MCAT content library / Chemistry & physics / Electrolysis: charge and product yield
Convert current and time into moles of electrons, then use the electrode half-reaction to predict how much product can form.
About 16 minutes, including the electroplating calculation and checks.
Useful foundations
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Oxidation and reduction keep their electrodes
In both galvanic and electrolytic cells, oxidation occurs at the anode and reduction occurs at the cathode. What changes is the energetic direction. A galvanic cell runs a spontaneous redox reaction to produce electrical work. An electrolytic cell uses an external power source to drive a nonspontaneous reaction.
Because of that external source, the sign labels differ from a galvanic cell: in a typical electrolytic cell, the anode is connected to the positive terminal and the cathode to the negative terminal. Do not use electrode sign as the definition of oxidation or reduction.
Concept: charge counts electrons
Current is charge flow per unit time, so the charge delivered by a constant current is
Q = It.
The Faraday constant gives the charge per mole of electrons: approximately 9.65 × 10⁴ C/mol e⁻. Therefore,
moles e⁻ = Q/F.
Once the electron amount is known, the balanced half-reaction converts electrons into moles of deposited metal, gas, or another product. This is a stoichiometry problem with charge as the starting measurement.
Worked application: copper electroplating
An original plating cell reduces copper(II) according to:
Cu²⁺ + 2e⁻ → Cu(s).
A current of 2.00 A flows for 965 s. Assume 100% current efficiency and use F = 96,500 C/mol e⁻.
The delivered charge is:
Q = (2.00 C/s)(965 s) = 1,930 C.
The electron amount is:
1,930 C ÷ 96,500 C/mol = 0.0200 mol e⁻.
Two moles of electrons deposit one mole of copper, so:
0.0200 mol e⁻ × (1 mol Cu / 2 mol e⁻) = 0.0100 mol Cu.
Using a molar mass of 63.55 g/mol gives:
0.0100 mol × 63.55 g/mol = 0.636 g Cu.
| Step | Result |
|---|---|
| Charge | 1,930 C |
| Electrons | 0.0200 mol e⁻ |
| Copper | 0.0100 mol Cu |
| Theoretical mass | 0.636 g Cu |
Pattern: current efficiency can explain a yield shortfall
Suppose the experiment produces only 0.508 g Cu. If the charge measurement is accurate, the ratio 0.508/0.636 = 0.80 indicates an 80% mass yield relative to the ideal electron-to-copper calculation. One explanation is that some current drove a competing reduction rather than copper deposition.
That conclusion should stay bounded. A low recovered mass could also result from product loss, incomplete collection, or another measurement problem. A current-efficiency claim is strongest when the electrode products are measured and the charge balance is accounted for.
Rule: rechargeable cells reverse reaction direction during charging
A rechargeable battery discharges galvanically: its spontaneous cell reaction drives current through an external circuit. Charging applies an external voltage to drive the net reaction in the reverse direction. The chemical species that are oxidized and reduced therefore switch roles relative to discharge.
The labels “anode” and “cathode” follow oxidation and reduction for the process currently occurring. In battery discussions, fixed physical-electrode names can sometimes create confusion, so read the stated charging or discharging direction before assigning the reaction.
ACTIVE PRACTICE · ELECTROCHEMICAL STOICHIOMETRY
Trace Electrical Charge Into Product Yield
For Cu²⁺ + 2e⁻ → Cu(s), start with 2.00 A for 965 s, F = 96,500 C/mol e⁻, and a copper molar mass of 63.55 g/mol. Predict the deposited copper mass at 100% current efficiency.
Charge Q = It: C. Electrons: mol. Theoretical Cu: g. Modeled Cu deposited: g.
100% of delivered charge used for copper deposition (assumed).
The efficiency slider *stipulates* the share of charge driving Cu²⁺ reduction with full product collection. A measured low mass alone cannot distinguish side reactions from product recovery losses; it does not establish current efficiency.
Check your reasoning
Choose an answer before opening its feedback.
Check 1
A constant current of 3.0 A flows for 200 s. How much charge passes?
A. 600 C
Correct. Q = It = 3.0 × 200.
B. 67 C
Review this choice. Current is multiplied by time, not divided by it.
C. 203 C
Review this choice. Adding current and time does not give charge.
D. 6,000 C
Review this choice. This is ten times the correct product.
Check 2
The cathode half-reaction is M³⁺ + 3e⁻ → M. If 0.30 mol e⁻ are delivered with 100% efficiency, how many moles of M form?
A. 0.10 mol
Correct. Three moles of electrons are required per mole of M.
B. 0.30 mol
Review this choice. This ignores the 3:1 electron-to-product stoichiometry.
C. 0.90 mol
Review this choice. The electron coefficient is a requirement, not a multiplier for product moles.
D. 3.0 mol
Review this choice. This is far larger than the electron amount permits.
Check 3
Which statement is always true for an electrochemical cell while it is operating in a specified direction?
A. Oxidation occurs at the anode.
Correct. This definition holds for galvanic and electrolytic operation.
B. The anode is always negative.
Review this choice. An electrolytic anode is typically positive.
C. Reduction occurs at the anode.
Review this choice. Reduction occurs at the cathode.
D. The reaction must be spontaneous.
Review this choice. Electrolysis drives a nonspontaneous reaction with external energy.
Continue learning
- Galvanic cells and cell potential
- Redox bookkeeping and balanced half-reactions
- Resistor circuits and measurement
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