On the basis of the following `E^(@)` values, the stongest oxidizing agent is `[Fe(CN)_(6)]^(4-) rarr [Fe(CN)_(6)]^(3-)+e^(-), E^(@) = -0.35 V`
`Fe^(2+) rarr Fe^(3+)+e^(-), E^(@) = -0.77 V`

Options

A. `[Fe(CN)_(6)]^(4-)`
B. `Fe^(2+)`
C. `Fe^(3+)`
D. `[Fe(CN)_(6)]^(3-)`

Explanation

To determine the strongest oxidizing agent based on the given standard electrode potentials (E° values), we will follow these steps:

### Step 1: Identify the half-reactions and their E° values
We have two half-reactions with their respective E° values:
1. \([Fe(CN)_6]^{4-} \rightarrow [Fe(CN)_6]^{3-} + e^{-}\), \(E° = -0.35 \, V\)
2. \(Fe^{2+} \rightarrow Fe^{3+} + e^{-}\), \(E° = -0.77 \, V\)

### Step 2: Convert oxidation potentials to reduction potentials
To find the reduction potential for each half-reaction, we need to change the sign of the given E° values:
1. For the first half-reaction:
\[ E°_{reduction} = -(-0.35 \, V) = +0.35 \, V \]
2. For the second half-reaction:
\[ E°_{reduction} = -(-0.77 \, V) = +0.77 \, V \]

### Step 3: Compare the reduction potentials
Now we compare the reduction potentials:
- For \([Fe(CN)_6]^{4-}\): \(E° = +0.35 \, V\)
- For \(Fe^{2+}\): \(E° = +0.77 \, V\)

### Step 4: Identify the strongest oxidizing agent
The stronger the oxidizing agent, the higher the reduction potential. In this case, \(Fe^{2+}\) has a higher reduction potential (\(+0.77 \, V\)) compared to \([Fe(CN)_6]^{4-}\) (\(+0.35 \, V\)). Therefore, \(Fe^{2+}\) is the strongest oxidizing agent.

### Conclusion
The strongest oxidizing agent among the given options is:
\[ \text{Answer: } Fe^{2+} \]

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