Questions on commercial cells, rechargeable cells and fuel cells

Question 1

Evaluate the use of hydrogen fuel cells in place of petrol engines to power cars.

Accepted Answer

Indicative content

the only product at the point of use is water, so the car emits no carbon dioxide, carbon monoxide or nitrogen oxides
a fuel cell converts a larger fraction of the fuel's energy into useful work than combustion in an engine, because it does not go through heat
the cell is refuelled in minutes, like a petrol car, and needs no electrical recharging from a supply
hydrogen is flammable and has a low density, so it must be stored at high pressure or as a liquid at very low temperature
most hydrogen is made from methane by steam reforming, releasing carbon dioxide, and electrodes need an expensive platinum catalyst
a judgement: the overall benefit depends on how the hydrogen is made, and is large only if it comes from electrolysis using renewable electricity

Levels

Level 3 (5–6 marks) advantages and disadvantages are both covered with the chemistry behind each: water as the only product at the point of use, the efficiency and quick refuelling, the storage problem, the carbon dioxide from steam reforming and the platinum cost, with a judgement that depends on how the hydrogen is made
Level 2 (3–4 marks) two advantages and a drawback are given with some chemistry, and a conclusion; where the hydrogen comes from, or how it is stored, is not raised, or the judgement does not follow
Level 1 (1–2 marks) fuel cells are called pollution-free because they make only water, with no thought for how the hydrogen is made or stored

Question 2

Calculate a value for the Avogadro constant, L, from the results in the table. A student passed a steady current between copper electrodes in copper(II) sulfate solution and weighed the cathode before and after. Charge on one electron = 1.60 × 10⁻¹⁹ C; Ar(Cu) = 63.5.

Accepted Answer

M1 Q = It = 0.200 × 3600 = 720 C M1 copper deposited = 0.236 g, so moles of Cu = 0.236 ÷ 63.5 = 3.72 × 10⁻³ mol M1 Cu²⁺ + 2e⁻ → Cu, so moles of electrons = 2 × 3.72 × 10⁻³ = 7.43 × 10⁻³ mol M1 number of electrons = Q ÷ e = 720 ÷ 1.60 × 10⁻¹⁹ = 4.50 × 10²¹, and L = 4.50 × 10²¹ ÷ 7.43 × 10⁻³ A1 L = 6.05 × 10²³ mol⁻¹, to three significant figures; allow 6.06 × 10²³ from rounded working

Question 3

Give the half-equation at each electrode of a lithium-ion cell during discharge, and at each electrode while it is being charged.

Accepted Answer

B1 discharge, negative electrode: Li → Li⁺ + e⁻ B1 discharge, positive electrode: Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻ (allow LiCoO₂) B1 charging, positive electrode: Li⁺[CoO₂]⁻ → Li⁺ + CoO₂ + e⁻ B1 charging, negative electrode: Li⁺ + e⁻ → Li; ignore state symbols

Question 4

Calculate the EMF of a lithium iron phosphate cell, write the overall equation for its discharge, and deduce the change in the oxidation state of iron. FePO₄(s) + Li⁺ + e⁻ → LiFePO₄(s), E = +0.41 V; Li⁺ + e⁻ → Li(s), E = −3.04 V.

Accepted Answer

B1 Li(s) + FePO₄(s) → LiFePO₄(s), the lithium half-equation reversed and the two added B1 iron goes from +3 in FePO₄ to +2 in LiFePO₄, so it is reduced at the positive electrode M1 the iron phosphate electrode is more positive and on the right: EMF = (+0.41) − (−3.04) A1 EMF = +3.45 V

Question 5

Complete the half-equations for the hydrogen–oxygen fuel cell with an acidic electrolyte, H₂(g) → ... and O₂(g) + ... → 2H₂O(l), and explain why this cell has the same EMF as the alkaline cell.

Accepted Answer

B1 H₂(g) → 2H⁺(aq) + 2e⁻ at the negative electrode B1 O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) at the positive electrode B1 both cells have the same overall reaction, 2H₂(g) + O₂(g) → 2H₂O(l) B1 changing from acid to alkali lowers both electrode potentials by the same 0.83 V, so their difference, the EMF, is 1.23 V in both

Question 6

Calculate the EMF of a hydrazine fuel cell with an alkaline electrolyte, and write the overall equation for its reaction. N₂(g) + 4H₂O(l) + 4e⁻ → N₂H₄(aq) + 4OH⁻(aq), E° = −1.16 V; O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq), E° = +0.40 V.

