Free MCAT-style practice, 4C
MCAT Electrochemistry Practice Questions: Galvanic and Electrolytic Cells
Nine free MCAT-style electrochemistry practice questions on galvanic and electrolytic cells, cell EMF, free energy, the Nernst equation and Faraday's law, with full explanations.
Quick concept primer
Every electrochemical cell has oxidation at the anode and reduction at the cathode, and electrons travel through the external wire from anode to cathode. A galvanic cell runs a spontaneous reaction: Ecell = Ecathode − Eanode is positive, ΔG = −nFE is negative, and the anode is the negative electrode. An electrolytic cell uses an outside power source to drive a nonspontaneous reaction, so ΔG is positive and the anode is the positive electrode. Away from standard conditions, the Nernst equation at 25 °C gives E = E° − (0.0592/n) log Q.
Faraday's law links charge to chemistry: moles of electrons = It/F, with F about 96,485 C per mole of electrons. The 4C electricity and electrochemistry pack adds flashcards and a larger question set on these cells and on circuits, and the free sampler lets you try sample cards first.
Traps: never multiply a reduction potential by a stoichiometric coefficient, anions in a salt bridge drift toward the anode, and "anode" always means oxidation even when its sign flips.
9 practice questions
Question 1
A student builds a galvanic cell from a silver electrode in Ag+ solution and a nickel electrode in Ni2+ solution. Take E° for Ag+/Ag as +0.80 V and for Ni2+/Ni as −0.25 V. What is the standard cell potential?
- 0.55 V
- 1.85 V
- 0.30 V
- 1.05 V
Show answer and explanation
Answer: D. 1.05 V
Silver has the higher reduction potential, so it is the cathode and nickel is oxidized at the anode. Ecell = 0.80 − (−0.25) = 1.05 V. The 0.55 V choice adds the two values with the wrong sign, 1.85 V doubles the silver potential because two Ag+ ions are reduced per Ni, which is wrong because potentials are intensive, and 0.30 V is not a valid combination.
Question 2
A cell reaction transfers 2 moles of electrons per mole of reaction and has E° = +1.05 V. Using F = 96,485 C/mol, what is ΔG° for the reaction?
- −101 kJ/mol
- −203 kJ/mol
- +203 kJ/mol
- −405 kJ/mol
Show answer and explanation
Answer: B. −203 kJ/mol
ΔG° = −nFE° = −(2)(96,485)(1.05) = −202,600 J/mol, about −203 kJ/mol. A positive potential always gives a negative free energy, so +203 kJ/mol has the wrong sign. The −101 kJ/mol choice uses n = 1, and −405 kJ/mol uses n = 4, double counting the electrons.
Question 3
A technician plates copper from a Cu2+ solution using a steady current of 2.50 A for 30.0 minutes. About what mass of copper is deposited? Use 63.5 g/mol for copper and F = 96,485 C/mol.
- 1.48 g
- 0.74 g
- 2.96 g
- 0.023 g
Show answer and explanation
Answer: A. 1.48 g
Charge is Q = It = (2.50 A)(1800 s) = 4500 C, which is 4500/96,485 = 0.0466 mol of electrons. Each Cu2+ needs 2 electrons, so 0.0233 mol of copper forms, which is 0.0233 × 63.5 = 1.48 g. The 2.96 g choice forgets to divide by 2, 0.74 g divides by 2 twice, and 0.023 g reports moles of copper, not grams.
Question 4
A concentration cell has two copper electrodes, one in 1.0 M Cu2+ and the other in 0.0010 M Cu2+, at 25 °C. Using E = E° − (0.0592/n) log Q, what is the cell potential?
- 0 V
- 0.030 V
- 0.089 V
- 0.178 V
Show answer and explanation
Answer: C. 0.089 V
In a concentration cell E° = 0. The dilute side is the anode, so Q = 0.0010/1.0 = 10−3 and E = −(0.0592/2)(−3) = 0.089 V. Zero would only hold if the concentrations were equal. The 0.178 V choice uses n = 1 instead of 2, and 0.030 V uses a factor of 1 instead of 3 for the logarithm.
Question 5
In a working galvanic cell with a potassium nitrate salt bridge, which statement is correct?
- Electrons flow through the external wire from the anode to the cathode.
- Nitrate ions in the salt bridge migrate toward the cathode.
- Oxidation takes place at the positive electrode.
- The mass of the anode increases as the cell runs.
Show answer and explanation
Answer: A. Electrons flow through the external wire from the anode to the cathode.
Oxidation at the anode releases electrons, which travel through the wire to the cathode, where reduction uses them. Nitrate anions move toward the anode to balance the positive charge building up there, so B is reversed. In a galvanic cell the anode is the negative electrode, and a metal anode usually loses mass as it dissolves into ions, so C and D are wrong.
Question 6
All of the following are true of an electrolytic cell EXCEPT:
- ΔG for the reaction being driven is positive.
- Reduction occurs at the cathode.
- The anode is connected to the positive terminal of the power supply.
- The cathode is the positive electrode.
Show answer and explanation
Answer: D. The cathode is the positive electrode.
In an electrolytic cell the power supply pushes electrons onto the cathode, making it the negative electrode, so D is the false statement. The driven reaction is nonspontaneous, so its ΔG is positive. Reduction always occurs at the cathode in any cell, and the anode is wired to the positive terminal, which pulls electrons out of it.
Question 7
Pure molten sodium chloride, with no water present, is electrolyzed between inert electrodes. What forms at each electrode?
- Sodium metal at the cathode and chlorine gas at the anode
- Chlorine gas at the cathode and sodium metal at the anode
- Hydrogen gas at the cathode and oxygen gas at the anode
- Hydrogen gas at the cathode and chlorine gas at the anode
Show answer and explanation
Answer: A. Sodium metal at the cathode and chlorine gas at the anode
The only species present are Na+ and Cl−. Na+ is reduced to sodium metal at the cathode and Cl− is oxidized to Cl2 at the anode. Choice B swaps oxidation and reduction. Hydrogen or oxygen gas would require water, as in aqueous brine, which this molten salt does not contain.
Question 8
A rechargeable cell is connected to a charger. Which statement best describes the cell while it is charging?
- It operates as a galvanic cell with a positive Ecell for the reaction occurring.
- It operates as an electrolytic cell, and the electrode that was the anode during discharge now undergoes reduction.
- The reaction occurring has a negative ΔG, so no outside energy is needed.
- Electrons flow through the external circuit in the same direction as during discharge.
Show answer and explanation
Answer: B. It operates as an electrolytic cell, and the electrode that was the anode during discharge now undergoes reduction.
Charging forces the discharge reaction to run backward, so the cell becomes electrolytic. The electrode that was oxidized during discharge is now reduced, which makes it the cathode during charging. The driven reaction is nonspontaneous, so its ΔG is positive and its Ecell is negative, ruling out A and C, and the electron flow reverses, ruling out D.
Question 9
A researcher measures standard reduction potentials for four invented couples:
| Half-reaction | E° (V) |
|---|---|
| W2+ + 2e− → W | −1.20 |
| X+ + e− → X | +0.15 |
| Y3+ + 3e− → Y | −0.40 |
| Z2 + 2e− → 2Z− | +1.05 |
Which species is the strongest reducing agent?
- Z2
- Z−
- W
- W2+
Show answer and explanation
Answer: C. W
A strong reducing agent is easily oxidized, which is the product side of the half-reaction with the most negative reduction potential. That is W, at −1.20 V. Z2 is the strongest oxidizing agent, not reducing agent. Z− is the weakest reducing agent in the table, and W2+ is an oxidized form that would need to gain electrons.
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