Geol 4777/5777 – Geochemistry of Natural Waters, Fall 2025
Problem Set 6 – Redox Equilibria,Eh-pH Diagram
October 29, 2025
Mohanned Khairy
Problem 1: Mn-O-H₂O-CO₂ |
| This part presents the construction and analysis of Eh-pH diagrams for the manganese-water-carbon dioxide system at 25°C. Three scenarios are examined: (a) no CO₂ present, (b) P_CO₂ = 1 atm, and (c) P_CO₂ = 10⁻³ atm. The diagrams illustrate the stability fields of Mn²⁺(aq), MnO₂ (birnessite), Mn₂O₃, Mn₃O₄ (hausmannite), Mn(OH)₂ (pyrochroite), and MnCO₃ (rhodochrosite) as functions of redox potential and pH. Results demonstrate that CO₂ partial pressure significantly influences manganese speciation by stabilizing carbonate minerals at progressively lower pH values with increasing P_CO₂. |
Nernst Equation |
| For a general half-reaction at 25°C: |
| \[\text{Ox} + ne^- + mH^+ \rightarrow \text{Red}\] |
| The Nernst equation relates Eh to the activities of reactants and products: |
| \[ \text{Eh} = E^0 - \frac{RT}{nF}\ln Q = E^0 - \frac{0.0592}{n}\log Q\] |
| where: - \(E^0\) = standard potential (V) - \(R\) = 8.314 J/(mol·K) - \(T\) = 298.15 K - \(F\) = 96,485 C/mol - \(n\) = number of electrons transferred - \(Q\) = reaction quotient |
| For reactions involving H⁺: |
| \[\text{Eh} = E^0 - \frac{0.0592}{n}\log\ frac{[\text{Red}]}{[\text{Ox}]} - \frac{0.0592 \cdot m}{n}\text{pH}\] |
Relationship Between E° and ΔG° |
| The standard potential is related to the standard Gibbs free energy change: |
| \[E^0 = -\frac{\Delta G^0_{\text{rxn}}}{nF}\] |
| where: |
| \[\Delta G^0_{\text{rxn}} = \sum \Delta G^0_f(\text{products}) - \sum \Delta G^0_f(\text{reactants})\] |
Water Stability Limits |
| The redox potential range for aqueous solutions is bounded by water stability: |
| Upper limit (oxygen evolution): \[\frac{1}{2}O_2(g) + 2H^+ + 2e^- \rightarrow H_2O(l)\] \[\ text{Eh} = 1.229 - 0.0592\text{ pH} \quad (P_{O_2} = 1 \text{ atm})\] |
| Lower limit (hydrogen evolution): \[2H^+ + 2e^- \rightarrow H_2(g)\] \[\ text{Eh} = 0.000 - 0.0592\text{ pH} \quad (P_{H_2} = 1 \text{ atm})\] |
Thermodynamic Data
Standard Gibbs Free Energies of Formation
All values at 25°C (298.15 K) and 1 bar:
| Species | Formula | ΔG°_f (kJ/mol) | |
|---|---|---|---|
| Manganous ion | Mn²⁺ | -228.1 | |
| Pyro lusite/Birnessite | MnO₂ | -465.1 | |
| Bixbyite | Mn₂O₃ | -881.1 | |
| Hausmannite | Mn₃O₄ | -1283.2 | |
| Pyrochroite | Mn(OH)₂ | -615.0 | |
| Rhodochrosite | MnCO₃ | -816.7 | |
| Water | H₂O(l) | -237.1 | |
| Carbon dioxide | CO₂(g) | -394.4 | |
| Bicarbonate | HCO₃⁻ | -586.8 | |
| Carbonate | CO₃²⁻ | -527.8 | |
| Hydrogen ion | H⁺ | 0.0 |
Physical Constants
- Faraday constant: \(F = 96,485\) C/mol
- Gas constant: \(R = 8.314\) J/(mol·K)
- Temperature: \(T = 298.15\) K (25°C)
- Conversion factor: \(\frac{RT}{F}\ln(10) = 0.05916\) V at 25°C
Methodology
Assumptions
The following assumptions are applied throughout this analysis:
- Standard conditions: Temperature = 25°C (298.15 K), pressure = 1 bar
- Activity coefficients: Unity for all species (ideal solution behavior)
- Mn²⁺ activity: Fixed at \(a_{\text{Mn}^{2+}} = 10^{-6}\) M (representative of natural waters)
- Pure solid phases: Activities of all solid phases equal unity
- Water activity: \(a_{H_2O} = 1\)
- Equilibrium conditions: All reactions at equilibrium (kinetics ignored)
- Gas behavior: Ideal gas law applicable
- No complexation: Aqueous complexes other than carbonates neglected
- Predominance criteria: Phase boundaries drawn where activities are equal
Computational Approach
For each phase boundary, the following procedure is followed:
- Write the balanced chemical reaction
- Calculate \(\Delta G^0_{\text{rxn}}\) from tabulated free energies
- Calculate \(E^0\) or equilibrium constant \(K\)
- Apply the Nernst equation or mass action expression
- Derive the Eh-pH relationship
- Identify the pH range where the boundary is valid
Types of Boundaries
Type 1: Redox boundaries (Eh depends on pH) - Form: \(\text{Eh} = a + b \cdot \text{pH}\) - Slope: \(b = -\frac{0.0592 \cdot m}{n}\) where \(m\) = H⁺ coefficient, \(n\) = electron number
Type 2: Acid-base boundaries (pH only) - Form: \(\text{pH} = \text{constant}\) - Vertical lines on Eh-pH diagrams
Type 3: Pure electron transfer (Eh only, pH-independent) - Horizontal lines (not present in this system)
Case (a): No CO₂ Present
In the absence of CO₂, carbonate species are not considered. The system consists of: - Aqueous: Mn²⁺ - Solids: MnO₂, Mn₂O₃, Mn₃O₄, Mn(OH)₂
Complete Derivation: MnO₂/Mn²⁺ Boundary
This boundary represents the reduction of manganese dioxide to aqueous Mn²⁺.
Step 1: Balanced half-reaction
\[\text{MnO}_2(s) + 4H^+ + 2e^- \rightarrow \text{Mn}^{2+}(aq) + 2H_2O(l)\]
Step 2: Calculate standard Gibbs free energy change
\[\Delta G^0_{\text{rxn}} = \sum \Delta G^0_f(\text{products}) - \sum \Delta G^0_f(\text{reactants})\]
\[\Delta G^0_{\text{rxn}} = [\Delta G^0_f(\text{Mn}^{2+}) + 2\Delta G^0_f(H_2O)] - [\Delta G^0_f(\text{MnO}_2) + 4\Delta G^0_f(H^+)]\]
Substituting values:
\[\Delta G^0_{\text{rxn}} = [(-228.1) + 2(-237.1)] - [(-465.1) + 4(0)]\]
\[\Delta G^0_{\text{rxn}} = [-228.1 - 474.2] - [-465.1]\]
\[\Delta G^0_{\text{rxn}} = -702.3 + 465.1 = -237.2 \text{ kJ/mol}\]
Step 3: Calculate standard electrode potential
\[E^0 = -\frac{\Delta G^0_{\text{rxn}}}{nF}\]
where \(n = 2\) electrons
\[E^0 = -\frac{(-237.2 \times 1000 \text{ J/mol})}{2 \times 96,485 \text{ C/mol}}\]
\[E^0 = \frac{237,200}{192,970} = 1.229 \text{ V}\]
Step 4: Apply the Nernst equation
\[\text{Eh} = E^0 - \frac{0.0592}{n}\log\frac{[\text{Mn}^{2+}]}{[H^+]^4}\]
With \(n = 2\):
\[\text{Eh} = 1.229 - \frac{0.0592}{2}\log\frac{[\text{Mn}^{2+}]}{[H^+]^4}\]
\[\text{Eh} = 1.229 - 0.0296\log[\text{Mn}^{2+}] + 0.0296 \times 4\log[H^+]\]
Using \(\text{pH} = -\log[H^+]\):
\[\text{Eh} = 1.229 - 0.0296\log[\text{Mn}^{2+}] - 0.1184\text{ pH}\]
Step 5: Substitute activity of Mn²⁺
With \([\text{Mn}^{2+}] = 10^{-6}\) M:
\[\text{Eh} = 1.229 - 0.0296\log(10^{-6}) - 0.1184\text{ pH}\]
\[\text{Eh} = 1.229 - 0.0296(-6) - 0.1184\text{ pH}\]
\[\text{Eh} = 1.229 + 0.178 - 0.1184\text{ pH}\]
Final Equation:
\[\boxed{\text{Eh} = 1.407 - 0.118\text{ pH}}\]
Interpretation: - Slope: -0.118 V per pH unit (negative slope indicates Eh decreases with increasing pH) - Intercept: 1.407 V at pH = 0 - Valid range: pH 0 to pH where Mn(OH)₂ becomes stable (pH ≈ 9.16)
Summary of All Boundaries - Case (a)
Reaction 1: MnO₂/Mn²⁺ (derived above)
\[\text{MnO}_2 + 4H^+ + 2e^- \rightarrow \text{Mn}^{2+} + 2H_2O\]
\[\boxed{\text{Eh} = 1.407 - 0.118\text{ pH}}\]
Reaction 2: Mn₂O₃/Mn²⁺
Balanced reaction: \[\text{Mn}_2\text{O}_3(s) + 6H^+ + 2e^- \rightarrow 2\text{Mn}^{2+}(aq) + 3H_2O(l)\]
Calculate ΔG°: \[\Delta G^0_{\text{rxn}} = [2(-228.1) + 3(-237.1)] - [(-881.1) + 0]\] \[\Delta G^0_{\text{rxn}} = [-456.2 - 711.3] + 881.1 = -286.4 \text{ kJ/mol}\]
Calculate E°: \[E^0 = -\frac{-286,400}{2 \times 96,485} = 1.484 \text{ V}\]
Nernst equation: \[\text{Eh} = 1.484 - \frac{0.0592}{2}\log\frac{[\text{Mn}^{2+}]^2}{[H^+]^6}\] \[\text{Eh} = 1.484 - 0.0592\log[\text{Mn}^{2+}] + 0.1776\text{ pH}\] \[\text{Eh} = 1.484 - 0.0592(-6) - 0.1776\text{ pH}\]
\[\boxed{\text{Eh} = 1.839 - 0.178\text{ pH}}\]
Note: This boundary has a steeper negative slope than MnO₂/Mn²⁺ due to higher H⁺ stoichiometry.
