For the following half reaction, E°_cell = 22.07 V:

AlF₆^3– + 3e^-  →  Al + 6F₂

Using data from Table 11.1, calculate the equilibrium constant at 25^oC for the reaction Al³⁺(aq) 1 6F₂(aq) ⇌ AlF₆^3–(aq)

Table 11.1

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Standard Reduction Potentials at 25°G (298 K) for Many Common Haf-reactions Half-reaction %" (V) Half-reaction go (v) Og + 2H2O + 4e-→ 4OH Cu2+ + 2e-→ Cu F2 + 2e-→ 2F- 0.34 0.27 0.22 0.20 1.99 AgCl+ e-→ Ag+ Cl- SO42-+ 4H' + 2e-→ H2SO3+ H2O Cu2++ e-→ Cu" MnO4-+ 4H7 + 3e--> MnO, + 2H20 10c + 2H+ + 2e-→ io,-+ H,O Mndh" + 8H+ + 5e-→ Mn2+ 4H2O Fe3++ 3e-→ Fe Pb2++ 2e-→ Pb Sn2++ 2e-→ Sn Ni2++ 2e-→ Ni -0.036 PbO2 + 4H + + 2e--+ Pb2+ + 2H20 Cl2+ 2e-→ 2Cl -0.35 Fe2eFe 0, + 4H+ + 4e-→ 2H2O MnO, 4H+ + 2e-→ Mn2+ + 2H2O 1.21 1.20 Br2 2e2Br Zn2++ 2e-→ Zn 2H20 2eH2 20H -0.76 0.99 0.96 Al3eAl H2 2e2H Mg2+ +2e-→ M 2.23 -2.37 -2.76 -2.92 0.30 La't ter→ La 0.80 Na e Na + e-→ A Fe3++ e-→ Fe2+ Ba2+ + 2e--+ Ba 0.56 Li++ e-→ Li Cu teCu 0.52