Accepted Answer

M1 the oxygen electrode has the more positive E°, so it is the positive electrode, on the right B1 the hydrazine half-equation is reversed and the two added; hydroxide ions and two waters cancel B1 N₂H₄(aq) + O₂(g) → N₂(g) + 2H₂O(l) A1 EMF = (+0.40) − (−1.16) = +1.56 V

Question 7

Calculate the mass of silver deposited at the cathode when a current of 0.250 A is passed through silver nitrate solution for 40.0 minutes. F = 96 500 C mol⁻¹; Ar(Ag) = 107.9.

Accepted Answer

M1 t = 40.0 × 60 = 2400 s, so Q = It = 0.250 × 2400 = 600 C M1 moles of electrons = Q/F = 600 ÷ 96 500 = 6.22 × 10⁻³ mol M1 Ag⁺(aq) + e⁻ → Ag(s), so n = 1 and moles of silver = 6.22 × 10⁻³ mol A1 mass = 6.22 × 10⁻³ × 107.9 = 0.671 g, to three significant figures; allow error carried forward

Question 8

Deduce the overall equation for the discharge of a lead–acid cell from its electrode half-equations, Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻ and PbO₂(s) + 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l), and explain why the density of the acid falls as the cell discharges.

Accepted Answer

B1 both half-equations carry two electrons, so they add directly B1 Pb(s) + PbO₂(s) + 4H⁺(aq) + 2SO₄²⁻(aq) → 2PbSO₄(s) + 2H₂O(l) B1 sulfate ions and hydrogen ions leave the solution, the sulfate as solid lead(II) sulfate on both plates B1 water is formed, so the sulfuric acid becomes more dilute and its density falls

Question 9

Calculate the EMF of a silver oxide–zinc button cell, and write the overall equation for its discharge. Ag₂O(s) + H₂O(l) + 2e⁻ → 2Ag(s) + 2OH⁻(aq), E° = +0.34 V; ZnO(s) + H₂O(l) + 2e⁻ → Zn(s) + 2OH⁻(aq), E° = −1.26 V.

Accepted Answer

M1 the silver oxide electrode is more positive, so it is the positive electrode, on the right M1 EMF = (+0.34) − (−1.26) B1 the zinc half-equation is reversed and the two added, giving Zn(s) + Ag₂O(s) → ZnO(s) + 2Ag(s) A1 EMF = +1.60 V

Question 10

Identify the product at the cathode and the product at the anode when each electrolyte in the table is electrolysed with inert electrodes.

Accepted Answer

B1 (a) sodium at the cathode and chlorine at the anode, as the melt holds only Na⁺ and Cl⁻ B1 (b) silver at the cathode and oxygen at the anode, since nitrate ions are not oxidised B1 (c) hydrogen at the cathode and bromine at the anode, from the concentrated bromide B1 (d) hydrogen at the cathode and oxygen at the anode, since Mg²⁺ and sulfate are not discharged

Question 11

Explain why electrolysis of concentrated sodium chloride solution with inert electrodes gives chlorine at the anode, whereas very dilute sodium chloride solution gives mostly oxygen. Include an equation for each anode reaction.

Accepted Answer

B1 chloride ions and water molecules compete to be oxidised at the anode B1 in concentrated solution [Cl⁻] is high and chloride is oxidised: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ B1 in dilute solution [Cl⁻] is low, so Cl₂ + 2e⁻ ⇌ 2Cl⁻ shifts right and chloride is harder to oxidise, while water is always present in large amount B1 water is oxidised instead: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻; allow 4OH⁻ → O₂ + 2H₂O + 4e⁻

Question 12

Describe two benefits and two risks to society of the use of lithium-ion cells.

Accepted Answer

B1 a high energy for its mass, which makes phones, laptops and electric cars practical B1 recharged hundreds of times, giving less waste than non-rechargeable cells B1 the electrolyte contains a flammable organic solvent, so a damaged or overcharged cell can overheat and catch fire B1 mining lithium and cobalt causes environmental damage and can involve unsafe working conditions, and spent cells are hard to recycle

Question 13

Compare a rechargeable cell with a fuel cell, and explain why a fuel cell does not need to be electrically recharged.