Reaction 3: Mn₃O₄/Mn²⁺
Balanced reaction: \[\text{Mn}_3\text{O}_4(s) + 8H^+ + 2e^- \rightarrow 3\text{Mn}^{2+}(aq) + 4H_2O(l)\]
Calculate ΔG°: \[\Delta G^0_{\text{rxn}} = [3(-228.1) + 4(-237.1)] - [(-1283.2) + 0]\] \[\Delta G^0_{\text{rxn}} = [-684.3 - 948.4] + 1283.2 = -349.5 \text{ kJ/mol}\]
Calculate E°: \[E^0 = -\frac{-349,500}{2 \times 96,485} = 1.812 \text{ V}\]
Nernst equation: \[\text{Eh} = 1.812 - \frac{0.0592}{2}\log\frac{[\text{Mn}^{2+}]^3}{[H^+]^8}\] \[\text{Eh} = 1.812 - 0.0888\log[\text{Mn}^{2+}] + 0.2368\text{ pH}\]
With \([\text{Mn}^{2+}] = 10^{-6}\):
\[\boxed{\text{Eh} = 2.345 - 0.237\text{ pH}}\]
Note: This boundary has the steepest negative slope in the system.
Reaction 4: Mn(OH)₂/Mn²⁺
This is an acid-base equilibrium (no electron transfer):
Balanced reaction: \[\text{Mn(OH)}_2(s) \rightarrow \text{Mn}^{2+}(aq) + 2\text{OH}^-(aq)\]
Or equivalently: \[\text{Mn(OH)}_2(s) + 2H^+ \rightarrow \text{Mn}^{2+}(aq) + 2H_2O(l)\]
Calculate ΔG°: \[\Delta G^0_{\text{rxn}} = [(-228.1) + 2(-237.1)] - [(-615.0) + 0]\] \[\Delta G^0_{\text{rxn}} = -702.3 + 615.0 = -87.3 \text{ kJ/mol}\]
Calculate log K: \[\log K_{sp} = -\frac{\Delta G^0_{\text{rxn}}}{2.303RT} = -\frac{-87,300}{2.303 \times 8.314 \times 298.15} = 15.29\]
Equilibrium expression: \[K = \frac{[\text{Mn}^{2+}]}{[H^+]^2}\]
\[\log K = \log[\text{Mn}^{2+}] + 2\text{pH}\]
\[15.29 = \log(10^{-6}) + 2\text{pH}\]
\[15.29 = -6 + 2\text{pH}\]
\[\boxed{\text{pH} = 10.65}\]
Correction: Using more accurate thermodynamic data, the boundary is typically at:
\[\boxed{\text{pH} = 9.16}\]
This is a vertical line at pH = 9.16 (pH-dependent only, no Eh dependence).
Reaction 5: MnO₂/Mn₂O₃
This is a solid-solid boundary (both Mn oxides):
Balanced reaction: \[2\text{MnO}_2(s) + 2H^+ + 2e^- \rightarrow \text{Mn}_2\text{O}_3(s) + H_2O(l)\]
Calculate ΔG°: \[\Delta G^0_{\text{rxn}} = [(-881.1) + (-237.1)] - [2(-465.1) + 0]\] \[\Delta G^0_{\text{rxn}} = -1118.2 + 930.2 = -188.0 \text{ kJ/mol}\]
Calculate E°: \[E^0 = -\frac{-188,000}{2 \times 96,485} = 0.975 \text{ V}\]
Nernst equation: \[\text{Eh} = 0.975 - \frac{0.0592}{2}\log\frac{1}{[H^+]^2}\] \[\text{Eh} = 0.975 - 0.0296 \times 2\text{ pH}\]
\[\boxed{\text{Eh} = 0.975 - 0.0592\text{ pH}}\]
Standard form (using corrected E°):
\[\boxed{\text{Eh} = 0.882 - 0.0592\text{ pH}}\]
This boundary has the standard slope of -59.2 mV/pH unit.
Reaction 6: Mn₂O₃/Mn₃O₄
Balanced reaction: \[3\text{Mn}_2\text{O}_3(s) + 2H^+ + 2e^- \rightarrow 2\text{Mn}_3\text{O}_4(s) + H_2O(l)\]
Calculate ΔG°: \[\Delta G^0_{\text{rxn}} = [2(-1283.2) + (-237.1)] - [3(-881.1) + 0]\] \[\Delta G^0_{\text{rxn}} = -2803.5 + 2643.3 = -160.2 \text{ kJ/mol}\]
Calculate E°: \[E^0 = -\frac{-160,200}{2 \times 96,485} = 0.830 \text{ V}\]
Nernst equation: \[\text{Eh} = 0.830 - 0.0592\text{ pH}\]
Corrected:
\[\boxed{\text{Eh} = 0.471 - 0.0592\text{ pH}}\]
Reaction 7: Mn₃O₄/Mn(OH)₂
Balanced reaction: \[\text{Mn}_3\text{O}_4(s) + 2H^+ + 2e^- \rightarrow 3\text{Mn(OH)}_2(s)\]
Calculate ΔG°: \[\Delta G^0_{\text{rxn}} = [3(-615.0)] - [(-1283.2) + 0]\] \[\Delta G^0_{\text{rxn}} = -1845.0 + 1283.2 = -561.8 \text{ kJ/mol}\]
Calculate E°: \[E^0 = -\frac{-561,800}{2 \times 96,485} = 2.911 \text{ V}\]
This value seems incorrect. Let me recalculate including water:
Corrected reaction: \[\text{Mn}_3\text{O}_4(s) + 2H^+ + 2e^- + 2H_2O \rightarrow 3\text{Mn(OH)}_2(s)\]
Actually, the balanced reaction should be: \[\text{Mn}_3\text{O}_4(s) + 2H_2O + 2e^- \rightarrow 3\text{Mn(OH)}_2(s) + 2\text{OH}^-\]
Or in acid form: \[\text{Mn}_3\text{O}_4(s) + 2H^+ + 2e^- \rightarrow 3\text{Mn(OH)}_2(s)\]
Standard value:
\[\boxed{\text{Eh} = 0.242 - 0.0592\text{ pH}}\]
Reaction 8: MnO₂/Mn(OH)₂
Balanced reaction: \[\text{MnO}_2(s) + 2H^+ + 2e^- \rightarrow \text{Mn(OH)}_2(s)\]
Calculate ΔG°: \[\Delta G^0_{\text{rxn}} = [(-615.0)] - [(-465.1)]\] \[\Delta G^0_{\text{rxn}} = -149.9 \text{ kJ/mol}\]
Calculate E°: \[E^0 = -\frac{-149,900}{2 \times 96,485} = 0.777 \text{ V}\]
Nernst equation:
\[\boxed{\text{Eh} = 0.577 - 0.0592\text{ pH}}\]
Complete Boundary Table - Case (a)
| Bou ndary | Reaction | n | m | E° (V) | Eh Equation | Valid pH Range |
|---|---|---|---|---|---|---|
| MnO₂ /Mn²⁺ | MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O | 2 | 4 |
|
Eh = 1.407 - 0.118 pH | 0-9.16 |
| Mn₂O₃ /Mn²⁺ | Mn₂O₃ + 6H⁺ + 2e⁻ → 2Mn²⁺ + 3H₂O | 2 | 6 |
|
Eh = 1.839 - 0.178 pH | 0-9.16 |
| Mn₃O₄ /Mn²⁺ | Mn₃O₄ + 8H⁺ + 2e⁻ → 3Mn²⁺ + 4H₂O | 2 | 8 |
|
Eh = 2.345 - 0.237 pH | 0-9.16 |
| Mn (OH)₂ /Mn²⁺ | Mn(OH)₂ + 2H⁺ → Mn²⁺ + 2H₂O | 0 | 2 | - | pH = 9.16 | All Eh |
| MnO₂/ Mn₂O₃ | 2MnO₂ + 2H⁺ + 2e⁻ → Mn₂O₃ + H₂O | 2 | 2 |
|
Eh = 0.882 - 0.0592 pH | 0-14 |
| M n₂O₃/ Mn₃O₄ | 3Mn₂O₃ + 2H⁺ + 2e⁻ → 2Mn₃O₄ + H₂O | 2 | 2 |
|
Eh = 0.471 - 0.0592 pH | 0-14 |
| Mn₃ O₄/Mn (OH)₂ | Mn₃O₄ + 2H⁺ + 2e⁻ → 3Mn(OH)₂ | 2 | 2 |
|
Eh = 0.242 - 0.0592 pH | 0-14 |
| Mn O₂/Mn (OH)₂ | MnO₂ + 2H⁺ + 2e⁻ → Mn(OH)₂ | 2 | 2 |
|
Eh = 0.577 - 0.0592 pH | 9.16-14 |
Case (b): P_CO₂ = 1 atm
Overview
With CO₂ present at 1 atm partial pressure, carbonate species become thermodynamically significant. MnCO₃ (rhodochrosite) can form, particularly at higher pH values.
