Electrochemistry MCQs — 40 Practice Questions with Answers & Explanations

Welcome to our free electrochemistry practice page. These 40 original multiple-choice questions cover electrolysis, half-equations, galvanic cells, standard electrode potentials and Faraday’s laws, and every question comes with a worked answer and explanation. They are written for GCSE, IGCSE, A-Level and AP Chemistry students, starting with core electrolysis ideas and finishing with A-Level / AP stretch calculations.

Quick formula recap: need the key equations first? Revise with our free formula sheets, then come back and test yourself here.

Electrolysis Basics

Q1. Which of the following substances is an electrolyte?

  • (A) Solid sodium chloride
  • (B) Molten sodium chloride
  • (C) Sugar solution
  • (D) Distilled water
Show Answer

Answer: B — In molten sodium chloride the ions are free to move to the electrodes. In the solid the ions are locked in a lattice, sugar forms no ions at all, and distilled water contains virtually no ions.

Q2. During electrolysis, the positive ions (cations) move towards the…

  • (A) Anode
  • (B) Positive terminal of the power supply
  • (C) Cathode
  • (D) Surface of the electrolyte
Show Answer

Answer: C — The cathode is the negative electrode in electrolysis, and opposite charges attract, so positive cations move towards it.

Q3. Molten lead(II) bromide is electrolysed using inert electrodes. The product formed at the cathode is…

  • (A) Lead
  • (B) Bromine
  • (C) Hydrogen
  • (D) Oxygen
Show Answer

Answer: A — Lead ions are the only positive ions present, and they gain electrons at the cathode: Pb2+ + 2e− → Pb.

Q4. In the electrolysis of molten lead(II) bromide, the product formed at the anode is…

  • (A) Lead
  • (B) Lead(II) oxide
  • (C) Hydrogen
  • (D) Bromine
Show Answer

Answer: D — Bromide ions lose electrons at the anode: 2Br− → Br2 + 2e−. Orange-brown bromine vapour is seen at the anode.

Q5. An ionic compound conducts electricity only when it is molten or dissolved in water because…

  • (A) The solid contains no ions at all
  • (B) The ions are free to move only in the liquid or aqueous state
  • (C) Water itself is an electrolyte
  • (D) The electrodes can only touch a liquid
Show Answer

Answer: B — The ions exist in the solid, but they are held in fixed positions in the lattice and cannot carry charge to the electrodes until the structure breaks down on melting or dissolving.

Q6. Graphite and platinum electrodes are described as inert electrodes because they…

  • (A) Do not conduct electricity
  • (B) Are always used as the cathode
  • (C) Do not react with the electrolyte or the products formed
  • (D) Melt at low temperatures
Show Answer

Answer: C — Inert electrodes conduct the current into the electrolyte but take no chemical part in the reaction, unlike active electrodes such as copper.

Q7. At the anode during electrolysis, negative ions are discharged by…

  • (A) Losing electrons, so they are oxidised
  • (B) Gaining electrons, so they are reduced
  • (C) Losing electrons, so they are reduced
  • (D) Gaining electrons, so they are oxidised
Show Answer

Answer: A — Negative ions give up their extra electrons to the positive anode. Loss of electrons is oxidation, so oxidation always happens at the anode.

Q8. In an electrolysis cell connected to a d.c. power supply, which statement is correct?

  • (A) The cathode is the positive electrode
  • (B) The anode is the negative electrode
  • (C) Both electrodes carry the same charge
  • (D) The cathode is the negative electrode and the anode is the positive electrode
Show Answer

Answer: D — The power supply pushes electrons onto the cathode, making it negative, and pulls them away from the anode, making it positive.

Q9. When molten potassium chloride is electrolysed, potassium metal forms at the cathode because each K+ ion…

  • (A) Loses one electron
  • (B) Gains one electron
  • (C) Shares a pair of electrons
  • (D) Reacts with hydroxide ions
Show Answer

Answer: B — K+ + e− → K. Gaining electrons is reduction, and reduction always happens at the cathode.

