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 find the value of 'X' in the reaction:

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

we will follow these steps:

### Step 1: Identify the charges on both sides of the equation.
- On the left-hand side (LHS):
- \( H_2O \) is neutral (charge = 0).
- \( SO_3^{2-} \) has a charge of -2.

Therefore, the total charge on the LHS is:
\[ 0 + (-2) = -2 \]

- On the right-hand side (RHS):
- \( SO_4^{2-} \) has a charge of -2.
- \( 2H^+ \) has a charge of +2.

Therefore, the total charge on the RHS is:
\[ -2 + 2 = 0 \]

### Step 2: Balance the charges.
To balance the charges, we need to ensure that both sides have the same total charge. Currently, the LHS has a charge of -2, and the RHS has a charge of 0.

### Step 3: Determine the number of electrons needed.
To balance the charges, we need to add electrons to the RHS. Let’s denote the number of electrons as 'n'. The equation for the charge on the RHS becomes:
\[ -2 + 2 + n = -2 + n \]

To balance the charges:
\[ -2 + n = -2 \]
This implies:
\[ n = -2 \]

Since we are adding electrons (which have a negative charge), we need to add 2 electrons to the RHS.

### Step 4: Write the final balanced equation.
Now, substituting 'X' with the number of electrons:
\[ H_2O + SO_3^{2-} \rightarrow SO_4^{2-} + 2H^+ + 2e^- \]

Thus, the value of 'X' is:
\[ X = 2e^- \]

### Final Answer:
The value of 'X' is \( 2e^- \).

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