All the following species are strong oxidizing agents. Their strength as oxidizing agents in acidic solution is such that

Options

A. `S_(2)O_(8)^(2-) gt Cr_(2)O_(7)^(2-) gt MnO_(4)^(-)`
B. `MnO_(4)^(-) gt Cr_(2)O_(4)^(2-) gt S_(2)O_(8)^(2-)`
C. `S_(2)O_(8)^(2-) gt MnO_(4)^(-) gt Cr_(2)O_(7)^(2-)`
D. `MnO_(4)^(-) gt S_(2)O_(8)^(2-) gt Cr_(2)O_(7)^(2-)`

Answer

To determine the order of strength of the given strong oxidizing agents in acidic solution, we need to analyze their oxidation states and reduction potentials. Here’s a step-by-step solution:

### Step 1: Identify the Species
The species mentioned are:
1. \( S_2O_8^{2-} \) (Peroxydisulfate)
2. \( Cr_2O_7^{2-} \) (Dichromate)
3. \( MnO_4^{-} \) (Permanganate)

### Step 2: Calculate Oxidation States
We will calculate the oxidation states of the central atoms in each species.

#### For \( S_2O_8^{2-} \):
Let the oxidation state of sulfur be \( x \).
- The equation is:
\[ 2x + 8(-2) = -2 \]
- Simplifying gives:
\[ 2x - 16 = -2 \implies 2x = 14 \implies x = +7 \]
- **Oxidation state of sulfur in \( S_2O_8^{2-} \) is +7.**

#### For \( Cr_2O_7^{2-} \):
Let the oxidation state of chromium be \( y \).
- The equation is:
\[ 2y + 7(-2) = -2 \]
- Simplifying gives:
\[ 2y - 14 = -2 \implies 2y = 12 \implies y = +6 \]
- **Oxidation state of chromium in \( Cr_2O_7^{2-} \) is +6.**

#### For \( MnO_4^{-} \):
Let the oxidation state of manganese be \( z \).
- The equation is:
\[ z + 4(-2) = -1 \]
- Simplifying gives:
\[ z - 8 = -1 \implies z = +7 \]
- **Oxidation state of manganese in \( MnO_4^{-} \) is +7.**

### Step 3: Compare Oxidation States
From the calculations:
- \( S_2O_8^{2-} \): +7
- \( Cr_2O_7^{2-} \): +6
- \( MnO_4^{-} \): +7

The higher the oxidation state, the stronger the oxidizing agent. Here, \( S_2O_8^{2-} \) and \( MnO_4^{-} \) both have the highest oxidation state of +7, while \( Cr_2O_7^{2-} \) has +6.

### Step 4: Compare Reduction Potentials
To further confirm the order, we can look at the standard reduction potentials:
- \( S_2O_8^{2-} \): +2.05 V
- \( MnO_4^{-} \): +1.51 V
- \( Cr_2O_7^{2-} \): +1.33 V

### Step 5: Determine the Order of Strength
Based on oxidation states and reduction potentials:
1. \( S_2O_8^{2-} \) (strongest oxidizing agent)
2. \( MnO_4^{-} \)
3. \( Cr_2O_7^{2-} \) (weakest oxidizing agent)

### Conclusion
The order of strength as oxidizing agents in acidic solution is:
\[ S_2O_8^{2-} > MnO_4^{-} > Cr_2O_7^{2-} \]