Q10. In the electrolysis of a molten metal halide, the non-metal element is formed at the…

  • (A) Cathode
  • (B) Negative terminal
  • (C) Anode
  • (D) Surface of the electrolyte
Show Answer

Answer: C — Halide ions are negative, so they move to the positive anode, where they lose electrons and form the non-metal (for example, Cl2 from chloride ions).

Electrolysis of Aqueous Solutions & Half-Equations

Q11. A dilute aqueous solution of sodium chloride is electrolysed using inert electrodes. The gas formed at the cathode is…

  • (A) Chlorine
  • (B) Oxygen
  • (C) Sodium vapour
  • (D) Hydrogen
Show Answer

Answer: D — Water provides H+ ions, and hydrogen is discharged in preference to sodium because sodium is far more reactive: 2H+ + 2e− → H2.

Q12. In the industrial electrolysis of concentrated sodium chloride solution (brine), the three useful products are…

  • (A) Hydrogen, chlorine and sodium hydroxide
  • (B) Sodium, chlorine and water
  • (C) Hydrogen, oxygen and sodium hydroxide
  • (D) Sodium, oxygen and hydrochloric acid
Show Answer

Answer: A — Hydrogen forms at the cathode, chlorine at the anode, and the sodium and hydroxide ions left behind form sodium hydroxide solution.

Q13. Copper(II) sulfate solution is electrolysed using graphite electrodes. The gas released at the anode is…

  • (A) Hydrogen
  • (B) Sulfur dioxide
  • (C) Oxygen
  • (D) Steam
Show Answer

Answer: C — Sulfate ions are not discharged, so hydroxide ions from the water lose electrons instead: 4OH− → O2 + 2H2O + 4e−.

Q14. When copper(II) sulfate solution is electrolysed with inert electrodes, the half-equation for the reaction at the cathode is…

  • (A) Cu(s) → Cu2+(aq) + 2e−
  • (B) Cu2+(aq) + 2e− → Cu(s)
  • (C) 2H+(aq) + 2e− → H2(g)
  • (D) Cu2+(aq) + e− → Cu+(aq)
Show Answer

Answer: B — Copper is less reactive than hydrogen, so Cu2+ ions are discharged in preference to H+ ions and a pink-brown layer of copper coats the cathode.

Q15. Copper(II) sulfate solution is electrolysed using copper electrodes, as in the purification of copper. Which observation is correct?

  • (A) The cathode loses mass and the blue colour fades
  • (B) The anode gains mass
  • (C) The blue colour of the solution becomes deeper
  • (D) The anode loses mass, the cathode gains mass, and the blue colour stays the same
Show Answer

Answer: D — Copper dissolves from the anode (Cu → Cu2+ + 2e−) at the same rate as it deposits on the cathode, so the concentration of Cu2+ — and the blue colour — stays constant.

Q16. Which half-equation correctly shows the discharge of chloride ions at the anode?

  • (A) 2Cl− → Cl2 + 2e−
  • (B) 2Cl− + 2e− → Cl2
  • (C) Cl2 + 2e− → 2Cl−
  • (D) 2Cl− → Cl2 + 4e−
Show Answer

Answer: A — Chloride ions lose electrons (oxidation at the anode), and the charge balances: −2 on the left, and 0 + 2 × (−1) = −2 on the right. Option D fails because the charges do not balance.

Q17. An aqueous solution contains both potassium ions and hydrogen ions. Hydrogen, not potassium, is discharged at the cathode because…

  • (A) Potassium ions carry a bigger charge
  • (B) Hydrogen is a non-metal and potassium is a metal
  • (C) Potassium is more reactive than hydrogen, so H+ ions are discharged in preference
  • (D) Potassium ions are too large to reach the electrode
Show Answer

Answer: C — At the cathode, the ion of the less reactive element is discharged. Metals above hydrogen in the reactivity series (such as potassium and sodium) stay in solution, and hydrogen is given off instead.

Q18. A steel key is to be electroplated with silver. The correct arrangement is…

  • (A) Key as the anode, silver rod as the cathode, silver nitrate solution as the electrolyte
  • (B) Key as the cathode, silver rod as the anode, silver nitrate solution as the electrolyte
  • (C) Key as the cathode, graphite anode, copper(II) sulfate solution as the electrolyte
  • (D) Key as the anode, graphite cathode, silver nitrate solution as the electrolyte
Show Answer

Answer: B — The object to be plated is always the cathode, where Ag+ + e− → Ag deposits a layer of silver. The silver anode dissolves to replace the Ag+ ions used up.

