The method and the results are on the electrolysis lab. Do the analysis yourself first, then check the answers below.
Quick answer: Electrolysis uses direct current to split a molten or dissolved ionic compound. These answers go with the Chemistry 9-10 lab on electrolysis: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.
The model answers below use the sample table. Your own numbers will differ a little, but the straight line through the origin, the gradient near 0.6 g/A and the value close to 10⁵ C per mole of electrons should appear in any careful set of results.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE C4
1. The points lie on a straight line through the origin. 2. The mass of copper deposited is directly proportional to the current. Mass ∝ current for a fixed time.
Question 2 Find the gradient
Gradient = 0.57 g ÷ 1.00 A = 0.57 g/A. It is the mass of copper deposited per ampere in 30 minutes. 0.57 g/A.
1. Q = 1.00 × 1800 = 1800 C. 2. 1800 ÷ 0.0178 = 1.01 × 10⁵ C/mol. 3. (101 000 − 96 500) ÷ 96 500 × 100 = 4.7%. About 1.0 × 10⁵ C/mol, within 5% of the accepted value.
Question 5 Anode and cathode
1. Cu → Cu²⁺ + 2e⁻. 2. Some copper falls off the anode as small particles without dissolving, and some of the deposit on the cathode may flake off or be lost when rinsing. Anode: Cu → Cu²⁺ + 2e⁻.
Question 6 Uncertainty
1. 0.01 ÷ 0.11 × 100 = 9.1%. 2. 0.01 ÷ 0.57 × 100 = 1.8%. 3. The 0.57 g result is more reliable because the same absolute uncertainty is a smaller fraction of a bigger mass. 9.1% and 1.8%.
Question 7 Conclusion and improvements
The mass of copper deposited is proportional to the current, and about two electrons are needed per copper atom. Improvements: run for longer so the masses are larger, and dry the plates in exactly the same way each time before weighing; also repeat each current and take a mean. Mass is proportional to current.
