On the basis of the folwing `E^@` values, the strongest oxidizing agent is
`[Fe(CN)_6 ]_4^(-) rarr [Fe(CN)_5]^(3-) +3^- , E^@ =- 0.35 V`
`Fe^(2+) rarr Fe^(3+) + e^(- : E^@ =- 0. 77 V`.

Options

A. ` Fe^(2+)`
B. ` Fe^(3+)`
C. ` [Fe(CN)_6]^(3-)`
D. ` [Fe(CN)_6]^(4-)`

Explanation

To determine the strongest oxidizing agent based on the given standard electrode potentials (E° values), we will analyze the provided half-reactions and their corresponding potentials.

### Step-by-Step Solution:

1. **Identify the Given Reactions and E° Values:**
- The first reaction is:
\[ [Fe(CN)_6]^{4-} \rightarrow [Fe(CN)_5]^{3-} + e^{-} \quad E° = -0.35 \, V \]
- The second reaction is:
\[ Fe^{2+} \rightarrow Fe^{3+} + e^{-} \quad E° = -0.77 \, V \]

2. **Convert to Reduction Potentials:**
- For the first reaction, we can write the reduction potential by reversing the reaction:
\[ [Fe(CN)_5]^{3-} + e^{-} \rightarrow [Fe(CN)_6]^{4-} \quad E° = +0.35 \, V \]
- For the second reaction, we also reverse it to find the reduction potential:
\[ Fe^{3+} + e^{-} \rightarrow Fe^{2+} \quad E° = +0.77 \, V \]

3. **Compare the Reduction Potentials:**
- Now we have the reduction potentials:
- For \([Fe(CN)_5]^{3-}\): \(E° = +0.35 \, V\)
- For \(Fe^{3+}\): \(E° = +0.77 \, V\)

4. **Determine the Strongest Oxidizing Agent:**
- The strongest oxidizing agent is the one with the highest reduction potential. Here, \(Fe^{3+}\) has a higher reduction potential (+0.77 V) compared to \([Fe(CN)_5]^{3-}\) (+0.35 V).
- Therefore, \(Fe^{3+}\) is the strongest oxidizing agent.

### Conclusion:
The strongest oxidizing agent among the given options is \(Fe^{3+}\).

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