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Problem Set IV Chemistry 3100 Winter 2011

1. The basal rate of consumption of O2 by a 70 kg human is about 16 mol of O2 per day. This O2 oxidizes food and is reduced to water, providing energy for the organism:

O2 + 4H+ + 4e- ( 2H2O

a) To what current does this respiration rate correspond?

(current = flow of electrons to O2)

b) Compare your answer with the current drawn by a refrigerator using 500 W at 115 V.

c) If the electrons flow from NADH to O2, they experience a potential drop of 1.1 V. What is the power output?

(Problem 13-3 in Harris 8th Ed.)

2. The space shuttles expendable booster engines derive their power from solid reactants:

6 NH4+ClO4-(s) + 10 Al(s) ( 3 N2(g) + 9 H2O(g) + 5 Al2O3(s) + 6 HCl(g)

a) What are the oxidation states of N, Cl, and Al in reactants and products? Which ones act as oxidants? Which ones as reducing agents?

b) The heat of reaction is 9334 kJ for every 10 mol of Al consumed. Express this as heat released per gram of total reactants.

(Problem 13-6 in Harris 8th Ed.)

3. Cyanide ion causes E for Fe(III) to decrease:

Fe3+ + e- ( Fe2+ E = 0.771 V

Fe(CN)63- + e- ( Fe(CN)64- E = 0.356 V

Which ion (Fe(II) or Fe(III)) is more stabilized by complexing with cyanide?

(Problem 13-13 in Harris 8th Ed.)

4. a) Write a line notation for the following cell:

b) Calculate the potential of each half-cell, and the cell voltage E. In which direction do the electrons flow through the circuit? Write the spontaneous cell reaction.

E (Al3+/Al) = -1.677 V; E (Br2(l)/Br-) = +1.078 Vc) The left cell was loaded with 14.3 mL of Br2(l) (density = 3.12 g/mL). The aluminum electrode contains 12.0 g of Al. Which element (Br2 or Al) is limiting?

d) If the cell is somehow operated under conditions in which it produces a constant voltage of 1.50 V, how much work will have been done when 0.231 mL of Br2(l) has been consumed?

e) If the potentiometer is replaced by a 1.20 k resistor, and if the heat dissipated by the resistor is 1.00 x 10-4 J/s, at what rate (g/s) is Al(s) dissolving?

(Problem 13-17 in Harris 8th Ed.)

5. Calculate E for the half-reaction: Pd(OH)2(s) + 2e- ( Pd(s) + 2 OH- given that Ksp for Pd(OH)2 is 3x10-28 and E = 0.915 V for the reaction Pd2+ + 2e- ( Pd(s)

(Problem 13-28 in Harris 8th Ed.)

6. The voltage of the cell shown here is 0.246 V. The RH cell contains the metal M2+ (E = -0.266 V). Calculate Kf for MY2-.

(Problem 13-38 in Harris 8th Ed.)

7. Silver forms a complex with ammonia: Ag+ + 2 NH3 ( Ag(NH3)2+

A solution was prepared containing 0.0410 M Ag(NH3)2+ and 0.0115 M NH3. A silver electrode was placed in the solution and connected via a salt bridge to a S.H.E. (the S.H.E. is the LH cell and acts as anode). The cell voltage was 0.529 V. Calculate the overall formation constant for the complex. E (Ag+/Ag) = +0.799 V. Answer: Kf = 1.1 x 107