Electrolysis is an important technique for extraction of metals, and each ion of the solution needs a minimum voltage to get discharged and this value is expressed in terms of discharge potential. For some metal ions the discharge potentials follow the order given below :
`Li^(+) gt K^(+) gt Ca^(2+) gt Na^(+) gt Mg^(2+) gt Al^(3+) gt Zn^(2+) gt Fe^(2+) gt Ni^(2+) gt H_(3)O^(+) gt Cu^(2+) gt Hg_(2)^(2+) gt Ag^(+) gt Au^(3+)`
For some anions the discharge potentials are in the order :
`SO_(4)^(2-) gt NO_(3)^(-) gt OH^(-) gt Br^(-) gt I^(-)`
When aqueous solution of cupric bromide is electrolyzed the product obtained at cathode will be :

Options

A. Cu
B. `H_(2)`
C. `Br_(2)`
D. `O_(2)`

Answer

To determine the product obtained at the cathode during the electrolysis of an aqueous solution of cupric bromide (CuBr₂), we need to analyze the ions present in the solution and their respective discharge potentials.

### Step-by-Step Solution:

1. **Identify the Ions in the Solution:**
When cupric bromide is dissolved in water, it dissociates into its constituent ions:
- Cu²⁺ (copper ions)
- Br⁻ (bromide ions)
- H₃O⁺ (hydronium ions)

2. **Determine the Cathode Reaction:**
The cathode is the negative electrode where reduction occurs. Therefore, we need to consider which positive ion will be reduced at the cathode:
- Cu²⁺ (copper ion)
- H₃O⁺ (hydronium ion)

3. **Refer to Discharge Potentials:**
From the given order of discharge potentials for metal ions:
- Cu²⁺ has a lower discharge potential compared to H₃O⁺.
- The order indicates that Cu²⁺ will be discharged (reduced) before H₃O⁺ because it requires less voltage to discharge.

4. **Conclusion on the Product at Cathode:**
Since Cu²⁺ has a lower discharge potential, it will be preferentially discharged at the cathode. Thus, copper (Cu) will be deposited at the cathode during the electrolysis process.

5. **Final Answer:**
The product obtained at the cathode when the aqueous solution of cupric bromide is electrolyzed is **copper (Cu)**.