Question

please answer the following questions

1- Assuming the partial pressure of oxygen gas is that of atmospheric oxygen 0.021 atm, rather than the 1.00 atm assumed in the derivation of Equation down, derive an equation describing the oxidizing pE limit of water as a function of pH.

equation is PE=20.75-PH

2- Using the equation derived in Problem 1 and Excel, plot logPO2 as a function of pE at pH 7.00.

3-Calculate the values of [Fe3+], pE and pH at the point in Figure 4.4 where Fe2+ is at a concentration of 1.00 x 10-5M, and Fe(OH)2 and Fe(OH)3 are in equilibrium.

Answer 1

(1)The limits of pE and pH in natural waters.

The limits on natural pE and pH in waters are the limits of stabilitly of water itself.

Water is stable if the availability of electrons is such that it is not all oxidized or reduced by these 1/2 rxns:

oxidizing limit 2H2O → O2 + 4H+ + 4e- Eo = 1.23V, pEo = 20.75

reducing limit 4H2O + 4e- → 2H2 + 4OH

only the oxidation state of H changes in this reaction, so this is equivalent to:

4H+ + 4e- → 2H2           Eo = 0V, pEo = 0

The oxidizing limit for the stability of water can be described as a function of Eo (or pE) and pH:

E = Eo + 0.059/n • log (PO2[H+ ] 4 /[H2O])

which reduces to this following, using n=4 and PO2 =1 atm

E = Eo + 0.059/4 • 4log (PO2[H+ ]/[H2O])

or E = Eo + 0.059log [H+ ]

or E = Eo - 0.059pH

E = - 0.059pH + 1.23V

or pE = - pH + 20.75 in terms of pE

Solving instead using PO2=0.2 atm (the value in the modern atmosphere instead of PO2=1 atm), the equation becomes:

E = - 0.059pH + 1.22V (small shift).

The corresponding pE/pH equation is also slightly shifted.