1. (10 points) Based on the Pourbaix Diagram for selenium (Se), develop answers to the following questions:

a. Which form of the element is the strongest oxidant? HSeO4-.

The strongest reductant? Se2-.

b. What is the slope of the double boundary between HSeO₄^- and H₂SeO₃ at pH between 0 and 2?

Slope = -0.059*#protons in half reaction/#electrons in half reaction = -0.059*3/2 = -0.0885

c. Estimate K_a for H₂Se. This is the pH at which the boundary between H2Se and HSe- occurs: 3-4

d. Estimate e ^o (the standard reduction potential) for the SeO₃^2-(aq)/Se(s) couple. Extrapolate to pH=0: About 1 Volt.

e. Using the diagram, estimate pK_b for Se^2-(aq). pKa for HSe- is about 14, so pKb for Se2- is 0.

2. (10 points) Consider the two reduction processes below, in acidic aqueous solution:

Cr₂O₇^2-/Cr³⁺ e ^o = 1.33 V

Ni²⁺/Ni(s) e ^o = -0.23 V

a. Write a balanced half reaction for each process.

Cr2O72- + 14H+ + 6e ® 2Cr3+ + 7H2O

Ni2+ + 2e ® Ni(s)

b. Indicate the oxidant and reductant in each half reaction.

Oxidants: Cr2O72-, Ni2+

Reductants: Cr3+, Ni

c. Combine the two half reactions to give a spontaneous, balanced redox process.

Cr2O72- + 3Ni + 14H+ ® 2Cr3+ + 7H2O + 3Ni2+

d. Calculate the standard potential for the redox process in c. 1.56 V

e. Calculate the (non-standard) potential for the redox process at pH = 14 (all other species 1M).

e = e o - 0.059/n log Q = 1.56 - 0.059/6 log [Cr³⁺]²*[Ni²⁺]³/[H⁺]¹⁴

= 1.56 - 0.059/6 log 10¹⁹⁶ = = 1.56 - 196*0.059/6 = -0.37 V