Hydrogen–Oxygen Fuel Cells Worked ExamplesIGCSE Chemistry 0620

(a) A hydrogen–oxygen fuel cell uses hydrogen and oxygen. Hydrogen enters at A, so oxygen (or air) enters at B. The only chemical product is water, which leaves at C.

(b) The cell produces electricity (an electric current), which drives the motor.

(c) Neither reactant contains carbon, so no carbon compound can be formed. Water is the only chemical product.

Statement 1 is correct: hydrogen and oxygen react to form water only. Statement 2 is wrong: there is no carbon in the reactants.

Statement 3 is correct: the energy released by the reaction of hydrogen with oxygen is given out as electrical energy.

Statement 4 is wrong: a fuel cell keeps producing electricity for as long as hydrogen and oxygen are supplied. It is refuelled with hydrogen, not recharged.

The answer is B. D is the tempting wrong answer: it treats the fuel cell like a rechargeable battery.

Advantage 1: the fuel-cell car releases no carbon dioxide, so it does not add to climate change while it is being driven.

Advantage 2: it releases no carbon monoxide (toxic) and no oxides of nitrogen (which cause acid rain and respiratory problems) — water is the only product.

Disadvantage 1: it travels a shorter distance on a full tank ($520$ km compared with $780$ km).

Disadvantage 2: there are far fewer refuelling stations ($14$ compared with $2600$). A further disadvantage in the table: hydrogen is a gas that must be stored at high pressure, which is harder than storing a liquid and is a hazard because hydrogen is flammable.

B is correct. Hydrogen has a very low density, so a large volume must be compressed into a strong tank; it is also flammable, so leaks are dangerous.

A is false: there is no carbon in hydrogen or oxygen, so no carbon monoxide forms. C is false: water is not acidic and does not cause acid rain.

D describes the energy change the wrong way round: a fuel cell converts chemical energy into electrical energy.

(a) Convert kg to g, then divide by $M_r(\text{H}_2) = 2$.

(b) The mole ratio $\text{H}_2 : \text{H}_2\text{O}$ is $2:2 = 1:1$, so $600$ mol of water forms. $M_r(\text{H}_2\text{O}) = 2 + 16 = 18$.

(c) The ratio $\text{H}_2 : \text{O}_2$ is $2:1$, so $300$ mol of oxygen reacts.

(a) Bonds broken (endothermic): two H–H and one O=O.

Bonds formed (exothermic): each water molecule has two O–H bonds, so four O–H bonds in total.

$\Delta H$ = energy to break bonds − energy released making bonds.

(b) The equation uses 2 mol of hydrogen.

(c) More energy is released making the new O–H bonds than is taken in breaking the H–H and O=O bonds, so overall energy is given out and $\Delta H$ is negative.

(a) Electrolysis of dilute sulfuric acid gives hydrogen and oxygen in a $2:1$ volume ratio, so L ($24\text{ cm}^3$) contains hydrogen and M ($12\text{ cm}^3$) contains oxygen. Test for oxygen: it relights a glowing splint.

(b) In position 2 the apparatus acts as a hydrogen–oxygen fuel cell: the hydrogen and oxygen on the platinum electrodes react to form water, and this produces electricity, which the voltmeter detects.

(c) Hydrogen and oxygen react in a $2:1$ mole ratio, and equal volumes of gases contain equal numbers of moles, so the volumes used are also $2:1$ ($8\text{ cm}^3 : 4\text{ cm}^3$).

(d) Position 1 (electrolysis): electrical energy → chemical energy. Position 2 (fuel cell): chemical energy → electrical energy.

(e) The hydrogen and oxygen are used up; once there is no fuel left, no more electricity is produced.

(a) Read the graph at a large distance for accuracy: at $400$ km the mass is $3.8$ kg.

(b) Divide the mass in the tank by the mass used per 100 km.

(c) Moles of hydrogen in $5.7$ kg ($5700$ g), with $M_r(\text{H}_2) = 2$:

$\text{H}_2 : \text{H}_2\text{O} = 1:1$, and $M_r(\text{H}_2\text{O}) = 18$.

(a) $M_r(\text{C}_8\text{H}_{18}) = (8 \times 12) + (18 \times 1) = 114$. Moles of octane in $4200$ g:

From the equation, $2$ mol octane → $16$ mol $\text{CO}_2$, so the ratio is $1:8$. $M_r(\text{CO}_2) = 44$.

(b) Divide by $100$ km.

(c) Zero: the fuel cell uses hydrogen, which contains no carbon, and water is the only chemical product.

Count the molecules: 4 $\text{H}_2$ and 3 $\text{O}_2$. The equation $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ shows that $\text{H}_2$ and $\text{O}_2$ react in the ratio $2:1$.

$4$ $\text{H}_2$ need only $2$ $\text{O}_2$ and form $4$ $\text{H}_2\text{O}$. (a) After the reaction the box contains 4 water molecules (each one pink circle joined to two white circles) and 1 oxygen molecule left over.

(b) Hydrogen is used up completely, so hydrogen is the limiting reactant (oxygen is in excess).

(c) Before: H atoms $= 4 \times 2 = 8$; O atoms $= 3 \times 2 = 6$. After: H atoms $= 4 \times 2 = 8$; O atoms $= (4 \times 1) + (1 \times 2) = 6$. Atoms are conserved.

(a) Moles of hydrogen in $1000$ g: $\dfrac{1000}{2} = 500$ mol. The ratio $\text{CO}_2 : \text{H}_2 = 1:4$.

(b) Carbon dioxide for the $0.95$ kg of hydrogen used per $100$ km:

(c) The claim is only partly true. The car itself releases no carbon dioxide (water is the only product), but making its hydrogen from methane releases about $52$ g/km — roughly 40% of the petrol car's $130$ g/km. So the fuel-cell car causes less carbon dioxide, but not none.

To cut this, make hydrogen by the electrolysis of water using electricity from a renewable source (such as solar or wind), which releases no carbon dioxide.

(a) Moles of hydrogen, then volume at r.t.p.

(b) Compare with the tank:

At room pressure the hydrogen would take up about 456 times the volume of the tank. To fit it in, the gas must be compressed: squeezing the molecules closer together needs a very high pressure, and so a strong, heavy tank.

(c) Hydrogen is flammable and forms an explosive mixture with air if it leaks; its small molecules leak easily; and high-pressure tanks are heavy and expensive. (Petrol is a liquid at room pressure and is easy to store.)