Oxides with a CaF2 crystal structure containing a high concentration of oxygen vacancies show oxide- — Physical Chemistry — Thermodynamics Chemistry Question
Oxide-ion conductors
Oxides with a CaF2 crystal structure containing a high concentration of oxygen vacancies show oxide-ion conduction when heated at high temperatures, and these oxides are often called solid electrolytes. A cell using a solid electrolyte with porous Pt electrodes on both sides can be applied to oxygen sensors, oxygen pumps and fuel cells etc.
In the oxygen pump, oxygen molecules are reduced to oxide ions at the cathode, and the oxide ions move to the anode and are oxidized to oxygen molecules at the anode, by an applied voltage. Meanwhile, when the oxygen partial pressures are different at two electrodes which are not short-circuited, an electromotive force is generated between both electrodes, and this phenomenon is used for the oxygen sensor.
Represent reaction formulas at the cathode and the anode during oxygen pumping.
Model Answer
Cathode: O2 + 4 e– → 2 O2–
Anode: 2 O2– → O2 + 4 e–
An electric current of 1.93 A was flowed for 500 s to move oxygen ions from the cathode to the anode. Calculate the volume of oxygen gas (cm3) produced at the anode at 800 ºC and a pressure of 1.01×105 Pa. Given your answer to two significant figures.
Model Answer
1/4 mol O2 moves when 1 mole of electrons flows.
Moved O2 : n = (1.93 × 500 / 96485) × (1/4) = 2.5×10–3 mol
V = nRT/p = (2.5×10^–3 * 8.314 * 1073) / (1.01×10^5) = 2.208×10^–4 m3 = 2.208×10^2 cm3
Answer: 2.2×102 cm3
Calculate the electromotive force (V) when the ratio of the oxygen partial pressures p1 and p2 at both electrodes is maintained to be p1 / p2 = 100 at 800 ºC. Electronic conduction of the solid electrolyte can be ignored.
Model Answer
In an oxygen concentration cell, ∆Gº and Eº are 0 and z = 4.
E = – (RT / 4F) * ln(p_ox / p_red) = – (RT / 4F) * ln(p_cathode / p_anode) = 2.303 * (RT / 4F) * lg(p1 / p2) = (2.303 * 8.314 * 1073 * 2) / (4 * 96485) = 0.1064 V
Answer: 1.1×10–1 V