Q.What will happen during the electrolysis of aqueous solution of $CuSO_4$ in the presence of $Cu$ electrodes? (Two or more than two options may be correct.)

The key idea is Faraday's laws of electrolysis applied to an electrolytic cell with active (copper) electrodes. Unlike inert electrodes, the copper anode itself can participate in the reaction.

1. Cathode reaction: $Cu^{2+}$ ions in solution gain electrons and deposit as copper metal:

   $Cu^{2+} + 2e^- \rightarrow Cu(s)$ — so copper deposits at the cathode.

2. Anode reaction: Since the anode is made of copper, it can oxidise (dissolve) more easily than any anion in solution (like $SO_4^{2-}$ or $OH^-$). The reaction is: …

This is a classic case of electrolysis with active electrodes. The key idea is that when the electrodes are made of the same metal as the cation in the electrolyte, the electrode reactions change completely — the anode is no longer inert, so it participates in the reaction.

Let’s break it down.

1. What are the species present in the solution?

Aqueous $CuSO_4$ dissociates into $Cu^{2+}$ and $SO_4^{2-}$ ions. Water itself is also present, providing $H^+$ and $OH^-$ ions (though in very low concentration).

So at the electrodes, we have competing possibilities:

• At cathode (reduction): $Cu^{2+}$ ions can be reduced to $Cu$ metal, or $H^+$ from water can be reduced to $H_2$ gas.
• At anode (oxidation): $SO_4^{2-}$ ions can be oxidised (very difficult), $OH^-$ from water can be oxidised to $O_2$ gas, or the copper metal of the anode itself can be oxidised to $Cu^{2+}$.

The standard reduction potentials tell us which reaction is favoured.

2. Why copper deposits at the cathode — not hydrogen

The reduction potentials (at 298 K, 1 M, vs SHE) are:

A more positive $E^\circ$ means a greater tendency to be reduced. $Cu^{2+}$ reduction is much more favourable than $H^+$ reduction. So at the cathode, copper ions are reduced to copper metal, which deposits on the cathode.

So option (i) Copper will deposit at cathode is correct.

3. What happens at the anode — the crucial difference

If the anode were inert (like platinum or graphite), the only possible oxidation would be of $OH^-$ to $O_2$:

But here the anode is copper metal. Copper can itself be oxidised:

The oxidation potential (the reverse of the reduction potential) is $+0.34\ \text{V}$ for copper dissolution, compared to $+0.40\ \text{V}$ for oxygen evolution. A lower oxidation potential means the reaction is easier. So copper metal from the anode dissolves into the solution as $Cu^{2+}$ ions, rather than oxygen being produced. …

Method: Electrode Potential Comparison for Electrolysis (Active Electrodes)

This method uses standard oxidation/reduction potentials, but with an added check: whether the electrode material itself can react (an "active" electrode) instead of only the ions in solution.

Steps:

1. List the species present.

   Aqueous $CuSO_4$ with copper electrodes: $Cu^{2+}$, $SO_4^{2-}$, water — and, crucially, the copper metal of the electrodes themselves.

2. At the cathode, compare reduction potentials:

   • $Cu^{2+} + 2e^- \rightarrow Cu(s)$; $E^\circ = +0.34\ \text{V}$
   • $2H^+ + 2e^- \rightarrow H_2(g)$; $E^\circ = 0.00\ \text{V}$

   $Cu^{2+}$ reduction is more favourable, so copper deposits at the cathode.

3. At the anode, first check whether the electrode itself can be oxidised.

   Because this anode is copper metal (an "active" electrode, not inert Pt/graphite), compare:

   • $Cu(s) \rightarrow Cu^{2+} + 2e^-$; $E^\circ_{\text{ox}} = +0.34\ \text{V}$
   • $4OH^- \rightarrow O_2 + 2H_2O + 4e^-$; $E^\circ_{\text{ox}} \approx +0.40\ \text{V}$ …

Here are the most common mistakes students make on this exact type of "active electrode electrolysis" question, along with how to avoid each.

1. Assuming Oxygen Is Released at the Anode, As With Inert Electrodes

The Mistake:

Students remember that with Pt/graphite electrodes, water is oxidised to $O_2$ at the anode, and wrongly apply the same conclusion here.

How to Avoid:

Always check the electrode material first. If the anode is made of the same metal as the cation in solution (copper anode in $CuSO_4$), the metal itself is oxidised in preference to water — no gas is evolved at all.

2. Forgetting to Compare the Electrode's Own Oxidation Potential

The Mistake:

Students only compare $SO_4^{2-}$ and $H_2O$ as candidates for anode oxidation, never considering that the copper electrode itself can be oxidised.

How to Avoid:

Write out $Cu(s) \rightarrow Cu^{2+} + 2e^-$ ($E^\circ_{\text{ox}} = +0.34\ \text{V}$) alongside $4OH^- \rightarrow O_2 + 2H_2O + 4e^-$ ($E^\circ_{\text{ox}} \approx +0.40\ \text{V}$). The lower oxidation potential (copper) is favoured, so the anode dissolves instead of evolving oxygen.

3. Thinking Copper Can Deposit at the Anode

The Mistake:

Students pick "copper deposits at anode" because copper is visibly involved at both electrodes.

How to Avoid:

Deposition is always a reduction process, and the anode is always where oxidation occurs. Copper metal is produced by oxidation (dissolves) at the anode, and consumed by reduction (deposits) only at the cathode — never the reverse.

4. Missing the Real-World Link to Electrorefining

The Mistake:

Students don't connect this setup to a named process, so they can't sanity-check their answer.

How to Avoid: …