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Chemistry · Ch 5 — Electrochemistry

Electrolysis of aqueous NaCl

5.5.2

Electrolysis of aqueous NaCl

Electrolysis of an aqueous NaCl solution can be carried out in the cell used for the electrolysis of molten NaCl, with inert electrodes (Fig. 5.4): the fused NaCl is replaced by a moderately concentrated aqueous solution of NaCl. The water involved in the electrolysis of aqueous NaCl leads to electrode reactions that differ from those of molten NaCl.

Reduction half reaction at cathode : At the cathode two reduction reactions compete. One is the reduction of Na+\mathrm{Na^+} ions, as in the case of molten NaCl:

i.    Na+ (aq)+e−⟶Na (s),    E0=−2.71 V\mathrm{i.}\;\; \mathrm{Na^+\,(aq) + e^- \longrightarrow Na\,(s)}, \;\; E^0 = -2.71\ \mathrm{V}

The other is the reduction of water to hydrogen gas:

ii.    2 H2O (l)+2e−⟶H2 (g)+2 OH− (aq),    E0=−0.83 V\mathrm{ii.}\;\; \mathrm{2\,H_2O\,(l) + 2e^- \longrightarrow H_2\,(g) + 2\,OH^-\,(aq)}, \;\; E^0 = -0.83\ \mathrm{V}

The standard potential (section 5.7.1) for the reduction of water is higher than that for the reduction of Na+\mathrm{Na^+}. This implies that water has a much greater tendency to get reduced than the Na+\mathrm{Na^+} ion. Hence reaction (ii) — the reduction of water — is the cathode reaction when aqueous NaCl is electrolysed.

Oxidation half reaction at anode : At the anode there is competition between the oxidation of Cl−\mathrm{Cl^-} ion to Cl2\mathrm{Cl_2} gas, as in the case of molten NaCl, and the oxidation of water to O2\mathrm{O_2} gas:

i.    2 Cl− (aq)⟶Cl2 (g)+2e−,    Eoxi0=−1.36 V\mathrm{i.}\;\; \mathrm{2\,Cl^-\,(aq) \longrightarrow Cl_2\,(g) + 2e^-}, \;\; E^0_{oxi} = -1.36\ \mathrm{V}

ii.    2 H2O (l)⟶O2 (g)+4 H+ (aq)+4e−,    Eoxi0=−0.4 V\mathrm{ii.}\;\; \mathrm{2\,H_2O\,(l) \longrightarrow O_2\,(g) + 4\,H^+\,(aq) + 4e^-}, \;\; E^0_{oxi} = -0.4\ \mathrm{V}

Note

The book prints reaction (ii) with "2e⊖"; the balanced equation requires 4e−4e^- (four electrons accompany the four H+\mathrm{H^+}), shown correctly above.

The standard electrode potential for the oxidation of water is greater than that of the Cl−\mathrm{Cl^-} ion — water has the greater tendency to undergo oxidation, so the anode half reaction would be expected to be the oxidation of water. The experiments have shown, however, that the gas produced at the anode is Cl2\mathrm{Cl_2} and not O2\mathrm{O_2}. This suggests that the anode reaction is the oxidation of Cl−\mathrm{Cl^-} to Cl2\mathrm{Cl_2} gas — because of the overvoltage, the discussion of which is beyond the scope of this book.

Note

It has been found experimentally that the actual voltage required for electrolysis is greater than that calculated using standard potentials. This additional voltage required is the overpotential.

Note

Do you know?

Refining of metal and electroplating are achieved by electrolysis.

Overall cell reaction — it is the sum of the electrode reactions:

2 Cl− (aq)⟶Cl2 (g)+2e−(oxidation at anode)\mathrm{2\,Cl^-\,(aq) \longrightarrow Cl_2\,(g) + 2e^-} \quad \text{(oxidation at anode)}

2 H2O (l)+2e−⟶H2 (g)+2 OH− (aq)(reduction at cathode)\mathrm{2\,H_2O\,(l) + 2e^- \longrightarrow H_2\,(g) + 2\,OH^-\,(aq)} \quad \text{(reduction at cathode)}

2 Cl− (aq)+2 H2O (l)⟶Cl2 (g)+H2(g)+2 OH− (aq)(overall cell reaction)\mathrm{2\,Cl^-\,(aq) + 2\,H_2O\,(l) \longrightarrow Cl_2\,(g) + H_2(g) + 2\,OH^-\,(aq)} \quad \text{(overall cell reaction)}

Results of electrolysis of aqueous NaCl

i. H2\mathrm{H_2} gas is liberated at the cathode.

ii. Cl2\mathrm{Cl_2} gas is released at the anode. …

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