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In the electrolysis of CuSO4, the reacti...

In the electrolysis of `CuSO_4`, the reaction:`Cu^(2+)+2e^(-) rarr Cu`,takes place at:

A

`(63.5)/(96.5)`

B

`(63.5)/(96500 xx 2)`

C

`(63.5 xx2)/(96500)`

D

`(96500)/(63.5 xx 2)`

Text Solution

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The correct Answer is:
To solve the question regarding the electrolysis of `CuSO4` and the reaction `Cu^(2+) + 2e^(-) → Cu`, we will follow these steps: ### Step-by-Step Solution: 1. **Understand the Electrolysis Process**: - Electrolysis involves the breakdown of a compound into its components using electrical energy. In this case, we are looking at the electrolysis of copper sulfate (`CuSO4`). 2. **Identify the Components**: - Copper sulfate dissociates in solution to give `Cu^(2+)` ions and `SO4^(2-)` ions. The relevant reaction for copper ions is `Cu^(2+) + 2e^(-) → Cu`. 3. **Identify the Electrodes**: - In electrolysis, there are two electrodes: the anode (positive electrode) and the cathode (negative electrode). - The cathode is where reduction occurs (gain of electrons), and the anode is where oxidation occurs (loss of electrons). 4. **Determine the Reaction Location**: - The given reaction `Cu^(2+) + 2e^(-) → Cu` indicates that copper ions are gaining electrons to form solid copper. This is a reduction reaction. - Since reduction occurs at the cathode, we can conclude that this reaction takes place at the cathode. 5. **Final Answer**: - Therefore, the reaction `Cu^(2+) + 2e^(-) → Cu` takes place at the **cathode** during the electrolysis of `CuSO4`.
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