Here is a galvanic cell: Zn (s) | Zn²⁺(aq) || Cd²⁺(aq) | Cd (s)

Write the cell equation and calculate E⁰_cell, ΔG_rxn and E_cell at 25°C.

Data: [Zn²⁺] = 1.00 x 10^-2M

[Cd²⁺] = 2.00 x 10^-2 M

E⁰_red (Zn²⁺/Zn) = - 0.762 V

E⁰_red (Cd²⁺/Cd) = - 0.403 V

Question

Here is a galvanic cell: Zn (s) | Zn²⁺(aq) || Cd²⁺(aq) | Cd (s)

Write the cell equation and calculate E⁰_cell, ΔG_rxn and E_cell at 25°C.

Data: [Zn²⁺] = 1.00 x 10^-2M

[Cd²⁺] = 2.00 x 10^-2 M

E⁰_red (Zn²⁺/Zn) = - 0.762 V

E⁰_red (Cd²⁺/Cd) = - 0.403 V

| General Chemistry 3

Accepted Answer

Zn (s) → Zn²⁺ (aq) + 2 e^-

Cd²⁺ (aq) + 2 e^- → Cd (s)

cell equation: Zn(s) + Cd²⁺ (aq) → Zn²⁺ (aq) + Cd (s)

E⁰_cell = E⁰_reduction + E⁰_oxidation

E⁰_cell = E⁰_red (Cd²⁺/Cd) - E⁰_red (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 $(\frac{10^{- 2}}{2 \times 10^{- 2}})$

ΔG_rxn = - 71.0 kJ.mol^-1

E_cell = -

\frac{∆ G_{rxn}}{ν_{e} F}

E_cell =

\frac{71.0 \times 10^{3}}{2 \times 96485}

E_cell = 0.368 V

Exercise 9 | Electrochemistry