Exercise 10 | Electrochemistry

E⁰_cell = - $\frac{∆{{\mathrm{G}}^0}_{\mathrm{rxn}}}{{\mathrm{\nu }}_{\mathrm{e}}\mathrm{F}}$

Determine ΔG⁰_rxn:

ΔG⁰_rxn = 2 x ΔG⁰_f [CO₂ (g)] + 3 x ΔG⁰_f [H₂O (l)] - ΔG⁰_f [CH₃CH₂OH (l)] – 3 x ΔG⁰_f [O₂ (g)]

ΔG⁰_rxn = 2 x (-394.4) + 3 x (-237.1) – (-174.8) – 3 x 0

ΔG⁰_rxn = - 1.325 x 10³ kJ.mol^-1

Determine ν_e:

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)

⇒ ν_e = 12

E⁰_cell = - $\frac{∆{{\mathrm{G}}^0}_{\mathrm{rxn}}}{{\mathrm{\nu }}_{\mathrm{e}}\mathrm{F}}$

E⁰_cell = $\frac{1.325 \times {10}^3}{12 \times 96485}$

E⁰_cell = 1.145 V