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S5 Chemistry: Chemical Cells and Fuel Cells (Paper 3: challenge)
S5 Chemistry: Chemical Cells and Fuel Cells (Paper 3: challenge)
Ten challenge items on the total energy of 1.0 kg of hydrogen, the water and oxygen involved, overall efficiency, the zinc used and time for 9650 C, power and energy of cells in series, matching silver oxide cells and the storage misconception.
Direct answers Answers in order: 143 MJ, 9.0 kg, 6.0 m3, 55.9 percent, 3.27 g, 5.36 hours, 0.30 W, 1080 J, suitable at 4.65 V, and a fuel cell stores no energy.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S5 Chemistry: Chemical Cells and Fuel Cells |
| Level | S5 |
02 Questions (10 items)
- FoundationA hydrogen fuel cell car carries 1.0 kg of hydrogen. How much energy in MJ is released if all of it is used, at 286 kJ per mole?
- FoundationThe same 500 mol of hydrogen reacts completely. What mass of water in kg is formed?
- FoundationWhat volume of oxygen in m3 is needed for 500 mol of hydrogen at 24 dm3/mol?
- Intermediate1.0 kg of hydrogen can release 143 MJ but only 80 MJ reaches the wheels. What is the overall efficiency?
- IntermediateA cell passes 9650 C. What mass of zinc is consumed at the negative electrode, zinc 65.4 and the Faraday constant 96500 C/mol?
- IntermediateHow many hours does 0.50 A take to pass 9650 C?
- ChallengeFour 1.5 V dry cells in series supply 0.050 A. What is the power output in W?
- ChallengeContinuing the last item, what electrical energy in J is delivered in one hour?
- ChallengeThree 1.55 V silver oxide cells are in series. Is this suitable for a bulb rated at 4.65 V?
- Error redoA student claims that a fuel cell must be charged before it can supply electricity. Why is this wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 143 MJ | 1.0 kg of hydrogen is 500 mol, and 500 times 286 gives 143000 kJ, which is 143 MJ. |
| 2 | 9.0 kg | 500 mol of water times 18 g gives 9000 g, which is 9.0 kg. |
| 3 | 6.0 m3 | 250 mol of oxygen is needed, and 250 times 24 gives 6000 dm3, which is 6.0 m3. |
| 4 | 55.9 percent | Dividing 80 by 143 and multiplying by 100 gives 55.9 percent. |
| 5 | 3.27 g | 9650 C corresponds to 0.100 mol of electrons, that is 0.050 mol of zinc, and 0.050 times 65.4 gives 3.27 g. |
| 6 | 5.36 hours | The time is 9650 divided by 0.50, which is 19300 seconds, and dividing by 3600 gives 5.36 hours. |
| 7 | 0.30 W | The total emf is 6.0 V, and power equals voltage times current, so 6.0 times 0.050 gives 0.30 W. |
| 8 | 1080 J | Energy equals power times time, so 0.30 times 3600 gives 1080 J. |
| 9 | Yes, the total emf is exactly 4.65 V | 1.55 times 3 gives 4.65 V, which matches the rated voltage of the bulb. |
| 10 | A fuel cell generates electricity from hydrogen and oxygen fed in continuously and stores no energy itself | It converts the chemical energy of the fuel directly into electrical energy and keeps working while fuel is supplied, so it needs no charging. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
For every item state the direction of electron transfer or the formula first, then substitute the numbers and give the unit.Tutor Registration →
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