Q19. In the chlor-alkali industry, the chlorine produced from brine is mainly used…

  • (A) To kill micro-organisms in water treatment, and to make bleach and PVC
  • (B) As a fuel for rockets
  • (C) To make margarine from vegetable oils
  • (D) To fill weather balloons
Show Answer

Answer: A — Chlorine sterilises drinking water and swimming pools and is a raw material for bleach and the plastic PVC. Making margarine is a use of the hydrogen from brine electrolysis, not the chlorine.

Q20. Dilute sulfuric acid is electrolysed with inert electrodes. Compared with the volume of oxygen collected at the anode, the volume of hydrogen collected at the cathode is…

  • (A) Half as much
  • (B) Exactly the same
  • (C) Four times as much
  • (D) Twice as much
Show Answer

Answer: D — Forming one H2 molecule takes 2 electrons (2H+ + 2e− → H2), while forming one O2 molecule releases 4 electrons (4OH− → O2 + 2H2O + 4e−). So for every 4 electrons, 2 H2 form but only 1 O2 — a 2 : 1 volume ratio.


Galvanic Cells & Standard Electrode Potentials

Q21. E°(Zn2+/Zn) = −0.76 V and E°(Cu2+/Cu) = +0.34 V. The standard cell potential of the Daniell cell, Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s), is…

  • (A) −1.10 V
  • (B) +0.42 V
  • (C) +1.10 V
  • (D) −0.42 V
Show Answer

Answer: C — E°cell = E°cathode − E°anode = (+0.34) − (−0.76) = +1.10 V. Copper is the positive electrode (cathode) and zinc is the anode.

Q22. A redox reaction is predicted to be feasible under standard conditions when the value of E°cell is…

  • (A) Negative
  • (B) Exactly zero
  • (C) More negative than −0.40 V
  • (D) Positive
Show Answer

Answer: D — A positive E°cell means the reaction is thermodynamically feasible (ΔG° is negative). A negative value means the reverse reaction is the feasible one.

Q23. The purpose of the salt bridge in a galvanic cell is to…

  • (A) Complete the electrical circuit by allowing ions to move between the two half-cells
  • (B) Mix the two electrolytes together thoroughly
  • (C) Increase the voltage produced by the cell
  • (D) Act as a catalyst for the electrode reactions
Show Answer

Answer: A — Ions flow through the salt bridge to balance the charges building up in each half-cell. Without it, charge would accumulate and the cell would stop working almost immediately.

Q24. E°(Fe2+/Fe) = −0.44 V and E°(Cu2+/Cu) = +0.34 V. For the reaction Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s), E°cell is…

  • (A) −0.78 V
  • (B) +0.78 V
  • (C) +0.10 V
  • (D) −0.10 V
Show Answer

Answer: B — E°cell = (+0.34) − (−0.44) = +0.78 V. The positive value confirms what the reactivity series predicts: iron displaces copper from copper(II) sulfate solution.

Q25. Standard electrode potentials are measured under standard conditions, which are…

  • (A) Solutions of 0.10 mol dm−3 at 273 K
  • (B) Solutions of any concentration at 298 K
  • (C) Solutions of 1.00 mol dm−3, a temperature of 298 K and a pressure of 100 kPa
  • (D) Solutions of 2.00 mol dm−3 at 373 K
Show Answer

Answer: C — Standard conditions for electrode potentials are 1.00 mol dm−3 for all solutions, 298 K (25 °C) and 100 kPa for any gases involved.

Q26. Which is the correct conventional cell diagram for the cell made from Zn2+/Zn (−0.76 V) and Cu2+/Cu (+0.34 V)?

  • (A) Cu(s)|Cu2+(aq)||Zn2+(aq)|Zn(s)
  • (B) Zn2+(aq)|Zn(s)||Cu(s)|Cu2+(aq)
  • (C) Cu2+(aq)|Cu(s)||Zn(s)|Zn2+(aq)
  • (D) Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
Show Answer

Answer: D — By convention the anode (oxidation, here Zn → Zn2+ + 2e−) is written on the left and the cathode on the right, in the order metal|ions of that metal || ions|metal, with || representing the salt bridge.