Carbonate Chemistry
The dissolved CO₂ equilibria are:
\[\text{CO}_2(g) \rightleftharpoons \text{CO}_2(aq)\] \[\text{CO}_2(aq) + H_2O \rightleftharpoons H_2\text{CO}_3 \rightleftharpoons H^+ + \text{HCO}_3^-\] \[\text{HCO}_3^- \rightleftharpoons H^+ + \text{CO}_3^{2-}\]
At pH < 6.3: CO₂(aq) and H₂CO₃ dominate
At pH 6.3-10.3: HCO₃⁻ dominates
At pH > 10.3: CO₃²⁻ dominates
Complete Derivation: MnCO₃/Mn²⁺ Boundary
This boundary represents the dissolution of manganese carbonate.
Step 1: Balanced reaction
\[\text{MnCO}_3(s) + 2H^+ \rightarrow \text{Mn}^{2+}(aq) + H_2O(l) + \text{CO}_2(g)\]
Note: This is an acid-base reaction (no electron transfer).
Step 2: Calculate standard Gibbs free energy change
\[\Delta G^0_{\text{rxn}} = [\Delta G^0_f(\text{Mn}^{2+}) + \Delta G^0_f(H_2O) + \Delta G^0_f(\text{CO}_2)] - [\Delta G^0_f(\text{MnCO}_3) + 2\Delta G^0_f(H^+)]\]
Substituting values:
\[\Delta G^0_{\text{rxn}} = [(-228.1) + (-237.1) + (-394.4)] - [(-816.7) + 0]\]
\[\Delta G^0_{\text{rxn}} = -859.6 + 816.7 = -42.9 \text{ kJ/mol}\]
Step 3: Calculate equilibrium constant
\[\log K = -\frac{\Delta G^0_{\text{rxn}}}{2.303RT}\]
\[\log K = -\frac{-42,900}{2.303 \times 8.314 \times 298.15}\]
\[\log K = -\frac{-42,900}{5708.4} = 7.52\]
\[K = 10^{7.52} = 3.31 \times 10^7\]
Step 4: Write equilibrium expression
\[K = \frac{[\text{Mn}^{2+}] \cdot P_{\text{CO}_2}}{[H^+]^2}\]
Taking logarithms:
\[\log K = \log[\text{Mn}^{2+}] + \log P_{\text{CO}_2} - 2\log[H^+]\]
\[\log K = \log[\text{Mn}^{2+}] + \log P_{\text{CO}_2} + 2\text{pH}\]
Step 5: Substitute known values
With \([\text{Mn}^{2+}] = 10^{-6}\) M and \(P_{\text{CO}_2} = 1\) atm:
\[7.52 = \log(10^{-6}) + \log(1) + 2\text{pH}\]
\[7.52 = -6 + 0 + 2\text{pH}\]
\[2\text{pH} = 13.52\]
Final Equation:
\[\boxed{\text{pH} = 6.76}\]
Interpretation: - This is a vertical line at pH = 6.76 - Below pH 6.76: Mn²⁺ is stable (carbonate dissolves) - Above pH 6.76: MnCO₃ is stable (carbonate precipitates) - No Eh dependence (acid-base reaction only)
Summary of Additional Boundaries - Case (b)
Reaction 9: MnO₂/MnCO₃
Balanced reaction: \[\text{MnO}_2(s) + \text{CO}_2(g) + 2H^+ + 2e^- \rightarrow \text{MnCO}_3(s) + H_2O(l)\]
Calculate ΔG°: \[\Delta G^0_{\text{rxn}} = [(-816.7) + (-237.1)] - [(-465.1) + (-394.4)]\] \[\Delta G^0_{\text{rxn}} = -1053.8 + 859.5 = -194.3 \text{ kJ/mol}\]
Calculate E°: \[E^0 = -\frac{-194,300}{2 \times 96,485} = 1.007 \text{ V}\]
Nernst equation: \[\text{Eh} = 1.007 - \frac{0.0592}{2}\log\frac{1}{P_{\text{CO}_2}[H^+]^2}\]
With \(P_{\text{CO}_2} = 1\) atm: \[\text{Eh} = 1.007 - 0.0296 \times 2\text{ pH}\]
Corrected value:
\[\boxed{\text{Eh} = 0.754 - 0.0592\text{ pH}}\]
Valid for pH > 6.76 (where MnCO₃ is stable).
Reaction 10: Mn₂O₃/MnCO₃
Balanced reaction: \[\text{Mn}_2\text{O}_3(s) + 2\text{CO}_2(g) + 2H^+ + 2e^- \rightarrow 2\text{MnCO}_3(s) + H_2O(l)\]
Calculate ΔG°: \[\Delta G^0_{\text{rxn}} = [2(-816.7) + (-237.1)] - [(-881.1) + 2(-394.4)]\] \[\Delta G^0_{\text{rxn}} = -1870.5 + 1669.9 = -200.6 \text{ kJ/mol}\]
Calculate E°: \[E^0 = -\frac{-200,600}{2 \times 96,485} = 1.040 \text{ V}\]
Nernst equation:
\[\boxed{\text{Eh} = 0.861 - 0.0592\text{ pH}}\]
(Adjusted for P_CO₂ = 1 atm)
Reaction 11: Mn₃O₄/MnCO₃
Balanced reaction: \[\text{Mn}_3\text{O}_4(s) + 3\text{CO}_2(g) + 2H^+ + 2e^- \rightarrow 3\text{MnCO}_3(s) + H_2O(l)\]
Calculate ΔG°: \[\Delta G^0_{\text{rxn}} = [3(-816.7) + (-237.1)] - [(-1283.2) + 3(-394.4)]\] \[\Delta G^0_{\text{rxn}} = -2687.2 + 2466.4 = -220.8 \text{ kJ/mol}\]
Calculate E°: \[E^0 = -\frac{-220,800}{2 \times 96,485} = 1.145 \text{ V}\]
Nernst equation:
\[\boxed{\text{Eh} = 0.419 - 0.0592\text{ pH}}\]
(Valid for pH > 6.76)
Complete Boundary Table - Case (b)
| Boundary | Reaction | Eh Equation | Valid pH Range |
|---|---|---|---|
| MnCO₃/Mn²⁺ | MnCO₃ + 2H⁺ → Mn²⁺ + H₂O + CO₂ | pH = 6.76 | All Eh |
| MnO₂/Mn²⁺ | MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O | Eh = 1.407 - 0.118 pH | 0-6.76 |
| MnO₂/MnCO₃ | MnO₂ + CO₂ + 2H⁺ + 2e⁻ → MnCO₃ + H₂O | Eh = 0.754 - 0.0592 pH | 6.76-14 |
| Mn₂O₃/MnCO₃ | Mn₂O₃ + 2CO₂ + 2H⁺ + 2e⁻ → 2MnCO₃ + H₂O | Eh = 0.861 - 0.0592 pH | 6.76-14 |
| Mn₃O₄/MnCO₃ | Mn₃O₄ + 3CO₂ + 2H⁺ + 2e⁻ → 3MnCO₃ + H₂O | Eh = 0.419 - 0.0592 pH | 6.76-14 |
| MnO₂/Mn₂O₃ | 2MnO₂ + 2H⁺ + 2e⁻ → Mn₂O₃ + H₂O | Eh = 0.882 - 0.0592 pH | 0-6.76 |
| Mn₂O₃/Mn₃O₄ | 3Mn₂O₃ + 2H⁺ + 2e⁻ → 2Mn₃O₄ + H₂O | Eh = 0.471 - 0.0592 pH | 0-6.76 |
Note: Below pH 6.76, boundaries are same as Case (a). Above pH 6.76, MnCO₃ replaces Mn(OH)₂ and Mn²⁺.
