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  1. Problems
  2. /A hydrogen and oxygen fuel cell delivers a steady 7.5 A at 0.62 V for 3.0 hours…
HardChemistryHigh School (11–12)Electrochemistry › Conductance, Electrolysis and BatteriesNumericalA-LevelAPIBNEET

A hydrogen and oxygen fuel cell delivers a steady 7.5 A at 0.62 V for 3.0 hours…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A hydrogen and oxygen fuel cell delivers a steady 7.5 A7.5\ \text{A}7.5 A at 0.62 V0.62\ \text{V}0.62 V for 3.03.03.0 hours. The anode reaction is 2H2+4OH−→4H2O+4e−2\text{H}_2 + 4\text{OH}^- \rightarrow 4\text{H}_2\text{O} + 4e^-2H2​+4OH−→4H2​O+4e−. Burning one mole of hydrogen would release 285.8 kJ285.8\ \text{kJ}285.8 kJ as heat. Find the moles of hydrogen used, the electrical energy delivered in kilojoules, and the efficiency as a percentage of the heat that burning that hydrogen would give.

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Work it out on paper, then type your number. The answer itself stays hidden.

mol
kJ
%
Need a reminder? The key ideaShowHide

One ampere-hour is 3600 C3600\ \text{C}3600 C. A lead atom gives 2 electrons in the lead storage battery, a zinc atom gives 2 in the dry cell and an iron atom gives 2 when it rusts. Energy is VItVItVIt. A fuel cell converts the energy of H2+12O2→H2O\text{H}_2 + \tfrac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}H2​+21​O2​→H2​O directly into electricity. Zinc, with a lower potential than iron, protects it.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Charge. Q=7.5×3.0×3600=81 000 CQ = 7.5 \times 3.0 \times 3600 = 81\,000\ \text{C}Q=7.5×3.0×3600=81000 C.

Step 2. Hydrogen. Electrons: 81 00096500=0.839 mol\dfrac{81\,000}{96500} = 0.839\ \text{mol}9650081000​=0.839 mol. Each H2\text{H}_2H2​ gives 2, so 0.420 mol0.420\ \text{mol}0.420 mol of hydrogen.

Step 3. Energy delivered. E=VIt=0.62×81 000=50 220 J=50.2 kJE = VIt = 0.62 \times 81\,000 = 50\,220\ \text{J} = 50.2\ \text{kJ}E=VIt=0.62×81000=50220 J=50.2 kJ.

Step 4. Energy in the fuel. 0.420×285.8=120 kJ0.420 \times 285.8 = 120\ \text{kJ}0.420×285.8=120 kJ.

Step 5. Efficiency. 50.2120×100=42 %\dfrac{50.2}{120} \times 100 = 42\ \%12050.2​×100=42 %.

Answer: 0.420 mol0.420\ \text{mol}0.420 mol of hydrogen, 50.2 kJ50.2\ \text{kJ}50.2 kJ of electrical energy and an efficiency of 42 %42\ \%42 %.

Check. The best possible is about 83 %83\ \%83 %, set by ΔG∘/ΔH∘\Delta G^\circ / \Delta H^\circΔG∘/ΔH∘. A cell delivering current at a voltage below its emf of 1.23 V1.23\ \text{V}1.23 V is well below that.

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