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What distinguishes a fuel cell from a galvanic cell such as a dry cell or lead-acid battery? - A fuel cell continuously converts chemical energy directly into electrical energy, whereas galvanic cells convert chemical energy into electrical energy but are typically used until the reactants are exhausted - VCE - SSCE Chemistry - Question 5 - 2006 - Paper 1

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What-distinguishes-a-fuel-cell-from-a-galvanic-cell-such-as-a-dry-cell-or-lead-acid-battery?----A-fuel-cell-continuously-converts-chemical-energy-directly-into-electrical-energy,-whereas-galvanic-cells-convert-chemical-energy-into-electrical-energy-but-are-typically-used-until-the-reactants-are-exhausted-VCE-SSCE Chemistry-Question 5-2006-Paper 1.png

What distinguishes a fuel cell from a galvanic cell such as a dry cell or lead-acid battery? - A fuel cell continuously converts chemical energy directly into elect... show full transcript

Worked Solution & Example Answer:What distinguishes a fuel cell from a galvanic cell such as a dry cell or lead-acid battery? - A fuel cell continuously converts chemical energy directly into electrical energy, whereas galvanic cells convert chemical energy into electrical energy but are typically used until the reactants are exhausted - VCE - SSCE Chemistry - Question 5 - 2006 - Paper 1

Step 1

What distinguishes a fuel cell from a galvanic cell such as a dry cell or lead-acid battery?

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Answer

A fuel cell continuously converts chemical energy directly into electrical energy, whereas galvanic cells convert chemical energy into electrical energy but are typically used until the reactants are exhausted. Fuel cells have reactants (like hydrogen and oxygen) that are supplied continuously, which allows them to operate under steady conditions, unlike conventional galvanic cells. The products of fuel cells (usually water) are not accumulated, allowing for efficient and ongoing reactions.

Step 2

On the diagram above i. in circles A and B, indicate the polarity of the cathode and anode

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Circle A represents the cathode and Circle B represents the anode.

Step 3

ii. show, by using an arrow, the direction of electron flow in the external circuit.

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Answer

Electrons flow from the anode (B) to the cathode (A) in the external circuit.

Step 4

Write an equation for each of the following reactions. i. the overall cell reaction

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Answer

The overall reaction is: 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O

Step 5

ii. the reaction at the cathode

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The reaction at the cathode is: O2+4e+2H2O4OHO_2 + 4e^- + 2H_2O \rightarrow 4OH^-

Step 6

i. How much electrical energy, in joules, would be provided by the cell?

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The energy provided is given by the formula: Energy=Current×Voltage×Time\text{Energy} = \text{Current} \times \text{Voltage} \times \text{Time} Thus, substituting the values: Energy=0.500×0.600×(10×60)=180 joules\text{Energy} = 0.500 \times 0.600 \times (10 \times 60) = 180 \text{ joules}

Step 7

ii. Calculate the charge, in coulombs, produced by the cell.

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Answer

The charge produced can be calculated by: Q=I×tQ = I \times t Substituting the given values: Q=0.500×(10×60)=300 coulombsQ = 0.500 \times (10 \times 60) = 300 \text{ coulombs}

Step 8

iii. If this particular cell operated at 60.0% efficiency, what amount of hydrogen gas (H₂), in mole, would be consumed by the fuel cell?

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Answer

First, calculate the effective energy output. Given the efficiency: Effective Energy=Energy×Efficiency\text{Effective Energy} = \text{Energy} \times \text{Efficiency} =180×0.60=108 joules= 180 \times 0.60 = 108 \text{ joules}

The molar energy of hydrogen gas is approximately 286 kJ/mol (or 286,000 J/mol). Thus, the amount of hydrogen consumed is: n=Effective Energy286000=108286000 moles of H2n = \frac{\text{Effective Energy}}{286000} = \frac{108}{286000} \text{ moles of H}_2

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