Questions: Electrolytic Cells and Non-Spontaneous Redox

Question 1

You want to electroplate copper onto a metal object using a copper sulfate solution. The object is placed in the solution as one electrode. Which role does the object play, and what occurs there?

• A. Anode — copper from the object dissolves into solution
• B. Cathode — copper ions from solution are reduced and deposited onto the object
• C. Cathode — the object loses mass as electrons are pulled away by the power supply
• D. Anode — copper ions are oxidized and plated onto the object

Answer: (B)

In electrolysis, the cathode is where reduction occurs: cations in solution gain electrons and deposit as metal. The object to be plated is connected to the negative terminal of the power supply (cathode), so Cu²⁺ ions are reduced onto its surface. The common confusion is thinking the anode receives material — it is the anode (connected to the positive terminal) where oxidation occurs, dissolving the copper metal electrode to replenish the solution.

Question 2

The standard cell potential for a galvanic cell is +0.80 V. You want to run the reverse reaction in an electrolytic cell. What is the minimum applied voltage required (ignoring overpotential)?

• A. 0 V — the reaction will proceed spontaneously once current begins to flow
• B. 0.40 V — half the galvanic potential because the cell is split into two half-reactions
• C. 0.80 V — you must supply at least the magnitude of the reversed cell potential
• D. 1.60 V — twice the galvanic potential to overcome thermodynamic barriers

Answer: (C)

The non-spontaneous (electrolytic) direction has ΔG > 0 equal in magnitude to the spontaneous direction. The minimum voltage to drive it equals the magnitude of the original cell potential, here 0.80 V. In practice, additional 'overpotential' is needed to overcome electrode kinetic barriers, but the thermodynamic minimum is the reverse cell potential. Applying less voltage simply fails to drive the reaction.

Question 3

In an electrolytic cell, the cathode is connected to the negative terminal of the external power supply.

• T. True
• F. False

Answer: True

This is the opposite convention from a galvanic cell, where the anode is the negative terminal. In electrolysis, the power supply forces electrons into the cathode (negative terminal), driving reduction of cations in solution. The anode is connected to the positive terminal, which pulls electrons away, forcing oxidation. Confusing the two conventions is the most common error when comparing galvanic and electrolytic cells.

Question 4

Electrolysis of molten sodium chloride and electrolysis of aqueous sodium chloride produce the same products at the cathode.

• T. True
• F. False

Answer: False

In molten NaCl, the only cations present are Na⁺, so sodium metal is produced at the cathode. In aqueous NaCl, water is also present, and H₂O is reduced preferentially at the cathode (lower overpotential), producing H₂ gas and OH⁻ rather than sodium metal. The competition between water and Na⁺ reduction in aqueous solution changes the product entirely — this is why the chlor-alkali industrial process (aqueous) produces H₂ and NaOH, not sodium metal.

Question 5

Why does an electrolytic cell require an external power source, and what thermodynamic condition distinguishes an electrolytic reaction from a galvanic one?

Model answer: An electrolytic cell drives a non-spontaneous redox reaction — one with ΔG > 0 and a negative standard cell potential. Because the reaction will not proceed on its own, energy must be supplied externally. A galvanic cell harnesses a spontaneous reaction (ΔG < 0, positive cell potential) to produce electrical energy. The external power supply in electrolysis does the reverse: it converts electrical energy into chemical energy, pushing electrons against the thermodynamically favored direction.

The key distinction is the sign of ΔG. Galvanic: ΔG < 0 (spontaneous, releases energy). Electrolytic: ΔG > 0 (non-spontaneous, requires energy input equal to at least |ΔG|). The minimum applied voltage corresponds to ΔG = −nFE, where E is the reverse cell potential. Understanding this thermodynamic framing makes sense of why industrial electrolysis (aluminum smelting, chlor-alkali, electroplating) is energy-intensive.