Identify the products formed at the electrodes and describe the observations made during the electrolysis of: (a) molten lead(II) bromide (b) concentrated aqueous sodium chloride (c) dilute sulfuric acid using inert electrodes made of platinum or car

Electrolysis of Different Substances

(a) Molten Lead(II) Bromide (PbBr₂)

⚡️ In the molten state, only ions are present. The inert electrodes (platinum or graphite) stay unchanged.

1. At the cathode (negative electrode) the lead ions gain electrons:
   

   \$Pb^{2+} + 2e^- \rightarrow Pb\$

2. At the anode (positive electrode) bromide ions lose electrons:
   

   \$2Br^- \rightarrow Br_2(g) + 2e^-\$

Overall reaction:

\$2PbBr2(l) \rightarrow 2Pb(l) + 2Br2(g)\$

Observations: bright yellow Br₂ gas bubbles at the anode and black lead deposits at the cathode. The molten mixture turns a bit darker as lead is removed.

(b) Concentrated Aqueous Sodium Chloride (NaCl)

💡 In a concentrated solution, water is still present but in smaller amounts. The electrodes are still inert.

1. At the cathode water is reduced:
   

   \$2H2O + 2e^- \rightarrow H2(g) + 2OH^-\$

2. At the anode chloride ions are oxidised:
   

   \$2Cl^- \rightarrow Cl_2(g) + 2e^-\$

Overall reaction (simplified):

\$2NaCl + 2H2O \rightarrow 2NaOH + Cl2(g) + H_2(g)\$

Observations: greenish‑yellow Cl₂ gas at the anode and bubbles of H₂ at the cathode. The solution becomes slightly basic because of the OH⁻ produced.

(c) Dilute Sulphuric Acid (H₂SO₄)

🔬 In dilute acid, water dominates the electrolysis. The inert electrodes remain unchanged.

1. At the cathode water is reduced:
   

   \$2H2O + 2e^- \rightarrow H2(g) + 2OH^-\$

2. At the anode water is oxidised:
   

   \$2H2O \rightarrow O2(g) + 4H^+ + 4e^-\$

Overall reaction:

\$2H2O \rightarrow 2H2(g) + O_2(g)\$ (the acid remains largely unchanged).

Observations: oxygen bubbles at the anode and hydrogen bubbles at the cathode. The solution stays acidic because the produced OH⁻ is quickly neutralised by H⁺ from the acid.