Exercice 9 | Électrochimie

Zn (s) → Zn²⁺ (aq) + 2 e^-

Cd²⁺ (aq) + 2 e^- → Cd (s)

équation de la cellule : Zn(s) + Cd²⁺ (aq) → Zn²⁺ (aq) + Cd (s)

E⁰_cell = E⁰_réduction + E⁰_oxydation

E⁰_cell = E⁰_rédox (Cd²⁺/Cd) - E⁰_rédox (Zn²⁺/Zn)

E⁰_cell = - 0.403 + 0.762

E⁰_cell = 0.359 V

ΔG⁰_rxn = - ν_eFE⁰_cell

ΔG⁰_rxn = - 2 x 96485 x 0.359

ΔG⁰_rxn = - 69.3 kJ.mol^-1

ΔG_rxn = ΔG⁰_rxn + RT ln Q

ΔG_rxn = - 69.3 x 10³ + 293 x 8.314 x ln $\left(\frac{{10}^{-2}}{2 \times {10}^{-2}}\right)$

ΔG_rxn = - 71.0 kJ.mol^-1

E_cell = - $\frac{∆{\mathrm{G}}_{\mathrm{rxn}}}{{\mathrm{\nu }}_{\mathrm{e}}\mathrm{F}}$

E_cell = $\frac{71.0 \times {10}^3}{2 \times 96485}$

E_cell = 0.368 V