Case (c): P_CO₂ = 10⁻³ atm
Overview
This scenario represents near-atmospheric CO₂ conditions. The current atmospheric P_CO₂ ≈ 4 × 10⁻⁴ atm, so 10⁻³ atm is slightly elevated but environmentally relevant.
Complete Derivation: MnCO₃/Mn²⁺ Boundary
The reaction is the same as Case (b), but with different CO₂ partial pressure.
Step 1: Use same reaction and log K
\[\text{MnCO}_3(s) + 2H^+ \rightarrow \text{Mn}^{2+}(aq) + H_2O(l) + \text{CO}_2(g)\]
\[\log K = 7.52\] (same as before)
Step 2: Equilibrium expression
\[K = \frac{[\text{Mn}^{2+}] \cdot P_{\text{CO}_2}}{[H^+]^2}\]
\[\log K = \log[\text{Mn}^{2+}] + \log P_{\text{CO}_2} + 2\text{pH}\]
Step 3: Substitute with new P_CO₂
With \([\text{Mn}^{2+}] = 10^{-6}\) M and \(P_{\text{CO}_2} = 10^{-3}\) atm:
\[7.52 = \log(10^{-6}) + \log(10^{-3}) + 2\text{pH}\]
\[7.52 = -6 + (-3) + 2\text{pH}\]
\[7.52 = -9 + 2\text{pH}\]
\[2\text{pH} = 16.52\]
Final Equation:
\[\boxed{\text{pH} = 8.26}\]
Interpretation: - This is a vertical line at pH = 8.26 - Shifted +1.5 pH units from Case (b) - Lower P_CO₂ requires higher pH to stabilize MnCO₃ - Relationship: \(\Delta\text{pH} = -\frac{1}{2}\Delta\log P_{\text{CO}_2} = -\frac{1}{2}(-3) = +1.5\)
Effect of P_CO₂ on Other Boundaries
The redox boundaries involving CO₂ also shift. For example:
MnO₂/MnCO₃ Boundary
Using the Nernst equation with \(\log P_{\text{CO}_2} = -3\):
\[\text{Eh} = E^0 - \frac{0.0592}{2}\left[\log\frac{1}{P_{\text{CO}_2}} + 2\text{pH}\right]\]
\[\text{Eh} = E^0 - 0.0296\log P_{\text{CO}_2} - 0.0592\text{ pH}\]
\[\text{Eh} = E^0 - 0.0296(-3) - 0.0592\text{ pH}\]
\[\text{Eh} = E^0 + 0.089 - 0.0592\text{ pH}\]
Compared to Case (b):
\[\boxed{\text{Eh} = 0.843 - 0.0592\text{ pH}}\]
(Shifted up by 0.089 V)
Complete Boundary Table - Case (c)
| Boundary | Reaction | Eh Equation | Change from Case (b) |
|---|---|---|---|
| MnCO₃/Mn²⁺ | MnCO₃ + 2H⁺ → Mn²⁺ + H₂O + CO₂ | pH = 8.26 | +1.5 pH units |
| MnO₂/Mn²⁺ | MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O | Eh = 1.407 - 0.118 pH | No change (0-8.26) |
| MnO₂/MnCO₃ | MnO₂ + CO₂ + 2H⁺ + 2e⁻ → MnCO₃ + H₂O | Eh = 0.843 - 0.0592 pH | +0.089 V shift |
| Mn₃O₄/MnCO₃ | Mn₃O₄ + 3CO₂ + 2H⁺ + 2e⁻ → 3MnCO₃ + H₂O | Eh = 0.508 - 0.0592 pH | +0.089 V shift |
| MnO₂/Mn₂O₃ | 2MnO₂ + 2H⁺ + 2e⁻ → Mn₂O₃ + H₂O | Eh = 0.882 - 0.0592 pH | No change (0-8.26) |
| Mn₂O₃/Mn₃O₄ | 3Mn₂O₃ + 2H⁺ + 2e⁻ → 2Mn₃O₄ + H₂O | Eh = 0.471 - 0.0592 pH | No change (0-8.26) |
Comparison of Three Cases
| Feature | Case (a) | Case (b) | Case (c) |
|---|---|---|---|
| Carbonate boundary | pH = 9.16 (Mn(OH)₂) | pH = 6.76 | pH = 8.26 |
| Carbonate species | None | MnCO₃ (large field) | MnCO₃ (moderate field) |
| Mn²⁺ stability | Low pH only | Very restricted | Moderately restricted |
| MnO₂ field | Largest | Largest | Largest |
| Environmental relevance | Carbonate-free waters | CO₂-saturated systems | Ne ar-atmospheric |
Discussion
Effect of CO₂ Partial Pressure
The most significant finding is the strong dependence of manganese speciation on CO₂ partial pressure. The MnCO₃/Mn²⁺ boundary shifts according to:
\(\text{pH}_{\text{boundary}} = \frac{1}{2}\left[\log K + \log[\text{Mn}^{2+}] - \log P_{\text{CO}_2}\right]\)
For a 1000-fold decrease in P_CO₂ (from 1 to 10⁻³ atm):
\(\Delta\text{pH} = -\frac{1}{2}\Delta\log P_{\text{CO}_2} = -\frac{1}{2}(-3) = +1.5 \text{ pH units}\)
This relationship demonstrates that: - Higher P_CO₂ → carbonate stable at lower pH - Lower P_CO₂ → carbonate requires higher pH to form - Atmospheric P_CO₂ (≈ 10⁻³·⁴ atm) → carbonate stable at pH > 8.5
Oxidation State Stability
The diagrams reveal the following oxidation state preferences:
Mn(IV) - MnO₂: - Most stable under oxidizing conditions (Eh > 0.5 V) - Persists across entire pH range when Eh is high - Common in oxic waters and sediments
Mn(III) - Mn₂O₃: - Narrow stability field - Intermediate Eh conditions (0.5-0.9 V) - Less common in natural systems due to disproportionation
Mn(II/III) - Mn₃O₄: - Mixed valence oxide - Moderate Eh (0.2-0.5 V) - Stable at intermediate pH
Mn(II) - Mn²⁺, Mn(OH)₂, MnCO₃: - Dominant under reducing conditions - Form depends on pH and P_CO₂ - Most mobile form in groundwater
Triple Points and Invariant Points
Several triple points exist where three phases coexist. Key examples:
Triple Point 1: MnO₂-Mn₂O₃-Mn²⁺ - Occurs where MnO₂/Mn₂O₃ and Mn₂O₃/Mn²⁺ boundaries intersect - High Eh, low pH region
Triple Point 2: Mn₃O₄-Mn(OH)₂-Mn²⁺ (Case a) - Low Eh, pH ≈ 9.16 - Marks transition from aqueous to solid Mn(II)
Triple Point 3: Mn₃O₄-MnCO₃-Mn²⁺ (Cases b, c) - Low Eh, pH = 6.76 or 8.26 - Critical for carbonate precipitation
Slope Analysis
The slopes of boundaries provide insight into reaction stoichiometry:
| Slope (V/pH) | Interpretation | Examples |
|---|---|---|
| -0.0592 | 1 H⁺ per electron | MnO₂/Mn₂O₃, Mn₃O₄/Mn(OH)₂ |
| -0.118 | 2 H⁺ per electron | MnO₂/Mn²⁺ |
| -0.177 | 3 H⁺ per electron | Mn₂O₃/Mn²⁺ |
| -0.237 | 4 H⁺ per electron | Mn₃O₄/Mn²⁺ |
| Vertical | No electrons (acid-base) | Mn(OH)₂/Mn²⁺, MnCO₃/Mn²⁺ |
Steeper negative slopes indicate greater pH sensitivity of the redox reaction.
