Exercise 10 | Electrochemistry

The combustion of ethanol in a fuel cell can be used to produce electricity.

CH3CH2OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l)

Calculate E0cell of the fuel cell at 25°C.


Data: ΔG0f [CH3CH2OH (l)] = - 174.8 kJ.mol-1

ΔG0f [CO2 (g)] = - 394.4 kJ.mol-1

ΔG0f [H2O (l)] = - 237.1 kJ.mol-1

E0cell = - G0rxnνeF

 

Determine ΔG0rxn:

ΔG0rxn = 2 x ΔG0f [CO2 (g)] + 3 x ΔG0f [H2O (l)] - ΔG0f [CH3CH2OH (l)] – 3 x ΔG0f [O2 (g)]

ΔG0rxn = 2 x (-394.4) + 3 x (-237.1) – (-174.8) – 3 x 0

ΔG0rxn = - 1.325 x 103 kJ.mol-1

 

Determine νe:


Unbalanced half-reactions:

Oxidation: CH3CH2OH (l) → 2 CO2 (g)

Reduction: O2 (g) → H2O (l)


Balanced half-reactions (see chap 24):

Oxidation: CH3CH2OH (l) + 3 H2O (l) → 2 CO2 (g) + 12 H+ (aq) + 12 e-

Reduction: O2 (g) + 4 H+ (aq) + 4 e- → 2 H2O (l)


3 O2 (g) + 12 H+ (aq) + 12 e- → 6 H2O (l)

⇒ νe = 12

 

E0cell = - G0rxnνeF

E0cell = 1.325 × 10312 × 96485

E0cell = 1.145 V