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

The key idea is Faraday’s laws of electrolysis combined with the electrochemical series — in an aqueous solution, the species that is easier to reduce (higher reduction potential) deposits at the cathode, and the species easier to oxidise (lower reduction potential) releases gas or dissolves at the anode.

1. Cathode: $Cu^{2+}$ has a reduction potential of $+0.34\ \text{V}$, while $H_2O$ (to $H_2$) has $-0.83\ \text{V}$. $Cu^{2+}$ is much easier to reduce, so copper metal deposits at the cathode.

   $Cu^{2+} + 2e^- \rightarrow Cu(s)$

2. Anode: Platinum is inert, so it does not dissolve. The competing oxidations are: …

This is a classic question on Faraday’s laws of electrolysis combined with the electrochemical series — you must decide which ions actually get discharged at each electrode when water is present. The key is to compare the reduction potentials of the possible reactions.

1. Identify all ions present in the solution

Aqueous CuSO₄ dissociates into:

• Cu²⁺ and SO₄²⁻ from the salt
• H⁺ and OH⁻ from water (autoionisation)

At the cathode (negative electrode), reduction occurs — the species that gains electrons. At the anode (positive electrode), oxidation occurs — the species that loses electrons.

2. What happens at the cathode?

Possible reduction reactions (with standard reduction potentials at 25°C):

• Cu²⁺ + 2e⁻ → Cu(s)  $E^\circ = +0.34\ \text{V}$
• 2H₂O + 2e⁻ → H₂(g) + 2OH⁻  $E^\circ = -0.83\ \text{V}$ (in neutral/alkaline conditions)

The more positive the reduction potential, the easier the reduction. Cu²⁺ has a much higher $E^\circ$ than water, so Cu²⁺ is preferentially reduced. This means copper metal deposits on the cathode.

Result at cathode: Copper deposits. Option (i) is correct.

3. What happens at the anode?

Platinum is inert — it does not participate in the reaction. So we look at oxidation of anions or water.

Possible oxidation reactions:

• 2H₂O → O₂(g) + 4H⁺ + 4e⁻  $E^\circ = +1.23\ \text{V}$ (oxygen evolution)
• 2SO₄²⁻ → S₂O₈²⁻ + 2e⁻  $E^\circ \approx +2.01\ \text{V}$ (very difficult)

The lower the oxidation potential (or the more negative the reduction potential of the reverse reaction), the easier the oxidation. Water oxidation at +1.23 V is far easier than sulphate oxidation at +2.01 V. So water is oxidised, producing oxygen gas at the anode. …

Method: Electrode Potential Comparison for Electrolysis

This method uses standard reduction/oxidation potentials to predict which species is discharged at each electrode when an inert electrode is used.

Steps:

1. List every species present in the electrolyte.

   Aqueous $CuSO_4$ with Pt electrodes: $Cu^{2+}$, $SO_4^{2-}$, and water ($H_2O$, giving trace $H^+$/$OH^-$).

2. At the cathode, compare reduction potentials — the highest wins:

   • $Cu^{2+} + 2e^- \rightarrow Cu(s)$; $E^\circ = +0.34\ \text{V}$
   • $2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-$; $E^\circ = -0.83\ \text{V}$

   $Cu^{2+}$ has the far more positive potential, so copper metal deposits at the cathode.

3. At the anode, compare oxidation potentials — the easier (lower-voltage) one wins.

   Since Pt is inert, only $H_2O$ or $SO_4^{2-}$ can be oxidised:

   • $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$; $E^\circ = +1.23\ \text{V}$
   • $2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-$; $E^\circ \approx +2.01\ \text{V}$

   Water oxidation needs far less voltage, so oxygen gas is evolved at the anode.

4. Check the electrode material itself. …

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

1. Assuming Water Is Never Involved Because a Salt Is Present

The Mistake:

Students see $CuSO_4$ (aq) and assume only $Cu^{2+}$ and $SO_4^{2-}$ can react, forgetting that water itself can be oxidised or reduced.

How to Avoid:

Always list every possible species at each electrode, including $H_2O$ (and the $H^+$/$OH^-$ it provides). With Pt electrodes, water frequently "wins" over a spectator anion like $SO_4^{2-}$.

2. Assuming Sulphate Gets Oxidised at the Anode

The Mistake:

Students pick "copper dissolves" or invent an $SO_4^{2-}$ discharge because sulphate is the only anion around.

How to Avoid:

Compare oxidation potentials: water oxidation to $O_2$ ($E^\circ = +1.23\ \text{V}$) is far easier than oxidising $SO_4^{2-}$ to peroxodisulphate ($E^\circ \approx +2.01\ \text{V}$). With inert electrodes, $SO_4^{2-}$ is essentially never discharged — oxygen evolves instead.

3. Forgetting That the Electrode Material Matters

The Mistake:

Students don't check whether the electrode itself (Pt here) can participate in the reaction, and wrongly pick "copper dissolves at anode."

How to Avoid:

Platinum is chemically inert under these conditions — it never dissolves. "Copper dissolves at the anode" is only true if the anode is made of copper metal (a different, active-electrode scenario — contrast with the next question in this set).

4. Reducing $H^+$ Instead of $Cu^{2+}$ at the Cathode

The Mistake:

Students assume $H^+$ from water is reduced first simply because it's present, ignoring the relative reduction potentials.

How to Avoid:

Compare: $Cu^{2+}+2e^-\to Cu$, $E^\circ=+0.34\ \text{V}$, versus $2H_2O+2e^-\to H_2+2OH^-$, $E^\circ=-0.83\ \text{V}$. The far more positive potential means $Cu^{2+}$ is reduced preferentially — copper deposits at the cathode, not hydrogen gas.

5. Forgetting This Is a "Select All That Apply" Question …