When the redox reaction
`SnO_(2)^(--)+Bi(OH)_(3)toSnO_(3)^(--)+Bi`
is balanced by ion electron method in basic medium what is correct coefficient of `SnO_(2)^(--)`

Answer

To balance the redox reaction \( \text{SnO}_2^{--} + \text{Bi(OH)}_3 \rightarrow \text{SnO}_3^{--} + \text{Bi} \) using the ion-electron method in a basic medium, follow these steps:

### Step 1: Identify Oxidation States
First, we need to determine the oxidation states of the elements in the reactants and products.

- For \( \text{SnO}_2^{--} \):
- Let the oxidation state of Sn be \( x \).
- The oxidation state of O is -2.
- Therefore, \( x + 2(-2) = -2 \) leads to \( x - 4 = -2 \), so \( x = +2 \).

- For \( \text{SnO}_3^{--} \):
- Let the oxidation state of Sn be \( y \).
- Therefore, \( y + 3(-2) = -2 \) leads to \( y - 6 = -2 \), so \( y = +4 \).

- For \( \text{Bi(OH)}_3 \):
- The oxidation state of Bi is +3.

- For elemental Bi:
- The oxidation state is 0.

### Step 2: Determine Changes in Oxidation States
- Sn changes from +2 in \( \text{SnO}_2^{--} \) to +4 in \( \text{SnO}_3^{--} \): **Oxidation** (loss of 2 electrons).
- Bi changes from +3 in \( \text{Bi(OH)}_3 \) to 0 in elemental Bi: **Reduction** (gain of 3 electrons).

### Step 3: Write Half-Reactions
1. **Oxidation half-reaction**:
\[ \text{SnO}_2^{--} \rightarrow \text{SnO}_3^{--} + 2e^- \]

2. **Reduction half-reaction**:
\[ \text{Bi(OH)}_3 + 3e^- \rightarrow \text{Bi} + 3\text{OH}^- \]

### Step 4: Equalize the Number of Electrons
To balance the number of electrons transferred, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:

1. **Oxidation half-reaction (multiplied by 3)**:
\[ 3\text{SnO}_2^{--} \rightarrow 3\text{SnO}_3^{--} + 6e^- \]

2. **Reduction half-reaction (multiplied by 2)**:
\[ 2\text{Bi(OH)}_3 + 6e^- \rightarrow 2\text{Bi} + 6\text{OH}^- \]

### Step 5: Combine Half-Reactions
Now, we can add the two half-reactions together:
\[ 3\text{SnO}_2^{--} + 2\text{Bi(OH)}_3 \rightarrow 3\text{SnO}_3^{--} + 2\text{Bi} + 6\text{OH}^- \]

### Step 6: Final Check
- **Sn**: 3 on both sides.
- **Bi**: 2 on both sides.
- **O**: 6 from \( \text{SnO}_2^{--} \) and 6 from \( \text{Bi(OH)}_3 \) on the left; 9 from \( \text{SnO}_3^{--} \) and 6 from \( \text{OH}^- \) on the right.

### Conclusion
The balanced equation shows that the coefficient of \( \text{SnO}_2^{--} \) is **3**.

### Final Answer
Therefore, the correct coefficient of \( \text{SnO}_2^{--} \) is **3**.

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