In the electrolysis of CuSo4(g) using platinum electrode the reaction at the anode is

Analysis of the Question

The question asks for the reaction occurring at the anode during the electrolysis of aqueous copper(II) tetraoxosulphate(VI) solution ($\text{CuSO}_4$) using platinum (inert) electrodes.

Concept Explanation

In the electrolysis of an aqueous solution, we must consider the ions present in the electrolyte and the nature of the electrodes:

1. Ions present: $\text{Cu}^{2+}$ and $\text{SO}_4^{2-}$ from the salt, and $\text{H}^+$ and $\text{OH}^-$ from the water.
2. The Anode (Positive Electrode): Anions ($\text{SO}_4^{2-}$ and $\text{OH}^-$) migrate to the anode. According to the electrochemical series, the hydroxide ion ($\text{OH}^-$) is preferentially discharged over the sulphate ion ($\text{SO}_4^{2-}$).
3. The Reaction: When $\text{OH}^-$ ions reach the anode, they lose electrons (oxidation) to form water and oxygen gas.

The balanced half-equation for the discharge of hydroxide ions is:

Rearranging this to show the loss of electrons (as requested in your options):

Or, moving the electrons to the other side:

Step-by-Step Solution

• Step 1: Identify the anions migrating to the anode: $\text{SO}_4^{2-}$ and $\text{OH}^-$.
• Step 2: Apply the preferential discharge rule. $\text{OH}^-$ is discharged more easily than $\text{SO}_4^{2-}$.
• Step 3: Write the oxidation half-equation for $\text{OH}^-$. The hydroxide ion loses an electron to become a neutral hydroxyl radical, which then combines to form water and oxygen gas.
• Step 4: Balance the equation. To produce one molecule of $\text{O}_2$, we need four $\text{OH}^-$ ions, which release four electrons.
• Step 5: Compare with the provided options. Option B represents the discharge of hydroxide ions correctly: $4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^-$. (Note: The notation in the option $4\text{OH}^- = 4e^- + 2\text{H}_2\text{O} + \text{O}_2$ is a standard way of expressing this in exam formats).

Conclusion

The correct reaction at the anode is the oxidation of hydroxide ions to produce oxygen gas and water.

Correct Option: B

Exam Tip

Remember the mnemonic "OH- Anode": In aqueous electrolysis, if the anion is not a halide (like $\text{Cl}^-$, $\text{Br}^-$, or $\text{I}^-$), the hydroxide ion ($\text{OH}^-$) from water will almost always be the one discharged at the anode, resulting in the evolution of Oxygen gas.