Electrochemistry — Questions & Answers

Sample questions from Electrochemistry

1. Q1. Two electrolytic cells, one containing AgNO3 solution and the other containing CuSO4 solution, are connected in SERIES (same current flows through both) and electrolyzed for the same time. If 5.4 g of silver (atomic mass 108) is deposited in the first cell, calculate the mass of copper (atomic mass 63.5) deposited in the second cell (Ag+ +e- to Ag, n=1, Cu2+ +2e- to Cu, n=2).

   • A.Mass of copper deposited is approximately 6.35 g
   • B.Mass of copper deposited is approximately 3.18 g
   • C.Mass of copper deposited is approximately 1.59 g✓
   • D.Mass of copper deposited is approximately 0.79 g
   Solution

   Since the two cells are connected in SERIES, the SAME quantity of electricity (charge, Q) passes through both -- by Faraday's second law, the masses of different substances deposited by the same charge are proportional to their EQUIVALENT WEIGHTS (atomic mass/n) -- moles of electrons for silver deposition: moles of Ag=5.4/108=0.05 mol, and since n=1 for Ag, moles of electrons=0.05 mol -- since the SAME moles of electrons (0.05 mol) pass through the copper cell, and Cu2+ requires n=2 electrons per atom, moles of Cu deposited=0.05/2=0.025 mol -- mass of Cu deposited=0.025 x 63.5=1.5875 g, approximately 1.59 g -- this illustrates Faraday's second law directly: the same charge deposits different masses of different metals, in proportion to their equivalent weights (here, Ag equivalent weight=108, Cu equivalent weight=63.5/2=31.75, and the mass ratio 5.4:1.59 approximately matches the equivalent weight ratio 108:31.75).

2. Q2. Using E degree(Ag+/Ag) = +0.799 V and E degree(AgCl/Ag,Cl-) = +0.222 V, the solubility product of AgCl at 298 K works out to

   • A.5.8 x 10^-5
   • B.1.7 x 10^-5
   • C.3.4 x 10^-20
   • D.1.7 x 10^-10✓
   Solution

   Combining AgCl + e- -> Ag + Cl- (0.222 V) with Ag -> Ag+ + e- (reverse of 0.799 V) gives AgCl -> Ag+ + Cl-, for which E degree = 0.222 - 0.799 = -0.577 V with n = 1. Then log Ksp = nE degree/0.0591 = -0.577/0.0591 = -9.76, so Ksp = 1.7 x 10^-10. Electrochemistry thus measures solubility products far too small to determine by weighing.

3. Q3. For the electrolysis of water, the thermodynamic decomposition voltage is 1.23 V, yet a practical electrolyser needs about 1.8-2.0 V. The extra voltage is used

   • A.to overcome the activation overpotentials of the H2 and O2 reactions plus the ohmic drop in the electrolyte and cell hardware, all of which appear as heat✓
   • B.to increase the value of E degree
   • C.to raise the concentration of H+ in solution
   • D.to compensate for the loss of Faradays constant at high current
   Solution

   The reversible 1.23 V is the minimum for zero rate. To drive a useful current, extra potential is required at each electrode to overcome the activation barrier of the electron-transfer step (largest for the four-electron oxygen evolution), plus an IR drop across the electrolyte, separator and contacts. All of this extra energy is degraded to heat, so the voltage efficiency of a practical electrolyser is roughly 1.23/1.9, about 65 per cent, and improving catalysts to lower the oxygen overpotential is the main research target.

4. Q4. In the lead storage (lead-acid) battery, what happens to the density of the sulphuric acid electrolyte during DISCHARGE?

   • A.It increases steadily as PbSO4 forms
   • B.It decreases, because H2SO4 is consumed and water is produced at both electrodes✓
   • C.It remains exactly constant throughout discharge
   • D.It decreases only at the cathode, not the anode
   Solution

   During discharge of the lead-acid battery, both electrode reactions consume H2SO4 and produce water: at the anode, Pb + SO4^2- to PbSO4 + 2e-; at the cathode, PbO2 + SO4^2- + 4H+ + 2e- to PbSO4 + 2H2O. Since sulphuric acid is consumed and water is generated, the density (and concentration) of the electrolyte progressively DECREASES, which is why battery testers measure electrolyte density (specific gravity) to estimate charge state.

5. Q5. Given lambda degree(Al3+) = 189 S cm2 mol-1 and lambda degree(SO4^2-) = 160 S cm2 mol-1, the limiting molar conductivity of Al2(SO4)3 is

   • A.349 S cm2 mol-1
   • B.698 S cm2 mol-1
   • C.480 S cm2 mol-1
   • D.858 S cm2 mol-1✓
   Solution

   Kohlrauschs law must be applied with the correct stoichiometric numbers: lambda_m degree = 2 lambda degree(Al3+) + 3 lambda degree(SO4^2-) = 2(189) + 3(160) = 378 + 480 = 858 S cm2 mol-1. Simply adding the two ionic values (349) ignores the fact that one formula unit releases five ions.

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