Consider the following reaction,
`xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+)toxMn^(2+)+2yCO_(2)+(z)/(2)H_(2)O`
The values of x,y and z in the reactions are, respectively

Options

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

Answer

To solve the problem, we need to balance the given redox reaction and find the values of x, y, and z. The 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 1: Identify the Oxidation States
- For \(\text{MnO}_4^{-}\), the oxidation state of Mn is +7.
- For \(\text{C}_2\text{O}_4^{2-}\), the oxidation state of carbon is +3.
- For \(\text{CO}_2\), the oxidation state of carbon is +4.
- For \(\text{Mn}^{2+}\), the oxidation state of manganese is +2.

### Step 2: Write Half-Reactions
1. **Reduction Half-Reaction**:
\[ \text{MnO}_4^{-} \rightarrow \text{Mn}^{2+} \]
- The change in oxidation state for Mn is from +7 to +2, which is a decrease of 5.

2. **Oxidation Half-Reaction**:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow \text{CO}_2 \]
- The change in oxidation state for carbon is from +3 to +4, which is an increase of 1 for each carbon atom. Since there are 2 carbon atoms, the total increase is 2.

### Step 3: Balance the Half-Reactions
To balance the number of electrons transferred:
- For the reduction half-reaction (Mn), we need to multiply it by 2 (to account for 2 electrons gained).
- For the oxidation half-reaction (C), we need to multiply it by 5 (to account for 5 electrons lost).

Thus, we have:
\[ 2 \text{MnO}_4^{-} + 5 \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 \]

### Step 4: Balance Oxygen and Hydrogen
Now, we need to balance the oxygen and hydrogen atoms:
- On the left side, we have:
- From \(2 \text{MnO}_4^{-}\): 8 O atoms
- From \(5 \text{C}_2\text{O}_4^{2-}\): 20 O atoms
- Total: 28 O atoms

- On the right side, we have:
- From \(10 \text{CO}_2\): 20 O atoms
- We need to add water molecules to balance the remaining 8 O atoms:
\[ 2 \text{MnO}_4^{-} + 5 \text{C}_2\text{O}_4^{2-} + 8 \text{H}_2\text{O} \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 16 \text{H}^{+} \]

### Step 5: Count and Assign Values
From the balanced equation:
- \(x = 2\) (for \(\text{MnO}_4^{-}\))
- \(y = 5\) (for \(\text{C}_2\text{O}_4^{2-}\))
- \(z = 16\) (for \(\text{H}^{+}\))

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