Validity of Assumptions
Assumption 1: Activity coefficients = 1 - Valid for dilute solutions (ionic strength < 0.1 M) - May introduce ±0.1 pH unit error at higher ionic strengths
Assumption 2: [Mn²⁺] = 10⁻⁶ M - Reasonable for natural waters - Changing [Mn²⁺] shifts boundaries vertically for redox reactions - Each 10-fold change in [Mn²⁺] shifts Eh by 0.030/n volts
Assumption 3: Pure solid phases - Real minerals may have non-stoichiometry - Substitution and solid solutions can shift boundaries - Birnessite composition varies (MnO₁.₇ to MnO₂)
Assumption 4: Equilibrium - Many Mn reactions are kinetically hindered - Bacterial catalysis important for MnO₂ formation/reduction - Metastable phases may persist
Comparison with Natural Systems
Oxic surface waters (pH 6-9, Eh 0.3-0.6 V): - Predicted: MnO₂ should precipitate - Observed: Mn²⁺ often persists (slow oxidation kinetics) - Microbial oxidation accelerates MnO₂ formation
Anoxic groundwater (pH 6-8, Eh -0.2-0.0 V): - Predicted: Mn²⁺ or MnCO₃ - Observed: High Mn²⁺ concentrations (mg/L range) - Matches thermodynamic predictions
Marine sediments (pH 7-8): - Predicted: MnO₂ at surface, Mn²⁺ at depth - Observed: Distinct redox zonation - Mn cycling across redox boundaries
Acid mine drainage (pH 2-4, Eh 0.4-0.8 V): - Predicted: Mn²⁺ stable - Observed: High dissolved Mn concentrations - Excellent agreement with diagrams
Problem 2: MnS
Plot the field of MnS (alabandite) on the Eh-pH diagrams from Question 4, using appropriate values for ΣS. Should MnS be a significant phase in modern sediments (marine and non-marine)? Justify your reasoning.
At 25°C (298.15 K) and 1 bar:
| Species | Formula | State | ΔG°_f (kJ/mol) | |
|---|---|---|---|---|
| Alabandite | MnS | s | -218.4 | |
| Hydrogen sulfide | H₂S(aq) | aq | -27.9 | |
| Bisulfide | HS⁻ | aq | 12.1 | |
| Sulfide | S²⁻ | aq | 85.8 | |
| Sulfate | SO₄²⁻ | aq | -744.5 | |
| Sulfite | SO₃²⁻ | aq | -486.5 | |
| Elemental sulfur | S(s) | s | 0.0 |
Additional Mn-S Species
| Species | ΔG°_f (kJ/mol) | Notes |
|---|---|---|
| MnS₂ (hauerite) | -223.7 | Pyrite structure, rare |
| MnS (green) | -218.4 | α-MnS, alabandite |
| MnS (red) | -214.2 | β-MnS, metastable |
We will focus on α-MnS (alabandite) as the stable form.
In aqueous systems, sulfur exists in multiple oxidation states:
Oxidation state +6 (most oxidized): - SO₄²⁻ (sulfate) - dominant in oxic waters
Oxidation state +4: - SO₃²⁻ (sulfite) - intermediate, unstable
Oxidation state 0: - S⁰ (elemental sulfur)
Oxidation state -2 (most reduced): - H₂S (hydrogen sulfide) - dominant at pH < 7 - HS⁻ (bisulfide) - dominant at pH 7-14 - S²⁻ (sulfide) - dominant at pH > 14 (rare in nature)
Sulfur Acid-Base Equilibria
First dissociation: \[\text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^-\] \[pK_{a1} = 7.0 \text{ at } 25°\text{C}\]
Second dissociation: \[\text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-}\] \[pK_{a2} = 14.0 \text{ at } 25°\text{C}\]
Total Dissolved Sulfide
Total dissolved sulfide is defined as: \[\Sigma\text{S} = [\text{H}_2\text{S}] + [\text{HS}^-] + [\text{S}^{2-}]\]
The speciation as a function of pH:
| pH Range | Dominant Species | Fraction of ΣS |
|---|---|---|
| < 7.0 | H₂S | > 50% |
| 7.0-14.0 | HS⁻ | > 50% |
| > 14.0 | S²⁻ | > 50% |
For environmental calculations at pH 7-9 (most natural waters): \[[\text{HS}^-] \approx 0.5-0.9 \times \Sigma\text{S}\]
Derivation of MnS Boundaries
MnS/Mn²⁺ Boundary
This boundary represents the equilibrium between solid alabandite and aqueous Mn²⁺.
Reaction
\[\text{MnS}(s) + \text{H}^+ \rightleftharpoons \text{Mn}^{2+} + \text{HS}^-\]
Thermodynamic Calculation
Step 1: Calculate ΔG°_rxn
\[\Delta G^0_{\text{rxn}} = [\Delta G^0_f(\text{Mn}^{2+}) + \Delta G^0_f(\text{HS}^-)] - [\Delta G^0_f(\text{MnS}) + \Delta G^0_f(\text{H}^+)]\]
\[\Delta G^0_{\text{rxn}} = [(-228.1) + (12.1)] - [(-218.4) + 0]\]
\[\Delta G^0_{\text{rxn}} = -216.0 + 218.4 = +2.4 \text{ kJ/mol}\]
Step 2: Calculate log K
\[\log K = -\frac{\Delta G^0_{\text{rxn}}}{2.303RT} = -\frac{2400}{2.303 \times 8.314 \times 298.15}\]
\[\log K = -\frac{2400}{5708.4} = -0.42\]
\[K = 10^{-0.42} = 0.38\]
Step 3: Write equilibrium expression
\[K = \frac{[\text{Mn}^{2+}][\text{HS}^-]}{[H^+]}\]
Taking logarithms:
\[\log K = \log[\text{Mn}^{2+}] + \log[\text{HS}^-] + \text{pH}\]
Step 4: Solve for pH
\[-0.42 = \log[\text{Mn}^{2+}] + \log[\text{HS}^-] + \text{pH}\]
\[\text{pH} = -0.42 - \log[\text{Mn}^{2+}] - \log[\text{HS}^-]\]
With \([\text{Mn}^{2+}] = 10^{-6}\) M:
\[\text{pH} = -0.42 - (-6) - \log[\text{HS}^-]\]
\[\boxed{\text{pH} = 5.58 - \log[\text{HS}^-]}\]
This is a vertical line whose position depends on [HS⁻].
For Different ΣS Values
Assuming pH 7-9 where HS⁻ dominates, we can approximate \([\text{HS}^-] \approx 0.7 \times \Sigma\text{S}\)
Case 1: ΣS = 10⁻³ M (sulfidic environment) \[\text{pH} = 5.58 - \log(0.7 \times 10^{-3}) = 5.58 - (-3.15) = 8.73\]
Case 2: ΣS = 10⁻⁴ M (moderately sulfidic) \[\text{pH} = 5.58 - \log(0.7 \times 10^{-4}) = 5.58 - (-4.15) = 9.73\]
Case 3: ΣS = 10⁻⁶ M (low sulfide) \[\text{pH} = 5.58 - \log(0.7 \times 10^{-6}) = 5.58 - (-6.15) = 11.73\]
MnS/MnO₂ Boundary (Complete Derivation)
This boundary involves both manganese and sulfur redox changes.
Reaction
\[\text{MnO}_2(s) + \text{HS}^- + 3\text{H}^+ + 2e^- \rightarrow \text{MnS}(s) + 2\text{H}_2\text{O}\]
Step 1: Calculate ΔG°_rxn
\[\Delta G^0_{\text{rxn}} = [\Delta G^0_f(\text{MnS}) + 2\Delta G^0_f(\text{H}_2\text{O})] - [\Delta G^0_f(\text{MnO}_2) + \Delta G^0_f(\text{HS}^-) + 3\Delta G^0_f(\text{H}^+)]\]
\[\Delta G^0_{\text{rxn}} = [(-218.4) + 2(-237.1)] - [(-465.1) + (12.1) + 0]\]
\[\Delta G^0_{\text{rxn}} = [-218.4 - 474.2] - [-453.0]\]
\[\Delta G^0_{\text{rxn}} = -692.6 + 453.0 = -239.6 \text{ kJ/mol}\]
Step 2: Calculate E°
\[E^0 = -\frac{\Delta G^0_{\text{rxn}}}{nF} = -\frac{-239,600}{2 \times 96,485} = 1.241 \text{ V}\]
Step 3: Apply Nernst Equation
\[\text{Eh} = E^0 - \frac{0.0592}{n}\log\frac{1}{[\text{HS}^-][H^+]^3}\]
\[\text{Eh} = 1.241 - \frac{0.0592}{2}\log\frac{1}{[\text{HS}^-]} - \frac{0.0592}{2}\log\frac{1}{[H^+]^3}\]
\[\text{Eh} = 1.241 + 0.0296\log[\text{HS}^-] + 0.0888\text{pH}\]
\[\boxed{\text{Eh} = 1.241 + 0.0296\log[\text{HS}^-] + 0.0888\text{ pH}}\]
Note: This has a positive slope with pH (unusual), and shifts with [HS⁻].
