Worked example 1
Electrolysis of molten NaCl. Write the electrode reactions and identify the sign of each electrode.
Try it first: Notice that with no water present, only Na⁺ and Cl⁻ can react.
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What you'll be able to do: Compare electrolytic and galvanic cells, assign electrode signs correctly, and predict the products of an electrolysis.
Best after: Standard Reduction Potentials and Cell Potential
A galvanic cell lets a favorable reaction pay you in electricity. An electrolytic cell does the reverse: you pay electricity to force an unfavorable reaction to run. Aluminum, chlorine, and every chrome bumper come from that bargain.
These are recommended, not required. You can start this lesson at any time.
If E°cell for a reaction is negative, the reverse process is spontaneous, but the forward one can still be forced by supplying electrical energy. That is electrolysis. Charging a rechargeable battery is the same idea: the cell operates galvanically while discharging and electrolytically while charging.
applied voltage > |E°cell|
Oxidation is at the anode and reduction is at the cathode in both cell types. What changes is polarity. In a galvanic cell the anode is negative because it produces electrons; in an electrolytic cell the anode is positive because the power supply is pulling electrons out of it.
Thermodynamics gives the minimum voltage; reality demands more. Extra voltage, called overpotential, is needed to overcome the slow kinetics of forming gas bubbles at an electrode surface. Oxygen evolution has a particularly large overpotential, which is a major reason chlorine, not oxygen, is produced when concentrated brine is electrolyzed.
In an aqueous solution the solvent can be reduced (2 H₂O + 2 e⁻ → H₂ + 2 OH⁻) or oxidized (2 H₂O → O₂ + 4 H⁺ + 4 e⁻). The species with the more positive reduction potential is reduced at the cathode, and the species easiest to oxidize reacts at the anode. Since Na⁺ reduction is at -2.71 V, far below water, aqueous sodium chloride yields hydrogen gas, not sodium metal.
2 H₂O + 2 e⁻ → H₂ + 2 OH⁻, E = -0.83 V at pH 7
To make sodium metal you must remove water entirely. In the Downs cell, molten NaCl is electrolyzed: sodium metal forms at the cathode and chlorine gas at the anode. Aluminum is produced the same way from alumina dissolved in molten cryolite, a process that consumes a large fraction of the electricity used by heavy industry.
2 NaCl(l) → 2 Na(l) + Cl₂(g)
Electroplating deposits a thin metal layer by making the object the cathode in a solution of the plating metal. Anodizing thickens the oxide layer on aluminum by making it the anode. Both are electrolysis with a decorative or protective purpose.
A cell in which an external source drives a nonspontaneous redox reaction.
Electrolytic: anode positive, cathode negative. Galvanic: the reverse.
Oxidation at the anode and reduction at the cathode in both cell types.
The species easiest to reduce plates out; the species easiest to oxidize reacts at the anode.
Extra voltage beyond the thermodynamic minimum, needed to overcome slow electrode kinetics.
V(applied) > |E°cell| + overpotential
2 H₂O + 2 e⁻ → H₂(g) + 2 OH⁻
2 H₂O → O₂(g) + 4 H⁺ + 4 e⁻
Electrolysis of molten NaCl. Write the electrode reactions and identify the sign of each electrode.
Try it first: Notice that with no water present, only Na⁺ and Cl⁻ can react.
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Why does electrolysis of aqueous NaCl produce hydrogen instead of sodium metal at the cathode?
Try it first: Compare the two possible cathode reactions on a potential scale.
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A reaction has E°cell = -1.23 V. What can be said about the voltage required to drive it?
Try it first: Separate the thermodynamic minimum from the practical requirement.
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Why it's wrong: The definitions never change; only the polarity does.
Check instead: Anode = oxidation, always.
Why it's wrong: The power supply forces electrons onto the cathode, making it negative.
Check instead: Follow the electrons from the supply.
Why it's wrong: Water is reduced far more easily than Na⁺.
Check instead: Always include water among the candidate reactions in aqueous electrolysis.
Why it's wrong: Overpotential from slow gas-forming kinetics adds a real extra requirement.
Check instead: Treat |E°cell| as a lower bound.
Why it's wrong: Energy is supplied from outside, so the overall process still obeys the second law.
Check instead: Account for the external work in ΔG.
No practice questions are available for this topic yet. You can still practice the whole unit.
An electrolytic cell uses an external power supply to drive a reaction with a negative E°cell and a positive ΔG°. Oxidation still occurs at the anode and reduction still occurs at the cathode, but the electrode signs reverse relative to a galvanic cell: in electrolysis the anode is positive and the cathode is negative, because the power supply pulls electrons away from the anode and pushes them onto the cathode. The applied voltage must exceed the magnitude of E°cell, and in practice an additional overpotential is needed to overcome kinetic barriers, especially for gas-forming reactions. When a solution contains several possible reactions, the species that is easiest to reduce plates at the cathode and the species that is easiest to oxidize reacts at the anode; in aqueous solution, water itself is often the competitor, which is why electrolyzing aqueous NaCl gives hydrogen and chlorine rather than sodium metal. Molten salts remove water from the competition, which is how reactive metals such as sodium and aluminum are produced industrially.
Access for free at openstax.org License
Access for free at openstax.org License