Use the data in Appendix 11 to rationalize quantitatively why:

(a) Mg liberates H₂ from dilute HCl, but Cu does not;

(b) Br₂ liberates I₂ from aqueous KI solution, but does not liberate Cl₂ from aqueous KCl solution;

(c) the role of Fe³⁺ ions as an oxidizing agent is influenced by the presence of certain ligands in solution;

(d) a method of growing Ag crystals is to immerse a zinc foil in an aqueous solution of AgNO₃.

Data from Appendix 11

The concentration of each aqueous solution is 1 mol dm⁻³ and the pressure of a gaseous component is 1 bar (10⁵ Pa). (Changing the standard pressure to 1 atm (101 300 Pa) makes no difference to the values of E^o at this level of accuracy.) Each half-cell listed contains the specified solution species at a concentration of 1 mol dm⁻³; where the half-cell contains [OH]⁻, the value of E^o refers to [OH⁻] = 1 mol dm⁻³, hence the notation E^o_{[OH⁻] = 1}

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Reduction half-equation Li (aq) +eLi(s) Cst (aq) +eCs(s) Rb+ (aq) + e Rb(s) K+ (aq) + eK(s) 2+ Ca²+ (aq) + 2e Ca(s) Na (aq) + e Na(s) 3+ La³+ (aq) + 3e = 2+ Mg²+ (aq) + 2e Y³+ (aq) + 3eY(s) Se³+ (aq) + 3e Sc(s) Al(s) Al³+ (aq) + 3e [HPO3)²(aq) + 2H₂O(1) + 2e [H₂PO₂] (aq) + 3[OH(aq) = = La(s) = Mg(s) Ti²+ (aq) + 2e Ti(s) Mn(OH) ₂ (s) +2e=Mn(s) + 2OH(aq) Mn²+ (aq) + 2e = Mn(s) 2+ Te(s) + 2e Te² (aq) = = V²+ (aq) + 2e = V(s) 2[SO3)²(aq) + 2H₂O(1) +2e=4[OH] (aq) + [S₂0₁1²(aq) [SO4²- (aq) + H₂O(1) +2e= [SO3)2(aq) + 2[OH]¯ (aq) Se(s) + 2e Se- (aq) C+ (aq) +2e Cr(s) 2[NO3] (aq) + 2H₂O(1) +2e=N₂O4(g) + 4[OH]¯(aq) 2H₂O(1) + 2e = H₂(g) + 2[OH]- (aq) E or E" (OH-]=1/V -3.04 -3.03 -2.98 -2.93 -2.87 -2.71 -2.38 -2.37 -2.37 -2.03 -1.66 -1.65 -1.63 -1.56 -1.19 -1.14 -1.13 -1.12 -0.93 -0.92 -0.91 -0,85 -0.82