Electrolytic Cells ⚡

An electrolytic cell uses an external battery to drive a non-spontaneous redox reaction. Picture two copper strips dipped in CuSO4 solution. When you connect a DC supply:

  • Cathode (–): \( \mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)} \) — shiny copper coats the strip 🥳:contentReference[oaicite:0]{index=0}
  • Anode (+): \( \mathrm{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-} \) — the strip dissolves.:contentReference[oaicite:1]{index=1}

This trick purifies impure copper on an industrial scale. You also win metals like Na, Mg and Al by electrolysing their fused salts or oxides when no chemical reducing agent works.:contentReference[oaicite:2]{index=2}

Faraday’s Fun Maths ⚖️

  • First Law: Mass of substance ∝ charge passed.
  • Second Law: For the same charge, liberated masses ∝ equivalent weights.:contentReference[oaicite:3]{index=3}

You measure charge with \( Q = I\,t \) (ampere × seconds).:contentReference[oaicite:4]{index=4}
One mole of electrons carries the famous Faraday constant: \( F = 9.6487 \times 10^{4}\,\text{C mol}^{-1} \) (≈ 96 500 C).:contentReference[oaicite:5]{index=5}

Example (Cu plating): 1.5 A for 10 min gives 900 C. The reaction needs 2 F per mole, so copper gained = \( \dfrac{63\times900}{2\times96487} = 0.294 \text{g} \). Easy marks! 💪:contentReference[oaicite:6]{index=6}

Need–to–know reductions:

  • \( \mathrm{Mg^{2+}(l) + 2e^- \rightarrow Mg(s)} \) (2 F):contentReference[oaicite:7]{index=7}
  • \( \mathrm{Al^{3+}(l) + 3e^- \rightarrow Al(s)} \) (3 F):contentReference[oaicite:8]{index=8}

What Decides the Products? 🧪

The outcome depends on three buddies:

  1. Ions present in the melt/solution.
  2. Electrode type (inert Pt/Au vs reactive metal).:contentReference[oaicite:9]{index=9}
  3. Electrode potentials \(E^\circ\) – the more positive reduction potential wins at the cathode; the more positive oxidation potential wins at the anode (allowing for “extra push” called overpotential).:contentReference[oaicite:10]{index=10}

Molten vs Aqueous NaCl 🔬

ScenarioCathode (reduction)Anode (oxidation)Main Products
Molten NaCl\( \mathrm{Na^{+}+e^- \rightarrow Na} \)\( \mathrm{Cl^- \rightarrow \tfrac12 Cl_2 + e^-} \)Na metal + Cl2 gas
Aqueous NaCl\( \mathrm{H^{+}+e^- \rightarrow \tfrac12 H_2} \) (favoured over Na+ because \(E^\circ=0\,\text{V}>-2.71\,\text{V}\))\( \mathrm{Cl^- \rightarrow \tfrac12 Cl_2 + e^-} \) (beats water due to O2 overpotential)H2, Cl2, and NaOH in solution

Overall: \( \mathrm{NaCl(aq) + H_2O(l) \rightarrow Na^{+}(aq) + OH^{-}(aq) + \tfrac12 H_2(g) + \tfrac12 Cl_2(g)} \).

Sulphuric Acid Electrolysis 💧

Anode choices:

  • \( \mathrm{2H_2O(l) \rightarrow O_2(g) + 4H^{+} + 4e^-} \) \(E^\circ = 1.23 \text{V}\)
  • \( \mathrm{2SO_4^{2-}(aq) \rightarrow S_2O_8^{2-}(aq) + 2e^-} \) \(E^\circ = 1.96 \text{V}\)

In dilute acid, water wins, giving oxygen. In concentrated acid, persulfate reigns.🧑‍🔬

High-Yield NEET Nuggets 🎯

  1. Master Faraday’s laws and the quick-fire formula \( m = \dfrac{Q\,M}{zF} \) for mass calculations.
  2. Compare standard potentials to predict which ion gets oxidised or reduced in a mixed solution.
  3. Remember the magic number \( F \approx 96\,500 \text{C mol}^{-1} \) to convert charge ↔ moles e⁻.
  4. Know which metals require electrolysis for extraction (Na, Mg, Al 👑).
  5. Watch for overpotential and electrode material effects when products seem surprising.

😊 Happy studying  |  You’ve got this!