In the reaction,`2CuCl_(2)+2H_(2)O+SO_(2)toA+H_(2)SO_(4)+2HCl,A ` is

Options

A. `Cu_(2)Cl_(2)`
B. `CuCl`
C. `CuSO_(4)`
D. `CuSO_(4)`

Answer

To solve the question, we need to analyze the given reaction step by step.

### Step 1: Write down the reaction
The reaction provided is:
\[ 2 \text{CuCl}_2 + 2 \text{H}_2\text{O} + \text{SO}_2 \rightarrow A + \text{H}_2\text{SO}_4 + 2 \text{HCl} \]

### Step 2: Identify the reactants and products
In this reaction, we have:
- Reactants: Copper(II) chloride (\( \text{CuCl}_2 \)), water (\( \text{H}_2\text{O} \)), and sulfur dioxide (\( \text{SO}_2 \)).
- Products: We need to determine compound \( A \), along with sulfuric acid (\( \text{H}_2\text{SO}_4 \)) and hydrochloric acid (\( \text{HCl} \)).

### Step 3: Determine the oxidation states
- In \( \text{CuCl}_2 \), copper (Cu) has an oxidation state of +2.
- In \( \text{CuCl} \) (which is the product we are looking for), copper (Cu) has an oxidation state of +1.
- In \( \text{SO}_2 \), sulfur (S) has an oxidation state of +4, and in \( \text{H}_2\text{SO}_4 \), sulfur has an oxidation state of +6.

### Step 4: Identify the changes in oxidation states
- Copper is reduced from +2 to +1 (gaining electrons).
- Sulfur is oxidized from +4 to +6 (losing electrons).

### Step 5: Identify the product \( A \)
Since copper is reduced to cuprous chloride (\( \text{CuCl} \)), we can conclude that:
\[ A = \text{CuCl} \]

### Step 6: Write the balanced reaction
The balanced reaction can now be written as:
\[ 2 \text{CuCl}_2 + 2 \text{H}_2\text{O} + \text{SO}_2 \rightarrow 2 \text{CuCl} + \text{H}_2\text{SO}_4 + 2 \text{HCl} \]

### Conclusion
Thus, the compound \( A \) is:
\[ A = \text{CuCl} \]

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