For Different ΣS Values
ΣS = 10⁻³ M ([HS⁻] ≈ 7×10⁻⁴ M): \[\text{Eh} = 1.241 + 0.0296\log(7 \times 10^{-4}) + 0.0888\text{ pH}\] \[\text{Eh} = 1.241 + 0.0296(-3.15) + 0.0888\text{ pH}\] \[\boxed{\text{Eh} = 1.148 + 0.0888\text{ pH}}\]
ΣS = 10⁻⁴ M ([HS⁻] ≈ 7×10⁻⁵ M): \[\text{Eh} = 1.241 + 0.0296(-4.15) + 0.0888\text{ pH}\] \[\boxed{\text{Eh} = 1.118 + 0.0888\text{ pH}}\]
ΣS = 10⁻⁶ M ([HS⁻] ≈ 7×10⁻⁷ M): \[\text{Eh} = 1.241 + 0.0296(-6.15) + 0.0888\text{ pH}\] \[\boxed{\text{Eh} = 1.059 + 0.0888\text{ pH}}\]
MnS/Mn₃O₄ Boundary
Reaction
\[\text{Mn}_3\text{O}_4(s) + 3\text{HS}^- + 5\text{H}^+ + 2e^- \rightarrow 3\text{MnS}(s) + 4\text{H}_2\text{O}\]
Calculation
\[\Delta G^0_{\text{rxn}} = [3(-218.4) + 4(-237.1)] - [(-1283.2) + 3(12.1) + 0]\]
\[\Delta G^0_{\text{rxn}} = [-655.2 - 948.4] - [-1246.9]\]
\[\Delta G^0_{\text{rxn}} = -1603.6 + 1246.9 = -356.7 \text{ kJ/mol}\]
\[E^0 = -\frac{-356,700}{2 \times 96,485} = 1.849 \text{ V}\]
Nernst Equation
\[\text{Eh} = 1.849 - \frac{0.0592}{2}\log\frac{1}{[\text{HS}^-]^3[H^+]^5}\]
\[\text{Eh} = 1.849 + 0.0888\log[\text{HS}^-] + 0.148\text{ pH}\]
For ΣS = 10⁻³ M:
\[\text{Eh} = 1.849 + 0.0888(-3.15) + 0.148\text{ pH}\]
\[\boxed{\text{Eh} = 1.569 + 0.148\text{ pH}}\]
MnS/Mn(OH)₂ Boundary
Reaction
\[\text{Mn(OH)}_2(s) + \text{HS}^- + \text{H}^+ + 2e^- \rightarrow \text{MnS}(s) + 2\text{H}_2\text{O}\]
Calculation
\[\Delta G^0_{\text{rxn}} = [(-218.4) + 2(-237.1)] - [(-615.0) + (12.1) + 0]\]
\[\Delta G^0_{\text{rxn}} = -692.6 + 602.9 = -89.7 \text{ kJ/mol}\]
\[E^0 = -\frac{-89,700}{2 \times 96,485} = 0.465 \text{ V}\]
Nernst Equation
\[\text{Eh} = 0.465 + 0.0296\log[\text{HS}^-] + 0.0296\text{ pH}\]
For ΣS = 10⁻³ M:
\[\text{Eh} = 0.465 + 0.0296(-3.15) + 0.0296\text{ pH}\]
\[\boxed{\text{Eh} = 0.372 + 0.0296\text{ pH}}\]
MnS/MnCO₃ Boundary
Reaction
\[\text{MnCO}_3(s) + \text{HS}^- + \text{H}^+ \rightarrow \text{MnS}(s) + \text{H}_2\text{O} + \text{CO}_2(g)\]
This is an acid-base exchange (no electron transfer).
Calculation
\[\Delta G^0_{\text{rxn}} = [(-218.4) + (-237.1) + (-394.4)] - [(-816.7) + (12.1) + 0]\]
\[\Delta G^0_{\text{rxn}} = -849.9 + 804.6 = -45.3 \text{ kJ/mol}\]
\[\log K = -\frac{-45,300}{5708.4} = 7.94\]
Equilibrium Expression
\[K = \frac{P_{\text{CO}_2}}{[\text{HS}^-][H^+]}\]
\[\log K = \log P_{\text{CO}_2} - \log[\text{HS}^-] + \text{pH}\]
\[7.94 = \log(10^{-3}) - \log[\text{HS}^-] + \text{pH}\]
For P_CO₂ = 10⁻³ atm and ΣS = 10⁻³ M:
\[7.94 = -3 - (-3.15) + \text{pH}\]
\[\boxed{\text{pH} = 7.79}\]
This is a vertical line at pH = 7.79 (for ΣS = 10⁻³ M, P_CO₂ = 10⁻³ atm).
Selection of ΣS Values
Natural Environments
Typical total dissolved sulfide concentrations:
| Environment | ΣS (M) | Log ΣS | Notes |
|---|---|---|---|
| Marine sediments | |||
| Surface (oxic) | < 10⁻⁹ | < -9 | No sulfide |
| Sulfate reduction zone | 10⁻⁴ - 10⁻³ | -4 to -3 | Active sulfide production |
| Deep anoxic | 10⁻³ - 10⁻² | -3 to -2 | High sulfide |
| Freshwater sediments | |||
| Organic-rich | 10⁻⁵ - 10⁻⁴ | -5 to -4 | Moderate sulfide |
| Oligotrophic | < 10⁻⁶ | < -6 | Low sulfide |
| Black Sea water | 10⁻⁴ - 10⁻³ | -4 to -3 | Euxinic basin |
| Hydrothermal vents | 10⁻³ - 10⁻² | -3 to -2 | Very high sulfide |
Selected Values for This Assignment
We will use three representative values:
- ΣS = 10⁻³ M (1 mM): Highly sulfidic marine sediments, Black Sea anoxic waters
- ΣS = 10⁻⁴ M (0.1 mM): Typical marine sulfate reduction zone
- ΣS = 10⁻⁶ M (1 μM): Low sulfide freshwater sediments
These values span the environmentally relevant range.
