Consider the following reaction :
`xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+) to xMn^(2+)+2yCO_(2)+(z)/(2)H_(2)O`
The value of x, y and z in the reaction are, respectively.

Options

A. 2,5 and 16
B. 5,2 and 8
C. 5,2 and 16
D. 2,5 and 8

Explanation

To solve the given redox reaction and find the values of \( x \), \( y \), and \( z \) in the equation:

\[ x \text{MnO}_4^{-} + y \text{C}_2\text{O}_4^{2-} + z \text{H}^{+} \rightarrow x \text{Mn}^{2+} + 2y \text{CO}_2 + \frac{z}{2} \text{H}_2\text{O} \]

we will follow these steps:

### Step 1: Determine the oxidation states
1. **Manganese in \( \text{MnO}_4^{-} \)**:
- Oxygen has an oxidation state of -2.
- For 4 oxygen atoms: \( 4 \times (-2) = -8 \).
- The overall charge is -1, so:
\[ x + (-8) = -1 \implies x = +7 \]
- Therefore, the oxidation state of Mn in \( \text{MnO}_4^{-} \) is +7.

2. **Manganese in \( \text{Mn}^{2+} \)**:
- The oxidation state is +2.

3. **Carbon in \( \text{C}_2\text{O}_4^{2-} \)**:
- Let the oxidation state of carbon be \( x \).
- For 2 carbon atoms: \( 2x \).
- For 4 oxygen atoms: \( 4 \times (-2) = -8 \).
- The overall charge is -2, so:
\[ 2x - 8 = -2 \implies 2x = 6 \implies x = +3 \]
- Therefore, the oxidation state of carbon in \( \text{C}_2\text{O}_4^{2-} \) is +3.

4. **Carbon in \( \text{CO}_2 \)**:
- The oxidation state of carbon in \( \text{CO}_2 \) is +4.

### Step 2: Identify changes in oxidation states
- **Manganese**: +7 to +2 (reduction, decrease of 5).
- **Carbon**: +3 to +4 (oxidation, increase of 1).

### Step 3: Balance the changes in oxidation states
- For manganese, the total decrease is \( 5x \) (where \( x \) is the number of MnO4^-).
- For carbon, the total increase is \( 1y \) (where \( y \) is the number of C2O4^{2-}).
- To balance the reaction, we need:
\[ 5x = 2y \]

### Step 4: Determine coefficients
1. From the balancing equation \( 5x = 2y \), we can express \( y \) in terms of \( x \):
\[ y = \frac{5}{2}x \]

2. To keep \( y \) as a whole number, let’s choose \( x = 2 \):
\[ y = \frac{5}{2} \times 2 = 5 \]

### Step 5: Balance oxygen and hydrogen
1. **Oxygen balance**:
- Left side: \( 8 \) from \( 2 \text{MnO}_4^{-} \) and \( 20 \) from \( 5 \text{C}_2\text{O}_4^{2-} \) gives \( 8 + 20 = 28 \) oxygen atoms.
- Right side: \( 10 \text{CO}_2 \) contributes \( 20 \) oxygen and \( 8 \text{H}_2\text{O} \) contributes \( 8 \) oxygen, giving \( 20 + 8 = 28 \) oxygen atoms.

2. **Hydrogen balance**:
- Left side: \( z \) from \( z \text{H}^{+} \).
- Right side: \( 8 \text{H}_2\text{O} \) contributes \( 16 \) hydrogen atoms.
- Therefore, \( z = 16 \).

### Final Coefficients
Thus, the values of \( x \), \( y \), and \( z \) are:
- \( x = 2 \)
- \( y = 5 \)
- \( z = 16 \)

### Conclusion
The final answer is:
\[ x = 2, \quad y = 5, \quad z = 16 \]