21) Determine E°cell for the reaction: 2 Al + 3 Zn2+ → 2 Al3+ + 3 Zn. The half reactions are: Al3+(aq) + 3 e- → Al(s)E° = -1.676 V              Zn2+(aq) + 2 e- → Zn(s)          E° = -0.763 V A) 0.913 V B) -2.439 V C) 2.439 V D) -1.063 V E) -0.913 V 22) Determine E°cell for the reaction: 2 Ag+ + Mg → 2 Ag + Mg2+. The half reactions are: Mg2+(aq) + 2 e- → Mg(s)E° = -2.356 V              Ag+(aq) + e- → Ag(s)              E° = 0.800 V A) 0.756 V B) -0.756 V C) 3.156 V D) -1.556 V E) 1.556 V 23) Determine E°cell for the reaction: Pb2+ + Zn → Pb + Zn2+. The half reactions are: Pb2+(aq) + 2 e- → Pb(s)E° = -0.125 V              Zn2+(aq) + 2 e- → Zn(s)E° = -0.763 V A) 0.638 V B) -0.638 V C) 0.888 V D) -1.276 V E) -0.888 V 24) Choose the INCORRECT statement. A) The work available in a cell = -zFEcell. B) ΔG° = -zFE°cell C) When Ecell < 0, the reaction is spontaneous. D) When Ecell = 0, the reaction is at equilibrium. E) Reversing a cell reaction changes the sign of Ecell. 25) For the reaction: Mg(s) + AgNO3(aq) → Ag(s) + Mg(NO3)2(aq) Ag+(aq) + e- → Ag(s)E° = 0.800 V Mg2+(aq) + 2 e- → Mg(s)E° = -2.356 V Is the reaction spontaneous and why? A) No, Mg(s) does not react with Ag+. B) No, E° is a positive value. C) No, E° is a negative value. D) Yes, E° is a positive value. E) Yes, E° is a negative value. 26) What is the cell diagram for the spontaneous cell involving the Fe3+ ? Fe2+ (0.771V) half cell and the Zn2+ ? Zn (-0.763 V) half cell? A) Fe2+(aq) ? Fe3+(aq) ? Zn2+(aq) ? Zn(s) B) Pt(s) ? Fe2+(aq), Fe3+(aq) ? Zn2+(aq) ? Zn(s) C) Zn(s) ? Zn2+(aq) ? Fe3+(aq) ? Fe2+(aq) ? Pt(s) D) Zn(s) ? Zn2+(aq) ? Fe3+(aq) ? Fe2+(aq) E) Zn(s) ? Zn2+(aq) ? Fe3+(aq) ? Fe2+(aq) ? Zn(s) 27) If the voltage of an electrochemical cell is negative then the cell reaction is: A) nonspontaneous B) slow C) exothermic D) spontaneous E) fast 28) For the reaction: Mg(s) + AgNO3(aq) → Ag(s) + Mg(NO3)2(aq) Ag+(aq) + e- → Ag(s)E° = 0.800 V Mg2+(aq) + 2 e- → Mg(s)E° = -2.356 V Estimate Keq at 25°C. A) 5.7 × 10106 B) 2.0 × 1053 C) 3.7 × 1052 D) 1.1 × 1027 E) 10-107 29) Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25°C: 2 e- + PbI2(S) → Pb(s) + 2 I-(aq)E° = -0.365 V 2 e- + Pb2+(aq) → Pb(s)E° = -0.126 V A) 8 × 10-9 B) 9 × 10-5 C) 2 × 10-17 D) 5 × 10-13 E) 6 × 10-5 30) In the cell Fe2+ ? Fe3+ ? Cu2+ ? Cu which will increase the cell voltage the most? A) Halve [Cu2+]. B) Halve [Fe2+]. C) Double [Cu2+]. D) Double [Fe2+]. E) Cut Cu electrode in half.