Exercise practice: Electrochemistry

5.1 How the exercise set is arranged

The chapter has 18 exercises, numbered 2.1 to 2.18. They fall into three groups. Electrode potentials, Nernst equation and thermodynamics cover 2.1 to 2.9. Conductivity and molar conductivity cover 2.10 and 2.11. Electrolysis, Faraday's laws, cells and products cover 2.12 to 2.18. The worked items below are the types that repeat in board papers.

5.2 Group 1: potentials and thermodynamics

Exercises 2.1 to 2.5 ask for standard potentials, cell reactions and feasibility. The rule is the same each time: write the two half reactions, identify the cathode and anode, compute E°cell and then decide the sign of ΔG°. Exercises 2.6 to 2.9 add the Nernst correction and the equilibrium constant. A cell with E°cell positive and log Kc large is feasible; the reverse cell is not. For the relation ΔG° = -nFE°, n is the number of electrons in the balanced cell reaction, not the number in either half reaction alone.

5.3 Group 2: conductivity

Exercise 2.10 gives conductivity values at several concentrations and asks for molar conductivity at each and a plot against c^1/2. Convert each conductivity with 1000 kappa divided by M, plot the straight line and read the intercept at c = 0 as the limiting value. Exercise 2.11 gives kappa for 0.00241 M acetic acid as 7.896 times 10^-5 S cm^-1. Molar conductivity is 32.8 S cm^2 mol^-1. With the limiting value 390.5 the degree of dissociation is 0.0840, and Ka follows as 0.00241 times 0.0840 squared divided by 0.916, which is 1.85 times 10^-5.

5.4 Group 3: electrolysis and cells

Exercise 2.12 asks for the charge needed for one mole of each species. Al3+ to Al needs 3F, Cu2+ to Cu needs 2F, and MnO4- to Mn2+ needs 5F, because manganese falls from +7 to +2. In coulombs those are 289500, 193000 and 482500 C.

Exercise 2.13 works in reverse. For 20 g of Ca from CaCl2, the amount is 0.5 mol and each mole needs 2F, so about 1F is needed. For 40 g of Al from Al2O3, the amount is 1.48 mol and each mole needs 3F, so about 4.44F is needed.

Exercise 2.15 deposits nickel from Ni(NO3)2 with a current of 5 A for 20 minutes. The charge is 5 times 1200, which is 6000 C. Nickel is divalent, so the moles deposited are 6000 divided by 2 times 96500, which is 0.0311 mol. With a molar mass of 58.7 the mass is about 1.83 g.

Exercise 2.16 has three cells in series, so the same charge passes through all of them. When 1.45 g of silver is deposited, the amount of silver is 0.01343 mol and the charge is 1296 C. The time at 1.5 A is 864 s, or about 14.4 minutes. The same charge deposits 0.00672 mol of copper and 0.00672 mol of zinc, which is about 0.43 g of copper and 0.44 g of zinc.

Exercise 2.17 tests feasibility from standard potentials. Fe3+ with I- is feasible because the potential difference is 0.23 V. Ag+ with Cu is feasible at 0.46 V. Fe3+ with Br- is not feasible because the difference is negative, Ag with Fe3+ is marginally not feasible, and Br2 with Fe2+ is feasible because bromine is the stronger oxidiser. Exercise 2.18 asks for the products of electrolysis: silver nitrate with silver electrodes dissolves the anode and deposits silver at the cathode, the same salt with platinum electrodes gives silver at the cathode and oxygen at the anode, dilute sulphuric acid gives hydrogen and oxygen, and copper chloride gives copper and chlorine.

5.5 Two more types worth practising

Exercises 2.6 to 2.9 combine the Nernst equation with thermodynamics. A typical item gives a cell reaction and asks for E°cell, then ΔG°, then the equilibrium constant. The sequence is fixed: E°cell first, then n, then ΔG° = -nFE°, then log Kc = nE°cell/0.0591. Exercises 2.14 to 2.18 add the electricity calculations. For a charge in coulombs, convert to Faradays by dividing by 96500, then use the electron count from the balanced half reaction to get moles of the species. In the questions about the products of electrolysis, decide which of the two competing reactions at each electrode has the more favourable potential before naming the product.

5.6 The thermodynamic link in practice

When an exercise gives standard potentials and asks for the equilibrium constant, the sequence is E°cell, then n, then log Kc equal to nE°cell divided by 0.0591. A large positive E°cell gives a very large Kc, which is why the answer is often reported in powers of ten. When the same exercise asks for ΔG°, multiply n, F and E°cell with the correct sign: a spontaneous cell has negative ΔG°. Keeping the electrons balanced across the two half reactions before using n avoids a factor of two or three in both answers.

Questions about the products of electrolysis ask which of the competing half reactions has the more favourable potential at each electrode. Compare the metal couple with the water couple at the cathode, and the anion couple with the water couple at the anode, taking overpotential into account. Then write the two electrode reactions and the overall cell reaction with the electrons balanced.