Current Efficiency in Electrolysis: Solved Problems
Electrolysis is a core topic in Class 12 electrochemistry, tested regularly in board examinations and competitive entrance tests like JEE Main, NEET, and CUET. In standard textbook questions, students assume that 100% of the electric current produces the desired substance. In real electrochemical cells, however, parasitic reactions and internal resistance consume part of the electrical charge. Current efficiency quantifies this loss, linking theoretical Faraday calculations to actual experimental yields.
Fundamental Principles and Faraday's Laws
Michael Faraday established the quantitative laws of electrolysis. Faraday's first law states that the mass of a substance deposited or liberated at an electrode is proportional to the total electric charge passed through the electrolyte:
m = Z × I × t
Here, I is electric current in amperes, t is time in seconds, and Z is the electrochemical equivalent:
Z = M / (n × F)
In this equation, M is the molar mass in grams per mole, n is the number of electrons transferred per ion (the n-factor), and F is the Faraday constant. The exact value of F is approximately 96485 C/mol, but school boards and entrance exams routinely use the rounded value of 96500 C/mol for manual calculations. Students should follow the value specified on their exam paper.
Current Efficiency: Definition and Formula
Current efficiency (symbol η) represents the percentage of total electrical charge that produces the intended chemical product. Because product mass is directly proportional to electric charge, current efficiency equals the ratio of actual yield to theoretical yield, or the ratio of useful current to total supplied current.
Current Efficiency Formulas and Variables:
η = (Actual mass obtained / Theoretical mass calculated) × 100%
η = (Useful charge / Total charge supplied) × 100% = (I_useful / I_total) × 100%
Where:
- η = Current efficiency in percent (%)
- Actual mass = Mass of product collected at the electrode in grams (g)
- Theoretical mass = Ideal mass calculated from Faraday's law in grams (g)
- I_useful = Current utilized for the target reaction in amperes (A)
- I_total = Total electric current supplied across the cell in amperes (A)
- t = Duration of electrolysis in seconds (s)
- n = Number of electrons exchanged per ion or molecule
- M = Molar mass of product in grams per mole (g/mol)
- F = Faraday constant (96500 C/mol or 96485 C/mol)
When calculating the total time or total charge needed to deposit a specified mass at less than 100% efficiency, divide the theoretical charge by the efficiency fraction, because a real cell requires more charge and more time.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 12, State Boards, JEE Main, NEET, CUET
- Topic: Current Efficiency in Electrolysis
- Difficulty: School to foundation competitive level
Step-by-Step Solved Examples
The following numerical problems demonstrate the practical application of current efficiency principles. Every arithmetic step is shown, followed by verification using a second calculation route.
Question: A steady current of 2.50 A is passed through an aqueous copper(II) sulfate (CuSO4) solution for 30.0 minutes. The mass of copper deposited at the cathode is 1.18 g. Calculate the percentage current efficiency of the process. (Atomic mass of Cu = 63.55 g/mol, F = 96500 C/mol).
Step 1: Convert time into seconds. t = 30.0 × 60 = 1800 s.
Step 2: Calculate total charge supplied. Q_total = I × t = 2.50 × 1800 = 4500 C.
Step 3: Determine the n-factor and equivalent mass. Cu2+(aq) + 2e- → Cu(s), so n = 2. Equivalent mass E = 63.55 / 2 = 31.775 g/equiv.
Step 4: Compute theoretical mass. m_theoretical = (31.775 × 4500) / 96500 = 142987.5 / 96500 = 1.4817 g.
Step 5: Compute current efficiency. η = (1.18 / 1.4817) × 100% = 79.64%.
Verification via second route: Moles of Cu = 1.18 / 63.55 = 0.018568 mol. Useful charge Q_useful = 0.018568 × 2 × 96500 = 3583.6 C. η = (3583.6 / 4500) × 100% = 79.64%.
Answer: 79.64%
Question: Find the time in minutes needed to liberate 1.12 L of Cl2 gas at STP during the electrolysis of molten NaCl using a current of 5.00 A at 80.0% current efficiency. Note: standard molar volume at STP is 22.4 L/mol in older curricula and 22.7 L/mol under current IUPAC standard pressure (1 bar, 273.15 K). Calculate the time for 22.4 L/mol and also state the result for 22.7 L/mol. (F = 96500 C/mol).
Step 1: Write the anode half-reaction. 2Cl-(l) → Cl2(g) + 2e-, so n = 2 electrons per mole of Cl2.
Step 2: Moles and charge using 22.4 L/mol. Moles of Cl2 = 1.12 / 22.4 = 0.0500 mol. Q_useful = 0.0500 × 2 × 96500 = 9650 C.
Step 3: Total charge at 80.0% efficiency. Q_total = 9650 / 0.800 = 12062.5 C.
Step 4: Duration in minutes. t = 12062.5 / 5.00 = 2412.5 s. Time in minutes = 2412.5 / 60 = 40.21 minutes.
Step 5: Moles and charge using 22.7 L/mol. Moles of Cl2 = 1.12 / 22.7 = 0.04934 mol. Q_useful = 0.04934 × 2 × 96500 = 9522.6 C. Q_total = 9522.6 / 0.800 = 11903.3 C. Time = 11903.3 / 5.00 = 2380.7 s = 39.68 minutes.
Verification via second route: Effective current = 5.00 × 0.800 = 4.00 A. Time = 9650 / 4.00 = 2412.5 s = 40.21 minutes.
Answer: 40.21 minutes (at 22.4 L/mol) or 39.68 minutes (at 22.7 L/mol)
Common Mistakes in Current Efficiency Calculations
Key Errors to Avoid:
- Multiplying instead of dividing when finding time: A lower current efficiency means the cell requires more charge and longer time to deposit a given mass, never less. Always divide theoretical charge by the efficiency fraction.
- Neglecting time unit conversion: Current in amperes is coulombs per second. Entering minutes directly into Q = I × t produces an error by a factor of 60. Always convert minutes to seconds.
- Incorrect n-factor for gases: One mole of diatomic Cl2 requires 2 moles of electrons, while 1 mole of O2 requires 4 moles of electrons. Never confuse atomic and molecular moles.
- Confusing theoretical and actual values: Actual mass is the experimentally measured yield. Theoretical mass is the ideal maximum calculated from Faraday's law. Efficiency cannot exceed 100% in pure electrolysis.
Exam Relevance and Question Patterns
| Exam | Question Format | Focus Areas |
|---|---|---|
| CBSE Class 12 & State Boards | Numerical short and long answer questions | Faraday's first law, computing deposited metal mass with given current efficiency |
| JEE Main | Multiple choice and numerical value questions | Series cells, gas volume calculation at STP, and current efficiency comparisons |
| NEET | Single correct multiple choice questions | Fast arithmetic questions connecting current, time, n-factor, and percentage yield |
| CUET (UG) | Objective concept and formula based questions | Direct formula substitutions, unit conversions, and fundamental electrochemistry definitions |
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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