Exercice 6 | Électrochimie

Half reactions: Ga (s) → Ga³⁺ (aq) + 3 e^- Ag⁺ (aq) + e^- → Ag (s) Nernst equation: Oxidation half reaction: E_ox = E⁰_ox (Ga³⁺/Ga) – $\frac{0.02570}{{\mathrm{\nu }}_{\mathrm{e}}}$ ln [Ga³⁺] E_ox = - E⁰_red (Ga³⁺/Ga) – $\frac{0.02570}{{\mathrm{\nu }}_{\mathrm{e}}}$ ln [Ga³⁺] E_ox = 0.549 – $\frac{0.02570}{3}$ x ln 10^-5 E_ox = 0.648 V Reduction half reaction: E_red = E⁰_red (Ag⁺/Ag) – $\frac{0.02570}{{\mathrm{\nu }}_{\mathrm{e}}}$ ln $\frac{1}{\left[{\mathrm{Ag}}^{+}\right]}$ E_red = 0.7996 + 0.02570 ln 10^-3 E_red = 0.622 V Standard cell voltage E_cell = E_ox + E_red = 1.27 V