Q27. In a Cu2+/Cu half-cell, the concentration of Cu2+ ions is increased to above 1.00 mol dm−3. The electrode potential becomes…

  • (A) More negative
  • (B) More positive
  • (C) Exactly zero
  • (D) Unchanged
Show Answer

Answer: B — Raising [Cu2+] shifts the equilibrium Cu2+ + 2e− ⇌ Cu to the right (Le Chatelier’s principle). The electrode takes up electrons more readily, so E becomes more positive.

Q28. Using these E° values — Cl2/Cl− = +1.36 V, Br2/Br− = +1.09 V, Fe3+/Fe2+ = +0.77 V, I2/I− = +0.54 V — the strongest oxidising agent is…

  • (A) Cl2
  • (B) Br2
  • (C) Fe3+
  • (D) I2
Show Answer

Answer: A — The most positive E° belongs to the species with the greatest tendency to gain electrons, and gaining electrons is what an oxidising agent does. Chlorine is therefore the strongest oxidising agent here — it can even oxidise Br− and I− to the halogens.

Q29. E°(Mg2+/Mg) = −2.37 V and E°(Cu2+/Cu) = +0.34 V. A magnesium–copper galvanic cell has a standard EMF of…

  • (A) +2.03 V
  • (B) −2.71 V
  • (C) +2.37 V
  • (D) +2.71 V
Show Answer

Answer: D — E°cell = E°cathode (Cu) − E°anode (Mg) = (+0.34) − (−2.37) = +2.71 V. Option A comes from subtracting the wrong way round (2.37 − 0.34).

Q30. In a working galvanic cell, electrons flow through the external wire…

  • (A) From the cathode to the anode
  • (B) Through the salt bridge from one half-cell to the other
  • (C) From the anode, where oxidation happens, to the cathode
  • (D) In both directions at the same time
Show Answer

Answer: C — Oxidation at the anode releases electrons, which travel through the wire to the cathode, where they are used in reduction. In a galvanic cell the anode is the negative electrode. Ions, not electrons, move through the salt bridge.

Faraday’s Laws & Quantitative Electrochemistry (A-Level / AP stretch)

For this section, use F = 96,500 C mol−1 and Q = I × t (with t in seconds).

Q31. A current of 2.00 A is passed through an electrolyte for 30.0 minutes. The total charge passed is…

  • (A) 3600 C
  • (B) 60.0 C
  • (C) 360 C
  • (D) 36,000 C
Show Answer

Answer: A — Q = I × t. Convert the time first: 30.0 min = 30.0 × 60 = 1800 s. So Q = 2.00 × 1800 = 3600 C. Option B is the trap for forgetting to convert minutes into seconds.

Q32. How many moles of electrons are transferred when 19,300 C of charge passes through a cell? (F = 96,500 C mol−1)

  • (A) 0.0200 mol
  • (B) 0.200 mol
  • (C) 2.00 mol
  • (D) 5.00 mol
Show Answer

Answer: B — Moles of electrons = Q / F = 19,300 / 96,500 = 0.200 mol.

Q33. A charge of 0.100 mol of electrons is passed through copper(II) sulfate solution (Cu2+ + 2e− → Cu; Ar(Cu) = 63.5). The mass of copper deposited at the cathode is…

  • (A) 6.35 g
  • (B) 1.59 g
  • (C) 12.7 g
  • (D) 3.18 g
Show Answer

Answer: D — Each Cu2+ ion needs 2 electrons, so moles of Cu = 0.100 / 2 = 0.0500 mol. Mass = 0.0500 × 63.5 = 3.18 g. Option A ignores the 2 electrons in the half-equation.

Q34. A current of 1.50 A is passed through silver nitrate solution for 20.0 minutes (Ag+ + e− → Ag; Ar(Ag) = 108). The mass of silver deposited is…

  • (A) 0.503 g
  • (B) 4.03 g
  • (C) 2.01 g
  • (D) 20.1 g
Show Answer

Answer: C — Q = 1.50 × (20.0 × 60) = 1800 C. Moles of electrons = 1800 / 96,500 = 0.0187 mol, and since Ag+ takes 1 electron, moles of Ag = 0.0187 mol. Mass = 0.0187 × 108 = 2.01 g.

