In balancing the half reaction, `S_(2)O_(3)^(2-)toS`, the number of electrons that must be added is :-

Options

A. 2 on the right
B. 2 on the left
C. 4 on the left
D. 4 on the right

Answer

To balance the half-reaction \( S_2O_3^{2-} \rightarrow S \), we will follow these steps:

### Step 1: Write the unbalanced half-reaction
The half-reaction we need to balance is:
\[ S_2O_3^{2-} \rightarrow S \]

### Step 2: Balance the sulfur atoms
On the left side, we have 2 sulfur atoms in \( S_2O_3^{2-} \). To balance the sulfur atoms, we need 2 sulfur atoms on the right side as well:
\[ S_2O_3^{2-} \rightarrow 2S \]

### Step 3: Balance the oxygen atoms
Next, we have 3 oxygen atoms on the left side from \( S_2O_3^{2-} \). To balance the oxygen atoms, we can add water molecules to the right side. Since each water molecule contains 1 oxygen atom, we will add 3 water molecules:
\[ S_2O_3^{2-} \rightarrow 2S + 3H_2O \]

### Step 4: Balance the hydrogen atoms
Now, we have 6 hydrogen atoms on the right side (from 3 water molecules). To balance the hydrogen atoms, we need to add \( 6H^+ \) ions to the left side:
\[ S_2O_3^{2-} + 6H^+ \rightarrow 2S + 3H_2O \]

### Step 5: Balance the charges
Now we need to balance the charges on both sides. The left side has a charge of:
- \( 2- \) from \( S_2O_3^{2-} \)
- \( +6 \) from \( 6H^+ \)

So the total charge on the left side is:
\[ 6 - 2 = +4 \]

The right side has no charge (neutral), so we need to add electrons to the left side to balance the charge. To make the left side's charge equal to zero, we need to add 4 electrons:
\[ S_2O_3^{2-} + 6H^+ + 4e^- \rightarrow 2S + 3H_2O \]

### Final Balanced Half-Reaction
The final balanced half-reaction is:
\[ S_2O_3^{2-} + 6H^+ + 4e^- \rightarrow 2S + 3H_2O \]

### Conclusion
The number of electrons that must be added is **4**.

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