Question

A hydrogen-oxygen fuel cell operates under standard conditions. The standard half-cell reduction potential for oxygen is E°(O₂/H₂O) = 1.20 V. For the complete reaction: 2H₂(g) + O₂(g) → 2H₂O(l) If 1 mole of O₂ is consumed in the fuel cell, the magnitude of the maximum electrical work obtained is x kJ. Find the value of x. (Nearest integer). (Given: 1 F = 96500 C mol⁻¹)

Answer 1

463

Answer 2

The maximum electrical work ($W_{max}$) obtained from an electrochemical cell is given by the magnitude of the standard Gibbs free energy change, $|\Delta G^\circ|$, which is related to the cell potential ($E^\circ_{cell}$) by the equation: $W_{max} = |\Delta G^\circ| = nFE^\circ_{cell}$ where:

• $n$ is the number of moles of electrons transferred in the balanced reaction.
• $F$ is Faraday's constant ($96500$ C mol⁻¹).
• $E^\circ_{cell}$ is the standard cell potential.

The given overall reaction is: $2H_2(g) + O_2(g) \longrightarrow 2H_2O(l)$

From the half-reactions: Oxidation: $2H_2(g) \longrightarrow 4H^+(aq) + 4e^-$ Reduction: $O_2(g) + 4H^+(aq) + 4e^- \longrightarrow 2H_2O(l)$ The number of moles of electrons transferred ($n$) is 4 for the consumption of 1 mole of $O_2$.

The standard cell potential ($E^\circ_{cell}$) is calculated as: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ Given $E^\circ(O_2/H_2O) = 1.20$ V (cathode). The standard reduction potential for the hydrogen electrode (anode) is $E^\circ(H^+/H_2) = 0.00$ V. So, $E^\circ_{cell} = 1.20 \text{ V} - 0.00 \text{ V} = 1.20 \text{ V}$.

Now, calculate the maximum electrical work when 1 mole of $O_2$ is consumed ($n=4$): $W_{max} = nFE^\circ_{cell}$ $W_{max} = (4 \text{ mol}) \times (96500 \text{ C mol}^{-1}) \times (1.20 \text{ V})$ $W_{max} = 463200 \text{ J}$

Convert Joules to kilojoules: $W_{max} = \frac{463200}{1000} \text{ kJ} = 463.2 \text{ kJ}$

The question asks for the value of $x$ in kJ, rounded to the nearest integer. $x = 463.2 \text{ kJ}$. Rounding to the nearest integer gives $x = 463$.