A fuel cell is set up based on the oxidation of methane - VCE - SSCE Chemistry - Question 6 - 2004 - Paper 1

From the anode half reaction, we see that the oxidation of one mole of methane produces 8 moles of electrons (8e^-). Each mole of electrons carries a charge of approximately 96485 coulombs (Faraday's constant).

Therefore, the total charge (Q) produced from 8 moles of electrons is calculated as follows:

$Q = n \times F$

where:

• $n = 8$ moles (number of moles of electrons)
• $F \approx 96485 \text{ C mol}^{-1}$

Substituting the values, we get:

$Q = 8 \times 96485 \text{ C} \approx 771880 \text{ C}$

This value can be expressed as approximately $8 \times 10^5$ C.