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Calculate the ideal efficiency of the fuel cell.

Biology

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Answer

82.87%\text{82.87\%}

Step-by-step solution

2 steps

Here's the solution for Q2.b.ii:

  1. Calculate the ideal efficiency of the fuel cell.

    The ideal efficiency (η\eta) of a fuel cell is given by the ratio of the Gibbs free energy change (ΔG\Delta G) to the enthalpy change (ΔH\Delta H).

    η=ΔGΔH\eta = \frac{\Delta G}{\Delta H}

    Given:
    ΔG=−237 kJ/mol\Delta G = -237 \text{ kJ/mol}
    ΔH=−286 kJ/mol\Delta H = -286 \text{ kJ/mol}

    Substitute the values:

    η=−237 kJ/mol−286 kJ/mol\eta = \frac{-237 \text{ kJ/mol}}{-286 \text{ kJ/mol}} η≈0.82867\eta \approx 0.82867

    Convert to percentage:

    η≈0.82867×100%\eta \approx 0.82867 \times 100\% η≈82.87%\eta \approx 82.87\%

    The ideal efficiency of the fuel cell is \boxed{\text{82.87%}}.

  2. Give two reasons why hydrogen is difficult to store.

    • Hydrogen has a very low volumetric energy density, meaning a large volume is required to store a significant amount of energy, even when compressed or liquefied. This necessitates high-pressure tanks or cryogenic temperatures, which are energy-intensive and costly.
    • Hydrogen is highly flammable and explosive when mixed with air, posing significant safety challenges for storage and transportation.
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