Use the tabulated values of ΔG∘f to calculate E∘cell for the fuel cell breathalyzer, which employs the reaction below.
CH3CH2OH(g)+O2(g)→HC2H3O2(g)+H2O(g)
Reactant/Product ΔG∘f(kJ/mol)
CH3CH2OH(g) -167.9
O2(g) 0.0
HC2H3O2(g) -374.2
H2O(g) -228.6
ELECTROCHEMISTRY HELP REQUIRED. Lost and confused.?
2016-06-07 5:24 am
回答 (1)
2016-06-07 7:48 am
Reduction half equation : CH₃CH₂OH(g) + H₂O(l) → HC₂H₃O₂(g) + 4H⁺(aq) + 4e⁻
Oxidation half equation : 4H⁺(aq) + O₂ + 4e⁻ → 2H₂O(l)
Overall equation : CH₃CH₂OH(g) + O₂(g) → HC₂H₃O₂(g) + H₂O(g)
ΔG° = - n F ΔE°
[(-374.2) + (-228.8) + (-167.9)] × 1000 = - 4 × 96500 × ΔE°
- 4 × 96500 × ΔE° = -434900
ΔE° = +434900 / (4 × 96500) V
ΔE° = +1.13 V
Oxidation half equation : 4H⁺(aq) + O₂ + 4e⁻ → 2H₂O(l)
Overall equation : CH₃CH₂OH(g) + O₂(g) → HC₂H₃O₂(g) + H₂O(g)
ΔG° = - n F ΔE°
[(-374.2) + (-228.8) + (-167.9)] × 1000 = - 4 × 96500 × ΔE°
- 4 × 96500 × ΔE° = -434900
ΔE° = +434900 / (4 × 96500) V
ΔE° = +1.13 V
收錄日期: 2021-04-18 15:03:32
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