ΔGº for the following reaction is:4Al + 3O2 + 6H2O + 4OH– → 4Al(OH)4–Eºcell = 2.73 VΔfGº(OH–) = − 157 kJ mol−1ΔfGº(H2O) = − 237 kJ mol−1

Engineering

Chemistry

Cell Potential Free Energy and Equilibrium Constant

Question

ΔGº for the following reaction is:

4Al + 3O₂+ 6H₂O + 4OH^– → 4Al(OH)₄^–

Eº_cell = 2.73 V

Δ_fGº_(OH^–) = − 157 kJ mol⁻¹

Δ_fGº(H₂O) = − 237 kJ mol⁻¹

−3.16 × 10³ kJ mol⁻¹

−0.79 × 10³ kJ mol⁻¹

−0.263 × 10³ kJ mol⁻¹

+0.263 × 10³ kJ mol⁻¹

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As we know,

Δ_rGº_cell = − n_cellFEº_cell = − 12 × 96500 × 2.73

= − 3.16 × 10³ kJ mol⁻¹

For n_cell, write the half cell reaction as:

Cathode : 2H₂O(l) + O₂(g) + 4e⁻ → 4H(aq)

Anode : Al(s) + 4H (aq) → Al(OH)º₄(aq) + 3e⁻

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