`Cl_(2)(g)+Xbar(O)H rarr YClO_(3)""^(-)+ZH_(2)O+10e^(-)`
The coefficient X, Y and Z are

Options

A. 6, 2, 2
B. 5, 1, 3
C. 12, 2, 6
D. 12, 1, 6

Answer

To solve the given redox reaction and find the coefficients X, Y, and Z, we will follow a systematic approach to balance the reaction.

The reaction is:
\[ \text{Cl}_2(g) + X \text{OH}^- \rightarrow Y \text{ClO}_3^- + Z \text{H}_2O + 10 e^- \]

### Step 1: Balancing Chlorine Atoms
On the left-hand side (LHS), we have 2 chlorine atoms from \(\text{Cl}_2\). To balance the chlorine atoms, we need to determine the value of Y in \(\text{ClO}_3^-\).

- Each \(\text{ClO}_3^-\) contains 1 chlorine atom.
- Therefore, to have 2 chlorine atoms on the right-hand side (RHS), we need:
\[ Y = 2 \]

### Step 2: Balancing Oxygen Atoms
Now, we need to balance the oxygen atoms.

- On the RHS, we have \(Y \times 3\) oxygen atoms from \(Y \text{ClO}_3^-\) and \(Z\) oxygen atoms from \(Z \text{H}_2O\).
- Since \(Y = 2\), the total number of oxygen atoms on the RHS is:
\[ 3Y + Z = 3(2) + Z = 6 + Z \]

On the LHS, we have \(X\) oxygen atoms from \(X \text{OH}^-\). Therefore, we can set up the equation:
\[ X = 6 + Z \] (Equation 1)

### Step 3: Balancing Hydrogen Atoms
Next, we balance the hydrogen atoms.

- On the LHS, we have \(X\) hydrogen atoms from \(X \text{OH}^-\).
- On the RHS, we have \(2Z\) hydrogen atoms from \(Z \text{H}_2O\).

Setting up the equation for hydrogen gives us:
\[ X = 2Z \] (Equation 2)

### Step 4: Solving the Equations
Now we have two equations:
1. \( X = 6 + Z \) (Equation 1)
2. \( X = 2Z \) (Equation 2)

We can substitute Equation 2 into Equation 1:
\[ 2Z = 6 + Z \]

Subtract \(Z\) from both sides:
\[ 2Z - Z = 6 \]
\[ Z = 6 \]

### Step 5: Finding X
Now that we have \(Z\), we can substitute it back into Equation 2 to find \(X\):
\[ X = 2Z = 2(6) = 12 \]

### Summary of Coefficients
Now we have:
- \(X = 12\)
- \(Y = 2\)
- \(Z = 6\)

### Final Answer
Thus, the coefficients are:
- \(X = 12\)
- \(Y = 2\)
- \(Z = 6\)