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

Answer

To solve the given redox reaction and find the values of x, y, and z, we will follow these steps:

### Step 1: Write the unbalanced equation
The unbalanced reaction is:
\[ x \text{MnO}_4^{-} + y \text{C}_2\text{O}_4^{2-} + z \text{H}^{+} \rightarrow x \text{Mn}^{2+} + 2y \text{CO}_2 + \frac{z}{2} \text{H}_2\text{O} \]

### Step 2: Identify oxidation states
- In \(\text{MnO}_4^{-}\), manganese (Mn) has an oxidation state of +7.
- In \(\text{Mn}^{2+}\), Mn has an oxidation state of +2.
- In \(\text{C}_2\text{O}_4^{2-}\), carbon (C) has an oxidation state of +3.
- In \(\text{CO}_2\), C has an oxidation state of +4.

### Step 3: Write half-reactions
1. **Reduction half-reaction** for Mn:
\[ \text{MnO}_4^{-} + 8 \text{H}^{+} + 5 e^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

2. **Oxidation half-reaction** for oxalate:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2 e^{-} \]

### Step 4: Balance the electrons
To balance the electrons, we need to multiply the oxidation half-reaction by 5:
\[ 5 \text{C}_2\text{O}_4^{2-} \rightarrow 10 \text{CO}_2 + 10 e^{-} \]

### Step 5: Combine the half-reactions
Now we can combine the half-reactions:
\[ 2 \text{MnO}_4^{-} + 5 \text{C}_2\text{O}_4^{2-} + 16 \text{H}^{+} \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]

### Step 6: Identify x, y, and z
From the balanced equation:
- \(x = 2\) (for Mn)
- \(y = 5\) (for oxalate)
- \(z = 16\) (for H+)

### Final Answer
The values of \(x\), \(y\), and \(z\) are:
\[ x = 2, \quad y = 5, \quad z = 16 \]