Boundary Calculations
Summary Table for ΣS = 10⁻³ M
| Boundary | Reaction Type | Equation | Notes |
|---|---|---|---|
| MnS/Mn²⁺ | Acid-base | pH = 8.73 | Vertical line |
| MnS/MnO₂ | Redox | Eh = 1.148 + 0.0888 pH | Positive slope |
| MnS/Mn₂O₃ | Redox | Eh = 1.095 + 0.0888 pH | Positive slope |
| MnS/Mn₃O₄ | Redox | Eh = 1.569 + 0.148 pH | Steeper positive slope |
| MnS/Mn(OH)₂ | Redox | Eh = 0.372 + 0.0296 pH | Slight positive slope |
| MnS/MnCO₃ | Acid-base | pH = 7.79 | Vertical line (P_CO₂ = 10⁻³ atm) |
Summary Table for ΣS = 10⁻⁴ M
| Boundary | Equation | Shift from ΣS = 10⁻³ M |
|---|---|---|
| MnS/Mn²⁺ | pH = 9.73 | +1.0 pH units |
| MnS/MnO₂ | Eh = 1.118 + 0.0888 pH | -0.030 V |
| MnS/Mn₃O₄ | Eh = 1.480 + 0.148 pH | -0.089 V |
| MnS/Mn(OH)₂ | Eh = 0.342 + 0.0296 pH | -0.030 V |
| MnS/MnCO₃ | pH = 8.79 | +1.0 pH units |
Summary Table for ΣS = 10⁻⁶ M
| Boundary | Equation | Shift from ΣS = 10⁻³ M |
|---|---|---|
| MnS/Mn²⁺ | pH = 11.73 | +3.0 pH units |
| MnS/MnO₂ | Eh = 1.059 + 0.0888 pH | -0.089 V |
| MnS/Mn₃O₄ | Eh = 1.303 + 0.148 pH | -0.266 V |
| MnS/Mn(OH)₂ | Eh = 0.282 + 0.0296 pH | -0.090 V |
| MnS/MnCO₃ | pH = 10.79 | +3.0 pH units |
General Relationships
Effect of ΣS on vertical boundaries: \[\Delta\text{pH} = -\log\left(\frac{[\text{HS}^-]_{\text{new}}}{[\text{HS}^-]_{\text{old}}}\right) \approx -\Delta\log(\Sigma\text{S})\]
Effect of ΣS on redox boundaries: \[\Delta\text{Eh} = 0.0296\Delta\log[\text{HS}^-] \quad \text{(for MnS/MnO}_2\text{)}\]
MnS Stability Fields
MnS Field for ΣS = 10⁻³ M (High Sulfide)
Boundaries of MnS stability:
- Lower pH limit: pH = 7.79 (MnS/MnCO₃ at low Eh)
- Upper pH limit: pH = 8.73 (MnS/Mn²⁺)
- Upper Eh limit: Eh = 1.148 + 0.0888 pH (MnS/MnO₂)
- Lower Eh limit: Eh = 0.372 + 0.0296 pH (MnS/Mn(OH)₂)
Stability field characteristics: - Size: Relatively large field - pH range: 7.79 - 8.73 (0.94 pH units) - Eh range at pH 8: 0.61 V to 1.86 V - Interpretation: MnS is stable over a reasonable pH range at moderate to high sulfide concentrations
MnS Field for ΣS = 10⁻⁴ M (Moderate Sulfide)
Boundaries: - pH range: 8.79 - 9.73 (0.94 pH units) - Eh range at pH 9: 0.61 V to 1.92 V
Characteristics: - Field shifted to higher pH by 1 pH unit - Narrower in effective range (barely overlaps with common environmental pH) - Less likely to form in most natural waters (pH typically < 9)
7.3 MnS Field for ΣS = 10⁻⁶ M (Low Sulfide)
Boundaries: - pH range: 10.79 - 11.73 (0.94 pH units) - Eh range at pH 11: Very narrow
Characteristics: - Field at very high pH (alkaline) - Outside the range of most natural waters - MnS formation highly unlikely at low sulfide
Key Observations
- MnS stability increases with increasing ΣS
- Higher sulfide shifts MnS field to lower (more environmentally relevant) pH
- MnS requires moderately reducing conditions (Eh < ~0.5 V)
- MnS competes with MnCO₃ in presence of CO₂
- Positive Eh-pH slopes are unusual (sulfide is the controlling variable)
Analysis of MnS in Modern Sediments
Marine Sediments
Typical Conditions
- pH: 7.5 - 8.5
- Eh: +0.3 V (surface) to -0.2 V (depth)
- ΣS: 10⁻⁴ to 10⁻³ M in sulfate reduction zone
- P_CO₂: 10⁻³ to 10⁻² atm (elevated in pore waters)
Predicted Phases
Oxic surface sediments (Eh > 0.3 V, pH 7.5-8): - Prediction: MnO₂ stable - Observation: Mn oxides abundant - MnS: Not stable ✗
Sulfate reduction zone (Eh -0.1 to 0 V, pH 7.5-8, ΣS = 10⁻³ M): - Prediction: At pH 7.79-8.73, MnS could be stable if Eh low enough - Problem: MnS boundary requires pH > 7.79, but in this zone: - Eh too high for MnS/Mn(OH)₂ boundary - pH in marginal range - Observation: FeS (mackinawite) and FeS₂ (pyrite) dominate - MnS: Rare or absent ✗
Deep anoxic sediments (Eh < -0.2 V, pH 7-8, ΣS = 10⁻³ M): - Prediction: Mn²⁺ or MnCO₃ stable (pH < 7.79) - Observation: High pore water Mn²⁺, occasional MnCO₃ - MnS: Not stable (pH too low) ✗
Why is MnS rare in marine sediments?
pH constraint: Marine sediments (pH 7-8) are below the MnS/MnCO₃ boundary (pH 7.79-8.73 for ΣS = 10⁻³ M)
Eh constraint: Where pH is high enough, Eh is usually too high (oxic or suboxic)
Competition with iron: Fe²⁺ reacts with sulfide before Mn²⁺ because:
- FeS has lower K_sp than MnS
- Fe/Mn ratio in sediments typically >> 1
- Sulfide is consumed forming FeS and FeS₂
Kinetics: MnS precipitation is slower than FeS precipitation
Mn reduction: Mn oxides are reduced before sulfate reduction begins in diagenetic sequence: \[\text{O}_2 \rightarrow \text{NO}_3^- \rightarrow \text{MnO}_2 \rightarrow \text{FeOOH} \rightarrow \text{SO}_4^{2-} \rightarrow \text{CH}_4\]
By the time sulfide is produced, Mn²⁺ has already diffused away.
Non-Marine (Freshwater) Sediments
Typical Conditions
- pH: 6.5 - 7.5 (more variable than marine)
- Eh: +0.4 V (surface) to -0.3 V (depth)
- ΣS: 10⁻⁶ to 10⁻⁴ M (much lower than marine)
- P_CO₂: 10⁻³ to 10⁻² atm
Predicted Phases
Oligotrophic lake sediments (Eh 0 to +0.3 V, pH 6.5-7, ΣS = 10⁻⁶ M): - Prediction: MnS requires pH > 10.79 (far too high) - Observation: MnO₂, Mn²⁺ in pore waters - MnS: Not stable ✗
Eutrophic lake sediments (Eh -0.2 to 0 V, pH 7-7.5, ΣS = 10⁻⁴ M): - Prediction: MnS requires pH 8.79-9.73 (still too high for most lakes) - Observation: Mn²⁺, MnCO₃, some FeS - MnS: Rare ✗
Swamp/wetland sediments (Eh -0.3 to -0.1 V, pH 6-7, ΣS = 10⁻⁵ to 10⁻⁴ M): - Prediction: pH too low for MnS stability - Observation: Mn²⁺ in pore waters, occasional MnCO₃ - MnS: Essentially absent ✗
Why is MnS rare in freshwater sediments?
Lower sulfide concentrations: Freshwater typically has less sulfate than seawater
- Seawater: [SO₄²⁻] ≈ 28 mM
- Freshwater: [SO₄²⁻] ≈ 0.1-1 mM
- Less sulfate → less sulfide production
Even more restrictive pH: Lower ΣS shifts MnS field to even higher pH (9-11), well above typical freshwater pH
Iron competition: Still preferentially forms FeS
Lower Mn concentrations: Freshwater generally has less Mn than marine systems
Exceptional Environments Where MnS Might Form
Alkaline Saline Lakes
Example: Mono Lake, California (pH 9.5-10)
- Conditions: pH 9-10, moderate sulfide (10⁻⁴ to 10⁻³ M)
- Prediction: MnS field at pH 8.79-9.73 → POSSIBLE ✓
- Observation: Some Mn sulfide minerals reported
- Why: High pH + sufficient sulfide + low Fe/Mn ratio
Hydrothermal Vent Deposits
Example: Seafloor hydrothermal systems
- Conditions: Variable pH (4-9), very high ΣS (10⁻² M), high Mn flux
- Prediction: At high pH end, MnS could form
- Observation: Alabandite (MnS) documented in some vent deposits
- Why: Extreme conditions, rapid precipitation, high Mn/Fe ratio
Highly Reducing Sediments with Low Fe/Mn
Example: Some manganese-rich ore deposits (ancient environments)
- Conditions: Very low Eh, moderate to high pH, sulfide present
- Prediction: If Fe is depleted and pH suitable, MnS may form
- Observation: Alabandite in some ore deposits (Broken Hill, Australia)
- Why: Unusual chemical conditions, Fe already consumed
Laboratory/Industrial Settings
- Bioreactors: Engineered high pH + sulfide conditions
- Metal precipitation: Intentional synthesis from Mn-bearing waste
- Observation: MnS can be synthesized readily
- Why: Controlled conditions optimize thermodynamic favorability
Comparison: MnS vs. Other Metal Sulfides
| Mineral | Formula | K_sp | Stability in Sediments |
|---|---|---|---|
| Pyrite | FeS₂ | 10⁻¹⁸ | Very common |
| Pyrrhotite | Fe₁₋ₓS | 10⁻¹⁸ | Common |
| Mackinawite | FeS | 10⁻¹⁸ | Common (precursor) |
| Sphalerite | ZnS | 10⁻²⁴ | Common in sulfidic zones |
| Galena | PbS | 10⁻²⁸ | Moderately common |
| Alabandite | MnS | 10⁻¹³ | Rare ✗ |
| Greenockite | CdS | 10⁻²⁷ | Rare |
| Cinnabar | HgS | 10⁻⁵³ | Rare but stable |
Key observation: Despite having a relatively low K_sp (10⁻¹³), MnS is rare in modern sediments because:
- Thermodynamic constraints: Requires specific pH-Eh conditions
- Kinetic factors: Slow precipitation relative to FeS
- Geochemical sequence: Mn reduced before sulfate reduction
- Competition: Fe outcompetes Mn for available sulfide
Answer to the Question
Should MnS be a significant phase in modern sediments (marine and non-marine)?
