Q.Predict the products of electrolysis in each of the following:

Concept: Faraday’s Laws of Electrolysis — the product at each electrode depends on the standard reduction potential and the overpotential for gas evolution.

(i) AgNO₃ (aq) with silver electrodes

• Cathode: Ag⁺ has a higher reduction potential than H₂O → Ag metal deposits. …

Let's break down each case by first recalling the core principle: in electrolysis, the cation (positive ion) gets reduced at the cathode, and the anion (negative ion) gets oxidised at the anode. But in aqueous solution, water ($H_2O$) also provides $H^+$ and $OH^-$ ions (in tiny amounts) that can be reduced or oxidised instead. The species that is easier to reduce (higher reduction potential) wins at the cathode; the species that is easier to oxidise (lower reduction potential, or more negative) wins at the anode.

(i) Aqueous $AgNO_3$ with silver electrodes

Cathode: Possible reductions are:

• $Ag^+ + e^- \rightarrow Ag(s)$ ($E^\circ = +0.80$ V)
• $2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-$ ($E^\circ = -0.83$ V at pH 7)

$Ag^+$ has a much higher reduction potential, so silver metal deposits on the cathode. No contest here.

Anode: Possible oxidations are:

• $Ag(s) \rightarrow Ag^+ + e^-$ (the reverse of the above, $E^\circ = -0.80$ V)
• $2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$ ($E^\circ = -1.23$ V)
• $NO_3^-$ is very hard to oxidise (nitrate ion is a poor reducing agent)

The silver electrode itself can oxidise. Its oxidation potential ($-0.80$ V) is less negative than that of water ($-1.23$ V), meaning silver is easier to oxidise than water. So the anode dissolves: $Ag(s) \rightarrow Ag^+ + e^-$.

Overall: Silver deposits on the cathode, and the silver anode dissolves. The $AgNO_3$ concentration remains constant (the $Ag^+$ removed at the cathode is replenished by the anode). This is the principle behind electrolytic refining of silver.

(ii) Aqueous $AgNO_3$ with platinum electrodes

Cathode: Same as above — $Ag^+$ reduction is still favoured. Silver deposits on the platinum cathode.

Anode: Now the electrode is inert (platinum does not oxidise easily). So we must oxidise either water or $NO_3^-$. Comparing:

• $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ ($E^\circ = -1.23$ V)
• $NO_3^-$ oxidation would require a much higher potential (nitrate is very stable)

Water oxidation is the only feasible option. So oxygen gas bubbles off at the anode.

Overall: Silver deposits at the cathode, oxygen gas evolves at the anode, and the solution becomes acidic (because $H^+$ is produced).

(iii) Dilute $H_2SO_4$ with platinum electrodes

This is essentially electrolysis of water with a little acid to make it conductive.

Cathode: Possible reductions:

• $2H^+ + 2e^- \rightarrow H_2(g)$ ($E^\circ = 0.00$ V, but in dilute acid $[H^+]$ is low, so the actual potential is slightly less — still much higher than water reduction)
• $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ ($E^\circ = -0.83$ V)

$H^+$ reduction is strongly favoured. Hydrogen gas evolves.

Anode: Possible oxidations:

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

Water oxidation is much easier. Oxygen gas evolves. …

Faraday's Laws of Electrolysis — Product Prediction Method

Method: Electrochemical Series & Ion Discharge Tendency

This method uses the standard reduction potential ($E^\circ$) to decide which ion gets discharged (reduced at cathode / oxidised at anode) first. The rule:

• At cathode (reduction): Higher $E^\circ$ → reduced first (more positive = easier to reduce).
• At anode (oxidation): Lower $E^\circ$ → oxidised first (more negative = easier to oxidise).

Steps for any electrolysis problem:

1. Identify all ions present in the solution (from solute + water).
2. List possible cathode reactions (cations + water reduction) and compare their $E^\circ$ values.
3. List possible anode reactions (anions + water oxidation) and compare their $E^\circ$ values.
4. Check electrode material — if the anode is active (e.g., Ag, Cu, Ni), it may itself oxidise instead of the anions.
5. Write the product at each electrode.

(i) Aqueous $AgNO_3$ with silver electrodes

Step 1 — Ions present:

$Ag^+$, $NO_3^-$, $H^+$, $OH^-$ (from water)

Step 2 — Cathode:

$Ag^+ + e^- \rightarrow Ag(s)$ ($E^\circ = +0.80$ V)

$2H^+ + 2e^- \rightarrow H_2(g)$ ($E^\circ = 0.00$ V)

→ $Ag^+$ has higher $E^\circ$, so silver metal deposits.

Step 3 — Anode (active Ag electrode):

Silver electrode itself can oxidise:

$Ag(s) \rightarrow Ag^+ + e^-$ ($E^\circ = -0.80$ V)

vs. $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ ($E^\circ = -1.23$ V)

→ Ag electrode oxidises (more negative potential means easier oxidation).

Products:

• Cathode: Silver metal ($Ag$) deposits.
• Anode: Silver electrode dissolves (goes into solution as $Ag^+$).

