`S_8 + aOH^- to bS^(2-) + cS_2O_3^(2-) + dH_2O`
In balanced equation what is the value of a.

Answer

To solve the given equation \( S_8 + aOH^- \rightarrow bS^{2-} + cS_2O_3^{2-} + dH_2O \) and find the value of \( a \), we will follow these steps:

### Step 1: Identify the oxidation states
In the reaction, we start with \( S_8 \) where sulfur has an oxidation state of 0. In the products, sulfur is present in two forms: \( S^{2-} \) (where sulfur has an oxidation state of -2) and \( S_2O_3^{2-} \) (where we need to calculate the oxidation state of sulfur).

### Step 2: Determine the oxidation states in \( S_2O_3^{2-} \)
For \( S_2O_3^{2-} \):
- Let the oxidation state of sulfur be \( x \).
- The equation for the oxidation state is:
\[ 2x + 3(-2) = -2 \]
\[ 2x - 6 = -2 \implies 2x = 4 \implies x = 2 \]
So, in \( S_2O_3^{2-} \), sulfur has an oxidation state of +2.

### Step 3: Write the half-reactions
We can separate the overall reaction into oxidation and reduction half-reactions.

**Reduction half-reaction:**
\[ S_8 + 16e^- \rightarrow 8S^{2-} \]
This shows that 8 moles of \( S^{2-} \) are produced from \( S_8 \) by gaining 16 electrons.

**Oxidation half-reaction:**
\[ S_8 + 4OH^- \rightarrow 4S_2O_3^{2-} + 12H_2O + 16e^- \]
This indicates that for every \( S_8 \), 4 moles of \( S_2O_3^{2-} \) are produced.

### Step 4: Balance the overall reaction
Now we can combine the half-reactions:
\[ S_8 + 2OH^- \rightarrow 8S^{2-} + 4S_2O_3^{2-} + 12H_2O \]

### Step 5: Count the hydroxide ions
From the balanced equation, we can see that \( 24 \) hydroxide ions are required to balance the reaction. Thus, \( a = 24 \).

### Final Answer
The value of \( a \) is 24.

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