Use the standard reduction potentials (Appendix M) for the half-reactions [AuCl₄]⁻(aq) + 3 e⁻ → Au(s) + 4 Cl⁻(aq) and Au³⁺(aq) + 3 e⁻ → Au(s) to calculate the value of K_formation for the complex ion [AuCl₄]⁻(aq).

Data given in Appendix M

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TABLE 21 Standard Reduction Potentials in Aqueous Solution at 25 °C Acidic Solution F₂(g) + 2 e 2 F-(aq) Co3+ (aq) + e Coz+(aq) Pb4+ (aq) + 2 e - Pb²+(aq) HzOz(aq) + 2 H*(aq) +2e → 2H2O NiO₂(s) + 4 H+ (aq) + 2 e→→→→→→ Ni+(aq) + 2 HO PbO₂ (s) + SO4² (aq) + 4 H+ (aq) + 2e → PbSO4(s) + 2 H₂O Au+ (aq) + e→→→→→ Au(s) 2 HCIO(aq) + 2 H+ (aq) + 2 e-- - Ce+(aq) + e→→→→ Ce³+ (aq) NaBiO;(s) + 6 H+ (aq) + 2 e- → MnO4 (aq) + 8 H+ (aq) + 5 e Au³+ (aq) + 3 e→→→→→ Au(s) CIO (aq) + 6 H+ (aq) +5 e→→→→→→ BrO3 (aq) + 6 H+ (aq) + 6 e- Cl₂(g) + 2 e 2 Cl-(aq) Cr₂0,² (aq) + 14 H*(aq) + 6 e 2 Cr³+ (aq) + 7 H₂O N₂H5+ (aq) + 3 H+ (aq) + 2 e2 NH4+ (aq) MnO₂ (s) + 4 H+ (aq) + 2 e O₂(g) + 4 H+(aq) + 4 e 2 H₂O Pt+ (aq) + 2 e →→→→→ Pt(s) 10- (aq) + 6 H+ (aq) + 5 e1₂(aq) + 3 H₂O CIO (aq) + 2 H+ (aq) + 2 e CIO₂ (aq) + H₂O Br₂() +2 e 2 Br(aq) AuCl(aq) + 3 e → Cl₂(g) + 2 H₂O → Bi³+ (aq) + Na+ (aq) + 3 H₂O Mn²+ (aq) + 4H₂O Cl₂(g) + 3 H₂O Br (aq) + 3 H₂O Mn²+ (aq) + 2 H₂O Au(s) + 4 CI-(aq) Standard Reduction Potential E (volts) 2.87 1.82 1.8 1.77 1.7 1.685 1.68 1.63 1.61 = 1.6 1.51 1.50 1.47 1.44 1.36 1.33 1.24 1.23 1.229 1.2 1.195 1.19 1.08 1.00