A volume of 12.5 mL of 0.05 M `SeO_(2)` reacts with 25 mL of 0.1 M `CrSO_(4)` which is oxidised to `Cr^(3+)`. To what oxidation state was the selenium converted by the reaction ?

Answer

To determine the oxidation state to which selenium was converted during the reaction, we can follow these steps:

### Step 1: Identify the initial oxidation state of selenium.
Selenium in selenium dioxide (SeO₂) has an oxidation state of +4. This is calculated based on the fact that each oxygen has an oxidation state of -2. Therefore, for two oxygen atoms, the total is -4, and to balance this in SeO₂, selenium must be +4.

**Hint:** Remember that the oxidation state of oxygen is typically -2.

### Step 2: Write the balanced reaction.
The reaction involves selenium dioxide (SeO₂) reacting with chromium sulfate (CrSO₄), where chromium is oxidized from +2 to +3. We can represent the reaction as:
\[ \text{SeO}_2 + \text{CrSO}_4 \rightarrow \text{Cr}^{3+} + \text{Se} \]
Let the oxidation state of selenium after the reaction be \( x \).

**Hint:** Identify the reactants and products clearly to understand the changes in oxidation states.

### Step 3: Determine the change in oxidation state for selenium.
The change in oxidation state for selenium is from +4 to \( x \). Therefore, the change in oxidation state is:
\[ \text{Change} = 4 - x \]

**Hint:** The change in oxidation state indicates how many electrons are gained or lost.

### Step 4: Calculate the number of equivalents for selenium dioxide.
To find the number of equivalents, we use the formula:
\[ \text{Number of equivalents} = \text{Molarity} \times \text{Volume (in L)} \times \text{n factor} \]
For selenium dioxide:
- Molarity = 0.05 M
- Volume = 12.5 mL = 0.0125 L
- n factor = \( 4 - x \)

Thus, the number of equivalents for SeO₂ is:
\[ \text{Equivalents of SeO}_2 = 0.05 \times 0.0125 \times (4 - x) \]

**Hint:** Ensure to convert mL to L when calculating volumes.

### Step 5: Calculate the number of equivalents for chromium sulfate.
For chromium sulfate:
- Molarity = 0.1 M
- Volume = 25 mL = 0.025 L
- n factor = 1 (since Cr is oxidized from +2 to +3)

Thus, the number of equivalents for CrSO₄ is:
\[ \text{Equivalents of CrSO}_4 = 0.1 \times 0.025 \times 1 = 0.0025 \]

**Hint:** Use the same approach to find equivalents for both reactants.

### Step 6: Set the number of equivalents equal to each other.
Since the number of equivalents of SeO₂ will equal the number of equivalents of CrSO₄ in the reaction, we can set up the equation:
\[ 0.05 \times 0.0125 \times (4 - x) = 0.0025 \]

### Step 7: Solve for \( x \).
Calculating the left-hand side:
\[ 0.000625 \times (4 - x) = 0.0025 \]
Dividing both sides by 0.000625:
\[ 4 - x = 4 \]
This simplifies to:
\[ x = 0 \]

**Hint:** Isolate \( x \) to find the final oxidation state.

### Conclusion:
Selenium is converted from an oxidation state of +4 in SeO₂ to an oxidation state of 0 in the product.

**Final Answer:** The oxidation state of selenium after the reaction is 0.