Accepted Answer

B1 both produce an electric current from a feasible redox reaction at two electrodes B1 a rechargeable cell holds its reactants inside and regenerates them when an external supply drives the reaction in reverse B1 in a fuel cell the reactants, a fuel and oxygen, are supplied continuously from outside the cell B1 the products leave and the electrodes and electrolyte are not used up, so it is refuelled and never recharged

Question 14

Calculate the EMF of a direct ethanol fuel cell with an acidic electrolyte, write its overall equation, and show that the twelve electrons match the change in the oxidation state of carbon. 2CO₂(g) + 12H⁺(aq) + 12e⁻ → C₂H₅OH(l) + 3H₂O(l), E° = +0.09 V; O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l), E° = +1.23 V.

Accepted Answer

B1 C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), with the oxygen half-equation tripled B1 carbon averages −2 in ethanol and is +4 in CO₂, a rise of 6 for each of two carbons, which is twelve electrons M1 the oxygen electrode is more positive and on the right: EMF = (+1.23) − (+0.09) A1 EMF = +1.14 V

Question 15

Give the half-equations for the hydrogen–oxygen fuel cell with an alkaline electrolyte, and the overall equation.

Accepted Answer

B1 negative electrode: H₂(g) + 2OH⁻(aq) → 2H₂O(l) + 2e⁻; reject H₂ → 2H⁺ + 2e⁻ B1 positive electrode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq) B1 overall: 2H₂(g) + O₂(g) → 2H₂O(l)

Question 16

Explain why a lithium-ion cell can be recharged but a zinc–carbon cell cannot.

Accepted Answer

B1 in a lithium-ion cell the products of discharge stay on the electrodes as solids, Li⁺[CoO₂]⁻ at the positive electrode B1 an external supply with a voltage greater than the EMF drives both electrode reactions in reverse and regenerates the reactants B1 in a zinc–carbon cell the zinc case is used up and the products spread away from the electrodes, so the reactants cannot be regenerated

Question 17

Suggest the half-equation at each electrode, and the overall equation, for a methanol fuel cell with an acidic electrolyte, in which methanol is oxidised to carbon dioxide and oxygen is reduced to water.

Accepted Answer

B1 negative electrode: CH₃OH(l) + H₂O(l) → CO₂(g) + 6H⁺(aq) + 6e⁻ B1 positive electrode: O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) B1 overall: 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)

Question 18

Suggest why the electrolyte in a lithium-ion cell is a lithium salt dissolved in an organic solvent and not an aqueous solution, and why its Li⁺ ions are written without a state symbol.

Accepted Answer

B1 lithium is a very strong reducing agent and reacts with water, releasing hydrogen B1 an aqueous electrolyte would react with the negative electrode, and a cell EMF of about 3.6 V would also decompose the water B1 the ions are dissolved in an organic solvent, not water, so (aq) would be wrong

Question 19

Identify the oxidation state of cobalt in CoO₂ and in Li⁺[CoO₂]⁻, and identify the electrode reaction in which cobalt is reduced and the one in which it is oxidised.

Accepted Answer

B1 cobalt is +4 in CoO₂ and +3 in Li⁺[CoO₂]⁻ B1 cobalt is reduced at the positive electrode during discharge, Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻ B1 cobalt is oxidised at the same electrode during charging, Li⁺[CoO₂]⁻ → Li⁺ + CoO₂ + e⁻

Question 20

Explain one advantage and two disadvantages of methanol compared with hydrogen as the fuel for a fuel cell.

Accepted Answer

B1 methanol is a liquid at room temperature, so it is stored and carried in an ordinary tank without high pressure or very low temperature B1 the cell releases carbon dioxide at the point of use, and most methanol is made from natural gas B1 carbon monoxide formed during the oxidation poisons the platinum catalyst, or methanol crossing the membrane lowers the voltage, or methanol is toxic

Measurement	Value
Current	0.200 A
Time	60.0 min
Mass of cathode before	25.412 g
Mass of cathode after	25.648 g