The method and the results are on the electrochemical cells lab. Do the analysis yourself first, then check the answers below.
Quick answer: A galvanic cell turns a spontaneous redox reaction into electricity: oxidation at the anode, reduction at the cathode. These answers go with the Chemistry 11-12 lab on electrochemical cells: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.
Model answers use the sample class data. Your own readings may differ slightly because of surface oxide, solution age and the salt bridge, but the order of the metals and the downward trend with dilution should match.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R3.2
Larger emf against copper means a stronger reducing agent: magnesium (2.60 V), then zinc (1.08 V), then iron (0.74 V), then tin (0.46 V).
Question 2 Electrode potentials
E°(M) = 0.34 − emf: Mg = 0.34 − 2.60 = −2.26 V; Zn = −0.74 V; Fe = −0.40 V; Sn = −0.12 V. These are close to the data-book values (−2.37, −0.76, −0.44, −0.14).
Three reasons: the solutions were 0.10 mol/dm³ rather than 1.00 mol/dm³; magnesium had an oxide layer; a filter-paper salt bridge has resistance (the temperature was also not 298 K and the voltmeter draws a small current).
Question 5 Concentration trend
The emf falls steadily as [Cu²⁺] falls. From 1.10 V to 1.02 V over three tenfold dilutions: 0.08 ÷ 3 = 0.027 V per dilution (about 0.03 V).
Question 6 Explaining the trend
Lowering [Cu²⁺] moves the equilibrium to the left, so reduction of Cu²⁺ at the cathode is less favourable. The copper electrode becomes less positive and the difference from zinc shrinks, so the emf is lower.
Question 7 Equipment and improvement
A high-resistance voltmeter draws almost no current, so the reading is close to the true emf and the concentrations do not change during the reading. To improve reliability, repeat each cell with fresh solutions and fresh salt bridges, and average the readings.
