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A fuel cell that can provide power for buses is the phosphoric acid fuel cell, PAFC - VCE - SSCE Chemistry - Question 8 - 2008 - Paper 1

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A fuel cell that can provide power for buses is the phosphoric acid fuel cell, PAFC. The electrolyte is concentrated phosphoric acid and the reactants are hydrogen a... show full transcript

Worked Solution & Example Answer:A fuel cell that can provide power for buses is the phosphoric acid fuel cell, PAFC - VCE - SSCE Chemistry - Question 8 - 2008 - Paper 1

Step 1

i. Give the equation for the half reaction that takes place at the anode of this cell.

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Answer

The half reaction at the anode of the phosphoric acid fuel cell involves the oxidation of hydrogen gas. The equation is:

ightarrow 2H^+(aq) + 2e^-$$

Step 2

ii. cathode of this cell.

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Answer

The half reaction at the cathode involves the reduction of oxygen. The equation is:

ightarrow 2H_2O(l)$$

Step 3

b. On the diagram of the fuel cell, draw an arrow to show the direction in which the H2PO4− ion moves as the cell delivers an electrical current.

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Answer

On the diagram, the H2PO4− ion moves from the anode side to the cathode side, indicated by an arrow pointing to the right.

Step 4

i. A particular cell operates at 0.92 V. How much energy, in kJ, is delivered per mole of hydrogen in this fuel cell?

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Answer

Using the formula for electrical energy:

E=QVE = QV

Where

  • Charge, QQ = n × F
  • nn = 2 mol (for 2 moles of electrons from the oxidation of 1 mole of hydrogen)
  • FF = 96500 C/mol

Calculating:

Q=2extmolimes96500extC/mol=193000extCQ = 2 ext{ mol} imes 96500 ext{ C/mol} = 193000 ext{ C}

Thus,

E=193000extCimes0.92extV=177560extJ=177.56extkJE = 193000 ext{ C} imes 0.92 ext{ V} = 177560 ext{ J} = 177.56 ext{ kJ}

Step 5

ii. By comparing the energy delivered per mole of hydrogen in the fuel cell and the heat of combustion of hydrogen, calculate the energy efficiency of this fuel cell.

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Answer

The heat of combustion of hydrogen is approximately 286 kJ/mol. The energy efficiency can be calculated using:

extEfficiency=Energy deliveredHeat of combustion×100 ext{Efficiency} = \frac{\text{Energy delivered}}{\text{Heat of combustion}} \times 100

Therefore:

Efficiency=177.56 kJ286 kJ×10062.1%\text{Efficiency} = \frac{177.56 \text{ kJ}}{286 \text{ kJ}} \times 100 \approx 62.1\%

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