Q35. Molten aluminium oxide is electrolysed to extract aluminium (Al3+ + 3e− → Al; Ar(Al) = 27.0). The charge that must pass to produce 5.40 g of aluminium is…

  • (A) 57,900 C
  • (B) 19,300 C
  • (C) 38,600 C
  • (D) 173,700 C
Show Answer

Answer: A — Moles of Al = 5.40 / 27.0 = 0.200 mol. Each Al3+ needs 3 electrons, so moles of electrons = 3 × 0.200 = 0.600 mol. Q = 0.600 × 96,500 = 57,900 C. Options B and C come from using 1 or 2 electrons instead of 3.

Q36. A charge of 3860 C is passed through dilute sulfuric acid. The volume of hydrogen formed at the cathode, measured at RTP (molar gas volume = 24.0 dm3 mol−1), is…

  • (A) 960 cm3
  • (B) 480 cm3
  • (C) 240 cm3
  • (D) 48.0 cm3
Show Answer

Answer: B — Moles of electrons = 3860 / 96,500 = 0.0400 mol. Since 2H+ + 2e− → H2, moles of H2 = 0.0400 / 2 = 0.0200 mol. Volume = 0.0200 × 24.0 = 0.480 dm3 = 480 cm3. Option A forgets the 2 electrons per H2 molecule.

Q37. A charge of 9650 C is passed through acidified water. Using the anode half-equation 4OH− → O2 + 2H2O + 4e−, the volume of oxygen formed at RTP (molar gas volume = 24.0 dm3 mol−1) is…

  • (A) 2400 cm3
  • (B) 1200 cm3
  • (C) 300 cm3
  • (D) 600 cm3
Show Answer

Answer: D — Moles of electrons = 9650 / 96,500 = 0.100 mol. One O2 molecule is released per 4 electrons, so moles of O2 = 0.100 / 4 = 0.0250 mol. Volume = 0.0250 × 24.0 = 0.600 dm3 = 600 cm3.

Q38. For the Daniell cell, E°cell = +1.10 V and 2 moles of electrons are transferred per mole of reaction (ΔG° = −nFE°). The standard Gibbs free energy change is…

  • (A) −106 kJ mol−1
  • (B) +212 kJ mol−1
  • (C) −212 kJ mol−1
  • (D) −424 kJ mol−1
Show Answer

Answer: C — ΔG° = −nFE° = −(2 × 96,500 × 1.10) = −212,300 J mol−1 ≈ −212 kJ mol−1. The negative sign matches the positive E°cell: the reaction is feasible. Option A uses n = 1, and option D uses n = 4.

Q39. A Zn2+/Zn half-cell is set up with [Zn2+] = 0.010 mol dm−3 instead of the standard 1.00 mol dm−3. Compared with E°, its electrode potential is…

  • (A) More negative, because the equilibrium Zn2+ + 2e− ⇌ Zn shifts to the left
  • (B) More positive, because the equilibrium shifts to the right
  • (C) Unchanged, because concentrations never affect E
  • (D) Exactly zero
Show Answer

Answer: A — Diluting the Zn2+ ions shifts the equilibrium to the left (Le Chatelier’s principle), so the zinc gives up electrons more readily and takes them up less readily. The potential becomes more negative than −0.76 V.

Q40. Copper is deposited at the cathode during the purification of copper (Cu2+ + 2e− → Cu; Ar(Cu) = 63.5). How long must a current of 2.00 A flow to deposit 6.35 g of copper?

  • (A) 4825 s
  • (B) 9650 s
  • (C) 19,300 s
  • (D) 241 s
Show Answer

Answer: B — Moles of Cu = 6.35 / 63.5 = 0.100 mol, so moles of electrons = 2 × 0.100 = 0.200 mol and Q = 0.200 × 96,500 = 19,300 C. Then t = Q / I = 19,300 / 2.00 = 9650 s (about 2 hours 41 minutes). Option A uses only 1 electron per Cu2+ ion.

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