KI and `CuSO_4` solutions when mixed give

Options

A. `CuI_2+K_2SO_4`
B. `Cu_2I_2+K_2SO_4`
C. `K_2SO_4+Cu_2I_2+I_2`
D. `K_2SO_4+CuI_2+I_2`

Explanation

To solve the question regarding the reaction between KI and CuSO4 solutions, we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are potassium iodide (KI) and copper(II) sulfate (CuSO4).

### Step 2: Understand the Reaction Type
This is a redox reaction where iodide ions (I⁻) are oxidized to elemental iodine (I2), and copper(II) ions (Cu²⁺) are reduced to copper(I) ions (Cu⁺).

### Step 3: Write the Half-Reactions
1. **Oxidation half-reaction**:
\[ 2 I^- \rightarrow I_2 + 2 e^- \]
2. **Reduction half-reaction**:
\[ 2 Cu^{2+} + 2 e^- \rightarrow 2 Cu^+ \]

### Step 4: Combine the Half-Reactions
Combining the half-reactions gives:
\[ 2 I^- + 2 Cu^{2+} \rightarrow I_2 + 2 Cu^+ \]

### Step 5: Write the Overall Reaction
Now, we need to include the sulfate ions from CuSO4 and the potassium ions from KI in the overall balanced equation:
\[ 2 KI + CuSO_4 \rightarrow I_2 + Cu_2I_2 (s) + K_2SO_4 \]

### Step 6: Identify the Products
The products of the reaction are:
- Copper(I) iodide (CuI), which precipitates as a solid.
- Potassium sulfate (K2SO4), which remains in solution.
- Elemental iodine (I2), which is formed.

### Step 7: Final Balanced Equation
The final balanced equation can be summarized as:
\[ 2 KI + CuSO_4 \rightarrow Cu_2I_2 (s) + K_2SO_4 + I_2 \]

### Conclusion
When KI and CuSO4 solutions are mixed, the products formed are copper(I) iodide (CuI), potassium sulfate (K2SO4), and elemental iodine (I2).

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