Chemistry 65.100 A, V
Midterm Test #3
January 29, 1999
ANSWERS

Part A

1.    oxidation: Al_(s) à Al⁺³_(aq) + 3 e^-
      reduction: 2 H⁺_(aq) + 2 e^- à H_2(g)
      overall: 2 Al_(s) + 6 H⁺_(aq) à 2 Al⁺³_(aq) + 3 H_2(g)

2.    I_2(s) is higher in the reduction series than Ni. Thus, I_2(s) is more easily reduced, and can therefore act as an oxidant to Ni_(s). Also, you could add the two half reaction potentials and show that the sum, E^o_cell, is positive.

3.    Water is more easily reduced (to H_2(g) and OH^-_(aq)) than Al⁺³_(aq) is to Al_(s). Thus, the water gets reduced and the Al⁺³_(aq) does not.

4.    1. The reaction goes to completion, and the reactants are used up.
       2. The zinc anode gets corroded in the acidic electrolyte.

5.    The sacrificial anode oxidizes, and supplies high potential electrons to the object being protected, thus preventing it from oxidizing.

6.    Diethyl ether can not form hydrogen bonds because the H atoms are bound to the C atoms. For an H-bond to form, the H must be bound to N, O or F.

Part B

1.    The reduction reaction is Au⁺³_(aq) + 3 e^- à Au_(s)

q = it here, I = 1.5 A = 1.5 C/s and t = 1 h = 3600 s

thus, q = 1.5 C/s x 3600 s = 5400 C

but, n = q/F = 5400 C / (96500 C/mol e^-) = 0.0560 mol e^-

moles Au = n/3 = 0.0187 mol Au

g Au = 0.0187 mol x 197 g/mol = 3.68 g

2a. oxidation: Pb_(s) + HSO₄^-_(aq) à PbSO_4(s) + H⁺_(aq) + 2 e^-

        reduction: PbO_2(s) + HSO₄^-_(aq) + 3 H⁺_(aq) + 2 e^- à PbSO_4(s) + 2 H₂O_(l)

        overall: Pb_(s) + PbO_2(s) + 2 H⁺_(aq) + 2 HSO₄^-_(aq) à 2 PbSO_4(s) + 2 H₂O_(l)

2b. Q = 1/([H⁺_(aq)]²[HSO₄^-_(aq)]²)

2c. E = E^o – (0.0592/n) log₁₀Q

pH = -1.00, therefore [H⁺_(aq)] = [HSO₄^-_(aq)] = 10¹ = 10 M

Q = 1/([H⁺_(aq)]²[HSO₄^-_(aq)]²) = 1/(10² 10²) = 10^-4

Thus, E = 1.924 – 0.0592/2 log₁₀(10^-4) = 1.924 – (-0.118) = 2.042 V

(You could also use the other form of the Nernst equation: E = E^o – (RT/nF) ln Q

= 1.924 V – (8.314 x 298)/(2 x 96487) ln(10^-4) = 2.042 V

3a. Ar_(g) – Ar_(g) dispersion only
3b. HF_(g) – H₂O_(g) dispersion, H-bonds
3c. Br^-_(aq) – H₂O_(l) ion-dipole, dipole-induced dipole, dispersion
3d. H₂ – H₂O dipole-induced dipole, dispersion

4.    ln(p₂) = ln(p₁) + DH_vap/R [1/T₁ – 1/T₂]

"normal boiling point" means the vapor pressure = 1 atm at this temperature. Thus, p₁ = 1 atm, T₁ = 100^oC = 373 K, T₂ = 85^oC = 358 K

thus, ln(p₂) = ln(1) + 40700/8.314 (1/373 – 1/358) = - 0.550

thus, p₂ = e^-0.550 = 0.58 atm