fe^{3+} + 3e^- \rightarrow Fe \]

### Step 3: Relate the potentials
Using the relationship between the potentials, we can express \( E^\circ_{(Fe^{3+}|Fe)} \) in terms of \( E^\circ_1 \) and \( E^\circ_2 \).

From the reaction:
\[ Fe^{3+} + 3e^- \rightarrow Fe \]
we can express the potential as:
\[ E^\circ_{(Fe^{3+}|Fe)} = E^\circ_{(Fe^{3+}|Fe^{2+})} + E^\circ_{(Fe^{2+}|Fe)} \]

### Step 4: Substitute the values
Now we substitute the known values:
\[ E^\circ_{(Fe^{3+}|Fe)} = E^\circ_{(Fe^{3+}|Fe^{2+})} + E^\circ_{(Fe^{2+}|Fe)} \]
\[ E^\circ_{(Fe^{3+}|Fe)} = 0.77 \, V + (-0.44 \, V) \]

### Step 5: Calculate the final value
Now, we perform the calculation:
\[ E^\circ_{(Fe^{3+}|Fe)} = 0.77 - 0.44 = 0.33 \, V \]

### Final Answer
Thus, the standard electrode potential \( E^\circ_{(Fe^{3+}|Fe)} \) is:
\[ \boxed{0.33 \, V} \]