JAMB - Chemistry (2006 - No. 41)
In the electrolysis of CuSO\(_4\)(g) using platinum electrodes, the reaction at the anode is
4H+ + 4e- → 2H2
4OH- + 4e- → 2H2O + O2
2OH - 2e- → 2OH
2OH- + 2OH- → 2H2O + O2
Explanation
During the electrolysis of an aqueous copper sulfate (CuSO\(_4\)) solution using platinum electrodes, copper ions (Cu\(^{2+}\)) are reduced to solid copper (Cu) at the cathode, and water (H\(_2\)O) is oxidized to oxygen gas (O\(_2\)) and hydrogen ions (H\(^+\)) at the anode.
Here's a more detailed explanation:
-
Cathode (Reduction): Copper ions (Cu\(^{2+}\)) from the solution migrate towards the cathode and gain two electrons (2e-) to form solid copper, which deposits on the electrode.
- Reaction: Cu\(^{2+}\)(aq) + 2e- → Cu(s)
-
Anode (Oxidation): Water molecules (H\(_2\)O) are oxidized at the anode, losing electrons and forming oxygen gas (O\(_2\)) and hydrogen ions (H+).
- Reaction: 4OH\(^-\) + 4e\(^-\)(l) → 2H\(_2\)O(g) + O\(_2\)
Comments (0)