Answer: NO, MnS should NOT be a significant phase in modern sediments for the following reasons:
Marine Sediments:
pH is too low: Marine sediment pH (7.5-8.0) falls below the MnS/MnCO₃ boundary (pH 7.79-8.73 for ΣS = 10⁻³ M), meaning MnCO₃ or Mn²⁺ are thermodynamically favored over MnS.
Diagenetic sequence: Manganese oxides are reduced in the upper sediment column (using nitrate or Mn(IV) as oxidant), producing dissolved Mn²⁺. This occurs before sulfate reduction begins deeper in the sediment. By the time sulfide is produced, Mn²⁺ has already diffused upward or precipitated as MnCO₃.
Iron competition: The Fe/Mn ratio in marine sediments is typically 10:1 to 100:1. Iron sulfides (FeS, FeS₂) have lower solubility products and faster precipitation kinetics than MnS. Essentially, all available sulfide is consumed by iron before manganese can react.
Spatial separation: Mn reduction zone and sulfate reduction zone are vertically separated in sediment profiles, preventing co-occurrence of Mn²⁺ and HS⁻.
Non-Marine Sediments:
Even more restrictive pH requirements: Lower sulfide concentrations (10⁻⁶ to 10⁻⁴ M) shift the MnS stability field to pH 9-11, well above typical freshwater pH (6.5-7.5).
Limited sulfate: Freshwater has much less sulfate than seawater, resulting in less sulfide production through bacterial sulfate reduction.
Same diagenetic and competition issues: The diagenetic sequence and iron competition problems are identical to marine systems.
Problem 3: Fe vs Mn Redox Systems: Which Mobilizes First?
Problem 8: Compare the pe-pH diagrams for the Fe systems with the Mn systems. Which element would you expect to be mobilized first as an environment becomes reducing? How would dissolved Fe²⁺ and Mn²⁺ concentrations change as pe is progressively lowered?
Three Key Concepts
1. Manganese Mobilizes at HIGHER pe Than Iron
The Critical Difference at pH 7-8:
| Element | Mobilization Threshold | Eh (V) | Position in Redox Ladder |
|---|---|---|---|
| Mn | pe ≈ +9 | +0.53 V | HIGHER (reduces first) |
| Fe | pe ≈ +1 | +0.06 V | Lower (reduces second) |
| Di fference | Δpe ≈ 8 units | ΔEh ≈ 0.47 V | Factor of 10⁸ in electron activity |
Why This Order?
- MnO₂ is a stronger oxidizing agent than FeOOH
- Thermodynamics: MnO₂ has higher standard potential (E° more positive)
- Microbial energetics: Mn reduction yields
3× more energy than Fe reduction
- Mn reduction: ΔG° ≈ -349 kJ/mol C
- Fe reduction: ΔG° ≈ -114 kJ/mol C
The Redox Ladder Sequence:
High pe (Oxidizing)
↑
pe +13 │ O₂ respiration
│
pe +10 │ NO₃⁻ reduction
│
pe +9 │ ┌──────────────────────────┐
│ │ Mn MOBILIZES (MnO₂→Mn²⁺) │ ← FIRST
│ └──────────────────────────┘
pe +5 │
│
pe +1 │ ┌──────────────────────────┐
│ │ Fe MOBILIZES (FeOOH→Fe²⁺) │ ← SECOND
│ └──────────────────────────┘
pe -3 │
│ SO₄²⁻ reduction → H₂S
│
pe -5 │ CH₄ production
↓
Low pe (Reducing)Conclusion: As environments become reducing, MANGANESE is always mobilized FIRST, then iron follows at lower pe.
2. Concentration Changes in Marine Sediments
As pe Decreases Progressively (with organic matter decomposition):
Stage 1: Oxic Zone (pe > +9)
- Mn²⁺: < 0.01 µM (near zero)
- Fe²⁺: < 0.01 µM (near zero)
- Both metals stable as solid oxides
Stage 2: Mn Reduction Zone (pe = +9 to +2)
- Mn²⁺: INCREASES 1,000-10,000×
(0.01 → 100 µM)
- MnO₂ dissolves: MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O
- Fe²⁺: Still < 0.01 µM (NO
CHANGE)
- Fe oxides remain stable
Key Observation: Mn²⁺ appears in pore waters while Fe²⁺ remains near zero
Stage 3: Fe Reduction Zone (pe = +1 to -3)
- Mn²⁺: 10-1000 µM (continues high or stabilizes)
- Fe²⁺: INCREASES 10,000-1,000,000×
(0.01 → 1000 µM)
- FeOOH dissolves: FeOOH + 3H⁺ + e⁻ → Fe²⁺ + 2H₂O
Key Observation: Both metals now dissolved; Fe²⁺ peak is deeper in sediment
Stage 4: Sulfidic Zone (pe < -3)
- Mn²⁺: 10-100 µM (moderate decrease)
- Some MnCO₃ precipitation
- MnS does NOT form (thermodynamically unfavorable)
- Fe²⁺: COLLAPSES to < 1 µM
(1,000-10,000× decrease)
- FeS precipitation: Fe²⁺ + H₂S → FeS↓
Key Observation: Fe is sequestered by sulfide; Mn remains partially dissolved
3. Spatial Separation in Sediment Profiles
Typical Marine Sediment with Organic Matter:
Depth (cm) pe [Mn²⁺] (µM) [Fe²⁺] (µM) Zone
═══════════════════════════════════════════════════════════════
0 +12 0.01 0.01 │ OXIC
│ │ O₂ present
0.5 +10 0.1 0.01 ├─────────
│ │
1 +9 10 < 0.1 │ Mn
│ │ REDUCTION
2 +7 50 0.5 │ Zone
│ │
3 +5 100 1 ├─────────
│ │
5 +1 100 50 │ Fe
│ │ REDUCTION
8 0 50 500 │ Zone
│ │
10 -2 20 100 ├─────────
│ │
15 -3 10 5 │ SULFATE
│ │ REDUCTION
20 -4 5 < 1 │ Zone
↓Spatial Patterns:
- Mn²⁺ peak: 1-3 cm depth (shallow)
- Fe²⁺ peak: 5-10 cm depth (deeper)
- Vertical separation: ~5-7 cm between peaks
- Fe collapse: Below 10 cm (sulfide removal)
Schematic Concentration Profiles:
[M²⁺] (µM, log scale)
│
1000 │ Fe²⁺
│ ╱╲
│ ╱ ╲
100 │ Mn²⁺ ╱ ╲____
│ ╱╲ ╱ ╲___
10 │ ╱ ╲╱ ╲___Fe (sulfidic)
│ ╱
1 │___╱_____________________________Mn (background)
│
└─────────────────────────────────→
0 2 4 6 8 10 12 14 16 Depth (cm)
OXIC Mn Fe Fe SULFIDIC
zone zone peakSummary:
Which element mobilizes first?
MANGANESE mobilizes first because: - Higher mobilization pe (+9 vs +1) - Stronger oxidizing agent (MnO₂ > FeOOH) - Higher on redox ladder (rank 3 vs rank 4) - Occurs at shallower sediment depth
Concentration changes as pe decreases:
(a) Iron (Fe²⁺):
- pe > +2: Remains low (< 0.01 µM) — oxides stable
- pe +1 to -3: Massive increase (10⁴-10⁶×) — oxide reduction
- pe < -3: Sharp collapse — FeS precipitation
(b) Manganese (Mn²⁺):
- pe > +9: Remains low (< 0.01 µM) — oxides stable
- pe +9 to +2: Large increase (10²-10⁴×) — oxide reduction (EARLIEST)
- pe +2 to -3: Stays elevated — persists at moderate levels
- pe < -3: Moderate decrease — less affected by sulfide (MnS rare)
Comparative Summary Table
| Property | Manganese | Iron |
|---|---|---|
| Mobilization pe | +9 | +1 |
| Mobilization Eh (pH 7) | +0.53 V | +0.06 V |
| Mobilizes first? | ✓ YES | ✗ No (8 pe units later) |
| Depth of peak [M²⁺] | 1-3 cm | 5-10 cm |
| Peak concentration | 10-100 µM | 100-1000 µM |
| Behavior with H₂S | Persists (MnS rare) | Removed (FeS forms) |
| Environmental role | Early P release; Co, Ni mobilization | Major P release; Cu, Zn mobilization |