J's answer to Jane's Secondary 4 Chemistry Singapore question.
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Copper is a reactive electrode.
For electrolysis, on the anode side (+) you will have Cu metal atoms of the anode being oxidised to Cu²+ ions, which will enter the electrolyte solution.
Cu(s) → Cu²+ (aq) + 2e
This increases the concentration of Cu²+ ions in the solution.
On the cathode side (-) you will have Cu²+ ions in solution being reduced to Cu metal and being deposited onto the copper cathode.
This decreases the concentration of Cu²+ ions in the solution.
The increase and decrease cancels out. The net concentration of Cu²+ ions in the solution is unchanged. The concentration of sulfate ions is unchanged as well since they do not react and are only spectator ions.
Further reading : https://www.docbrown.info/page01/ExIndChem/electrochemistry04.htm
For electrolysis, on the anode side (+) you will have Cu metal atoms of the anode being oxidised to Cu²+ ions, which will enter the electrolyte solution.
Cu(s) → Cu²+ (aq) + 2e
This increases the concentration of Cu²+ ions in the solution.
On the cathode side (-) you will have Cu²+ ions in solution being reduced to Cu metal and being deposited onto the copper cathode.
This decreases the concentration of Cu²+ ions in the solution.
The increase and decrease cancels out. The net concentration of Cu²+ ions in the solution is unchanged. The concentration of sulfate ions is unchanged as well since they do not react and are only spectator ions.
Further reading : https://www.docbrown.info/page01/ExIndChem/electrochemistry04.htm
Date Posted:
3 years ago
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