CEE 680

Fall 2011

Water Chemistry

Homework #10

Consider a deep lake sediment in equilibrium with 10-4 atm of oxygen. Like many large lakes, this one has substantial deposits of manganese in its sediments.

Use the following thermodynamic data. Assume the system is buffered at pH 7.0.

Equ#	Half Cell Reaction	DE^o (Volts)
1	O2(g) + 4H+ + 4e- = 2H2O	+1.23
2	Mn^+3 + e- = Mn^+2	+1.51
3	Mn^+4 + e- = Mn^+3	+1.65
4	MnO4^- + 8H+ + 5e- = Mn^+2 + 4H2O	+1.49
5	Fe^+3 + e- = Fe^+2	+0.77

Equ#	Equilibrium	Log K
6	Mn^+2 + H2O = MnOH+ + H^+	-10.6
7	Mn^+2 + 3H2O = Mn(OH)3^- + 3H^+	-35
8	Mn^+2 + 4H2O = Mn(OH)4^-2 + 4H^+	-48.3
9	Mn(OH)2 (s) = Mn^+2 + 2OH^-	-12.8

Problem #1

Write a balanced equation for the oxidation of Mn(+II) to Mn(+III), and then on to Mn(+IV) by oxygen.

Solution to #1

Combine half reactions with the appropriate multiplier so that the electrons cancel out.

First Step

Mult.	Reaction	E^o (Volts)
1 X	Mn^+2 = Mn^+3 + e^-	-1.51
¼ X	O2(g) + 4H+ + 4e- = 2H2O	+1.23
	Mn^+2 + ¼ O2(g) + H^+ = Mn^+3 + ½ H2O	-0.28

Second Step

Mult.	Reaction	E^o (Volts)
1 X	Mn^+3 = Mn^+4 + e^-	-1.65
¼ X	O2(g) + 4H+ + 4e- = 2H2O	+1.23
	Mn^+3 + ¼ O2(g) + H^+ = Mn^+4 + ½ H2O	-0.42

Combined

Mult.	Reaction			E^o (Volts)
1^st step			Mn^+2 + ¼ O2(g) + H^+ = Mn^+3 + ½ H2O		-0.28
2^nd step			Mn^+3 + ¼ O2(g) + H^+ = Mn^+4 + ½ H2O		-0.42
		Mn^+2 + ½ O2(g) + 2H^+ = Mn^+4 + H2O			-0.35

Note that since this is a set of sequential balance redox reactions, the E^mpton value is deteremined by taking the average of the two single step values, as it represents the average per electron for both steps.

Problem #2

Determine the stoichiometric requirements of oxygen for the complete conversion of Mn(+II) to Mn(+IV) in mg-oxygen/mg-manganese.

Solution to #2

The stoichiometry for both steps is the same: 0.5 moles of oxygen (O₂) per mole of manganese. Since the GFW's for oxygen and manganese are 32 and 55, respectively, this translates to:

0.29 mg-O₂/mg-Mn

Problem #3

Determine the Log K for this reaction.

Solution to #3

First Step

Log K₁ = (1/0.059)(-0.28) = -4.75

K = 10^-4.75

Second Step

Log K₂ = (1/0.059)(-0.42) = -7.12

K = 10^-7.12

Combined

Log K = Log K₁ + Log K₂

K = 10^-11.87

Problem #4

Based on the partial pressure of oxygen, determine pe for this system.

Solution to Problem #4

For the half reaction: O₂(g) + 4H+ + 4e- = 2H₂O

pe = pe^o - (1/n)log (1/[H⁺]⁴ p_O2)

pe = (1.23/0.059) - (1/4)log (1/[10^-7]⁴ [10^-4])

pe = 20.8 - 8

pe = 12.8

Problem #5

Now calculate the ratios of the various free-Mn species (i.e., Mn⁺², Mn⁺³, Mn⁺⁴) in the pore water.

Solution to #5

Using the same equation, we get:

First Step

pe = pe^o - (1/n)log([Mn⁺²]/[Mn⁺³])

12.8 = (+1.51/0.059) - log([Mn⁺²]/[Mn⁺³])

- log([Mn⁺²]/[Mn⁺³]) = -12.8

[Mn⁺³]/[Mn⁺²] = 10^-12.8

Second Step

pe = pe^o - (1/n)log([Mn⁺³]/[Mn⁺⁴])

12.8 = (+1.65/0.059) - log([Mn⁺³]/[Mn⁺⁴])

- log([Mn⁺³]/[Mn⁺⁴]) = -15.2

[Mn⁺⁴]/[Mn⁺³] = 10^-15.2

This gives an overall ratio for Mn⁺⁴ to Mn⁺³ to Mn⁺² of:

10^-28/10^-12.8/1

Problem #6

Assume the lake contains large deposits of Mn(OH)2(s) . What will the total concentration of soluble manganese be in the pore water, if all species are in redox equilibrium?

Solution to #6

1. Determine concentration of Mn⁺² and total soluble Mn(+II) species in equilibrium with manganese hydroxide solid.

2. Then apply the ratios of the various free Mn species to the Mn⁺² concentration to get the Mn⁺³ and Mn⁺⁴,

3. Finally, add these up to get the total dissolved manganese for all oxidation states. (N.B. This ignores the possible precipitation and hydrolysis of the Mn⁺³ and Mn⁺⁴ species. In fact, Mn⁺⁴ readily forms manganese dioxide which will precipitate).

From Kso, we have:

Note that the divalent hydroxide is highly soluble and would certainly completely dissolve, but let’s see what the answer is for the more oxidized species.

[Mn⁺³]/[Mn⁺²] = 10^-12.8

[Mn⁺³] = 10^-12.810^+1.2

[Mn⁺³] = 10^-11.6

[Mn⁺⁴]/[Mn⁺³] = 10^-15.2

[Mn⁺⁴] = 10^-15.210^-11.6

[Mn⁺⁴] = 10^-26.8

And the total is essentially equal to the divalent form:

Mn_T ~ 10^+1.2