`CuSO_(4) + KI = Cu_(2) I_(2) + I_(2) + K_(2) SO_(4)`

Explanation

To balance the redox reaction \( \text{CuSO}_4 + \text{KI} \rightarrow \text{Cu}_2\text{I}_2 + \text{I}_2 + \text{K}_2\text{SO}_4 \), we will follow these steps:

### Step 1: Assign Oxidation States
- In \( \text{CuSO}_4 \), copper (Cu) has an oxidation state of +2, and sulfur (S) has an oxidation state of +6. The sulfate ion (\( \text{SO}_4^{2-} \)) has a net charge of -2.
- In \( \text{KI} \), potassium (K) has an oxidation state of +1, and iodine (I) has an oxidation state of -1.
- In \( \text{Cu}_2\text{I}_2 \), copper (Cu) is +1, and iodine (I) is 0 in \( \text{I}_2 \).
- In \( \text{K}_2\text{SO}_4 \), potassium (K) is +1, and sulfur (S) is +6.

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