(ii) Aqueous $AgNO_3$ with platinum electrodes

Step 1 — Ions same: $Ag^+$, $NO_3^-$, $H^+$, $OH^-$

Step 2 — Cathode: Same as above → $Ag$ deposits.

Step 3 — Anode (inert Pt):

$NO_3^-$ is very hard to oxidise (sulphate/nitrate ions are not discharged in aqueous solution).

Water oxidation: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ ($E^\circ = -1.23$ V)

$OH^-$ oxidation: $4OH^- \rightarrow O_2 + 2H_2O + 4e^-$ ($E^\circ = -0.40$ V)

→ $OH^-$ oxidises (more negative = easier).

Products:

• Cathode: Silver metal ($Ag$).
• Anode: Oxygen gas ($O_2$).

(iii) Dilute $H_2SO_4$ with platinum electrodes

Step 1 — Ions: $H^+$, $SO_4^{2-}$, $OH^-$, $H^+$ (from water)

Step 2 — Cathode:

$2H^+ + 2e^- \rightarrow H_2$ ($E^\circ = 0.00$ V)

Water reduction: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ ($E^\circ = -0.83$ V)

→ $H^+$ discharges (higher $E^\circ$).

Step 3 — Anode (inert Pt):

$SO_4^{2-}$ not discharged. Water oxidation:

$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ ($E^\circ = -1.23$ V)

→ $O_2$ evolves.

Products:

• Cathode: Hydrogen gas ($H_2$).
• Anode: Oxygen gas ($O_2$). (This is essentially electrolysis of water with $H_2SO_4$ as electrolyte.)

(iv) Aqueous $CuCl_2$ with platinum electrodes …

Here are the common mistakes students make when predicting electrolysis products, specifically for the four cases you listed, along with how to avoid each.

Mistake 1: Forgetting the Electrode Material Matters (Inert vs. Active)

The Error:

Students treat silver electrodes (Ag) the same as platinum (Pt) in $AgNO_3$ solution. They assume the anode reaction is always oxidation of water or the anion.

Why it’s wrong:

Silver is an active electrode — it can itself undergo oxidation. Platinum is inert — it does not participate chemically.

How to avoid:

• Always check the electrode material first.
• If the anode is made of a metal like Ag, Cu, Zn (which can oxidise), that metal will likely dissolve instead of water or the anion.
• For inert electrodes (Pt, graphite), focus only on the ions in solution.

Example:

In (i) $AgNO_3$ with Ag electrodes:

• Anode: Ag(s) → Ag⁺(aq) + e⁻ (not water or nitrate).
• Cathode: Ag⁺(aq) + e⁻ → Ag(s) (silver deposits). So the net effect is transfer of silver from anode to cathode.

Mistake 2: Ignoring the Concentration of the Electrolyte

The Error:

For dilute $H_2SO_4$, students write the same products as for concentrated $H_2SO_4$ — e.g., expecting $SO_4^{2-}$ to be oxidised at the anode.

Why it’s wrong:

In dilute $H_2SO_4$, the concentration of $SO_4^{2-}$ is very low. Water is more easily oxidised than sulfate ions.

How to avoid:

• Use the electrochemical series to compare standard electrode potentials.
• For dilute solutions, water’s oxidation ($E^\circ = +1.23$ V) is easier than sulfate oxidation ($E^\circ \approx +2.01$ V).
• Rule of thumb: In dilute aqueous solutions, water often gets oxidised/reduced instead of the ions.

Example:

In (iii) dilute $H_2SO_4$ with Pt electrodes:

• Anode: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$
• Cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ (Or simply $2H^+ + 2e^- \rightarrow H_2$ if pH is low.)

Mistake 3: Misidentifying Which Ion Gets Reduced at the Cathode

The Error:

In $CuCl_2$ solution, students think $Cl^-$ gets oxidised at the anode (correct) but then assume $H^+$ from water gets reduced at the cathode instead of $Cu^{2+}$.

Why it’s wrong:

$Cu^{2+}$ has a higher reduction potential ($E^\circ = +0.34$ V) than $H^+$ ($E^\circ = 0.00$ V). So $Cu^{2+}$ is reduced first.

How to avoid:

• Compare standard reduction potentials of all cations present.
• The one with the higher (more positive) reduction potential gets reduced at the cathode.
• For $CuCl_2$: cations are $Cu^{2+}$ and $H^+$ (from water). $Cu^{2+}$ wins.

Example:

In (iv) $CuCl_2$ with Pt electrodes:

• Cathode: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$ (copper deposits).
• Anode: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$ (chlorine gas).

Mistake 4: Forgetting to Check If the Anion Can Be Oxidised Before Water

The Error:

In $AgNO_3$ with Pt electrodes, students write $NO_3^-$ oxidation at the anode.

Why it’s wrong:

Nitrate ion ($NO_3^-$) is very stable — it has a very high oxidation potential (hard to oxidise). Water is oxidised much more easily.

How to avoid:

• For oxyanions like $NO_3^-$, $SO_4^{2-}$, $CO_3^{2-}$ — they are not oxidised in aqueous solution.
• Instead, water gets oxidised to oxygen gas.
• Memorise: Only halides ($Cl^-, Br^-, I^-$) and $OH^-$ are commonly oxidised before water. …