What distinguishes a fuel cell from a galvanic cell such as a dry cell or lead-acid battery? A fuel cell operates continuously, converting chemical energy from fuel directly into electrical energy, whereas a galvanic cell relies on a chemical reaction that occurs within a closed system and depletes reactants over time - VCE - SSCE Chemistry - Question 5 - 2006 - Paper 1

The polarity of the cathode at point A is positive, and the polarity of the anode at point B is negative.

Electrons flow from the anode (B) to the cathode (A) in the external circuit, which can be indicated by an arrow pointing from B to A.

The overall cell reaction can be represented as:

$2H_2 + O_2 \rightarrow 2H_2O$

The reaction at the cathode is:

$O_2 + 4e^- + 2H_2O \rightarrow 4OH^-$

The electrical energy (E) can be calculated using the formula:

$E = V \times I \times t$

Substituting the given values:

$E = 0.600 \text{ V} \times 0.500 \text{ A} \times (10.0 \text{ min} \times 60 \text{ s/min}) = 180 \text{ J}$

The charge (Q) can be calculated using the formula:

$Q = I \times t$

Substituting the values:

$Q = 0.500 \text{ A} \times (10.0 \text{ min} \times 60 \text{ s/min}) = 300 \text{ C}$

To find the moles of hydrogen gas consumed, we first need the total energy used:

Given the efficiency (η) is 60%:

$\text{Useful Energy} = 0.60 \times 180 \text{ J} = 108 \text{ J}$

The molar energy of hydrogen can be calculated using:

$\text{Molar Energy} = \frac{E}{n}$

As per the stoichiometry of the cell reaction:

$2 moles \text{ H}_2 \text{ produces } 286 \text{ kJ}$

From this, we can find:

$1 \text{ J} = \frac{1}{286000} \text{ moles H}_2 \rightarrow 108 \text{ J} \rightarrow n = \frac{108}{286000} \text{ moles H}_2$

Calculating gives the result for moles of H2 consumed.