In the following reaction the value of 'X' is
`H_(2)O+SO_(3)^(2-)rarr SO_(4)^(2-)+2H^(+)+X`

Options

A. `4e^(-)`
B. `3e^(-)`
C. `2e^(-)`
D. `1e^(-)`

Answer

To solve the given redox reaction and find the value of 'X', we need to balance the charges on both sides of the reaction. The reaction is:

\[ H_2O + SO_3^{2-} \rightarrow SO_4^{2-} + 2H^+ + X \]

### Step 1: Identify the charges on both sides of the equation
- On the left side, we have:
- Water (H2O) is neutral, so it has a charge of 0.
- Sulfite ion (SO3^2-) has a charge of -2.

Total charge on the left side = 0 + (-2) = -2.

- On the right side, we have:
- Sulfate ion (SO4^2-) has a charge of -2.
- Two hydrogen ions (2H+) have a charge of +2.

Total charge on the right side = -2 + 2 + X.

### Step 2: Set up the equation for charge balance
To balance the charges, we set the total charge on the left side equal to the total charge on the right side:

\[ -2 = -2 + 2 + X \]

### Step 3: Solve for 'X'
Now, we can simplify the equation:

\[ -2 = 0 + X \]

This implies:

\[ X = -2 \]

### Step 4: Identify 'X' in terms of electrons
In redox reactions, 'X' often represents the number of electrons transferred. Since we have found that \(X = -2\), it indicates that 2 electrons are involved in the reaction.

### Conclusion
Thus, the value of 'X' is -2, which represents the transfer of 2 electrons.

### Final Answer
The value of 'X' is -2.

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