Question

Using a graphing paper, construct the pE-pH diagram for the Cr-O-H2O system at 250C

and 1 atmosphere; assuming that the activity of chromium (Cr) is 10-6 M, and that the

chromium species shown below are the only prevalent species formed in this aqueous

system.

???+, ??(??)

?+, ?????, ?????

− ??? ????

?−

Use the textbook and any other relevant documents from peer reviewed literature

to accomplish the following:

1. Find necessary thermodynamic data from standard Tables THIS SECTION SHOULD BE

TYPED.

2. Use the step-by-step approach discussed in the lecture for construction of a pE-pH

diagram (include the steps defining the boundaries for stability of water). In this step,

show your work and a summary Table cataloging all the pairs of species used and the

determined equations to be plotted on the pE-pH diagram. THIS SECTION SHOULD BE

TYPED.

3. Plot on graphing paper and scan to upload to canvas. Make sure the document is upright

and readable

4. For each of the species and based on their distribution on your obtained pE-pH diagram,

name the types of natural aquatic environments where would one expect each of the

chromium species to occur. THIS SECTION SHOULD BE TYPED.

5. Based on your constructed pE-pH diagram, what combination of pH and pE conditions

would favor the prevalence of the most toxic species of chromium? For this last question,

support your conclusions with references from peer reviewed literature. THIS SECTION

SHOULD BE TYPED.1

Answer 1

pE-pH diagrams show in a comprehensive way how protons (pH) and electrons (pE) simultaneously shift reaction equilibria under various conditions indicating which species predominate under a given condition of pE and pH.

Two equations are used to produce the diagrams: pE = pE° +1/n log [{oxid}^o/ {reduc}^r]

pE° = 1/n log K

Oxidizing limit of diagrams: ¼ O ₂ + H ⁺ + e - → ½ H ₂ O

pE° = +20.75 (ref 1)

n = 1

pH = -log {H ⁺}

Set limit: {O ₂} = 1

pE = pE° +1/n log [({H⁺}{O2}^1/4)/ {H₂O}^1/2]

pE = 20.75 - pH

Reducing limit of diagrams: H ₂O + e ^- → ½ H ₂ + OH^-

OH^- + H ⁺ → H ₂ O

H ⁺ + e ^- → ½ H ₂

pE° = +0.0 (ref 1)

n = 1

pH = -log {H ⁺}

Set limit: {H ₂} = 1

pE = pE° +1/n log [{H⁺}/ {H₂}^1/2]

pE = - pH

ref 1: Aquatic Chemistry, W stumm and J J Morgan, p. 318

ref 2: Trans. Non-ferrous Met. Soc. China 20(2010) s26-s31