Q.Describe the electrolysis of molten NaCl using inert electrodes.

Step 1. The electrolytic cell uses a cylindrical steel cathode and a graphite anode, both dipped in molten NaCl, connected to an external DC power supply. Once the circuit closes, the supply forces electrons onto the cathode and pulls them from the anode.

Step 2. At the cathode (reduction), Na⁺ ions migrate toward it, accept electrons, and are reduced to liquid sodium metal: $\text{Na}^{+}\text{(l)}+e^{-} \rightarrow \text{Na(l)}$, $E^{o}=-2.71$V.

Step 3. At the anode (oxidation), Cl⁻ ions migrate toward it, lose electrons, and are oxidised to chlorine gas: $2\text{Cl}^{-}\text{(l)} \rightarrow \text{Cl}_2\text{(g)}+2e^{-}$, $E^{o}=-1.36$V.

Step 4. Combining the two half-reactions gives the overall electrolysis: $2\text{Na}^{+}\text{(l)}+2\text{Cl}^{-}\text{(l)} \rightarrow 2\text{Na(l)}+\text{Cl}_2\text{(g)}$, with overall $E^{o}=-2.71+(-1.36)=-4.07$V. …

State the cathode half-reaction (Na+ reduced), the anode half-reaction (Cl- oxidised), combine them into the overall equation, and note the negative overall E-degre …

Forgetting that the electrode SIGNS are reversed compared with a galvanic cell (here cathode is negative, anode positive), or omitting the negative …