# 5.3  Electron Activity, $pe$

## Defining $pe$

Aqueous solutions do not contain free protons ($\ce{H+}$) and free electrons ($\ce{e-}$), but it is nevertheless possible to define relative and  activities. The  $p\ce{H}$ ($=-\log\{\ce{H+}\}$), measures the relative abundance of protons. Similarly, we can define an equally convenient parameter for the redox intensity $pe$ ($=-\log\{e^-\}$).  $pe$ gives the (hypothetical) electron activity at equilibrium and measures the relative tendency of a solution to accept or transfer electrons. In a highly reducing solution the tendency to donate electrons, that is, electron activity, is relatively large. Just as the activity of protons is very low at high $p\ce{H}$, the activity of hypothetical electrons is very low at high $pe$. Thus, a high $pe$ indicates a relatively high tendency for oxidation. In equilibrium equations, $\ce{H+}$ and $\ce{e-}$ are treated in an analogous way.

All redox reactions, including half-reactions follow the same principles of thermodynamics (e.g., spontaneity, etc.) and can be described using chemical equilibrium principles (e.g., law of mass action, Le Châtelier's principle, etc.) Using these equilibrium principles allow us to explore oxidation and reduction problems effectively.

```{dropdown} Example: What is $pe$? 

Consider the $\ce{Fe}$ reduction half-reaction as shown below:

$$\ce{
Fe^3+ + e- -> Fe^2+ \quad\quad $K$
}$$


Note that this reaction has an equilibrium constant, $K$. Using the law of mass action, we can write the equilibrium expression as

$$
K = \frac{\{\ce{Fe^2+}\}}{\{\ce{Fe^3+}\}\{e^-\}}
$$ 

Rearrange equilibrium expression, 

$$
\{e^-\}  = \left[
\frac{1}{K}\frac{\{\ce{Fe^2+}\}}{\{\ce{Fe^3+}\}}
\right]$$ 

Apply $-\log_{10}$ on both sides, 

$$
-\log \{e^-\} = p\text{e} 
= \left[ \log K - \log \frac{\{\ce{Fe^2+}\}}{\{\ce{Fe^3+}\}} \right]
$$
```


```{dropdown} Example: What does the $pe$ value signify? 

Let's explore what the $pe$ value represents using the $\ce{Fe}$ reduction example.

$$\ce{
Fe^3+ + e- -> Fe^2+ \qquad $K$
} $$

When $\ce{e-}$ activity is lowered ($pe$ value is high), the above reaction shifts to the left, based on our understanding of chemical equilibrium principles. That is, the conditions lead to more oxidized form of $\ce{Fe}$ ($\ce{Fe^3+}$). Therefore, high $pe$ values imply strong oxidizing conditions.

When $\ce{e-}$ activity increases ($pe$ value is low), the above reaction shifts to the right, leading to the reduced form of $\ce{Fe}$ ($\ce{Fe^2+}$). Therefore, low $pe$ values imply strong reducing conditions.
```

```{dropdown} Example: Calculating $pe$ in $\ce{Fe}$ system 

Let's calculate $pe$ in the above $\ce{Fe}$ system, if $\ce{Fe^3+} = \pu{e-5 M}$, $\ce{Fe^2+} = \pu{e-3 M}$, & $K=\pu{e13}$

$$ \ce{
Fe^3+ + e- -> Fe^2+ \qquad $K$
} $$

Substitute known data in the $K$ expression: 

$$\begin{align}
\{e^-\}  &= \left[
\frac{1}{K}\frac{\{\ce{Fe^2+}\}}{\{\ce{Fe^3+}\}}
\right] = \frac{1}{\pu{e13}}\frac{\pu{e-3}}{\pu{e-5}} = \pu{e-11}\\
\therefore pe &= 11
\end{align}$$
```


```{dropdown} Example: Calculating $pe$ in $\ce{Mn}$ system 

Pyrolusite $\ce{MnO2 (s)}$ reduces in the environment as shown in the half-reaction below:

$$ \ce{
MnO2(s) + 4 H+ + 2 e- -> Mn^2+ + 2 H2O(l)
} $$

Let's write the $K$ expression for this reaction.

$$K = \frac{ \{\ce{Mn^2+}\} \{\ce{H2O(l)}\}^2}{\{\ce{MnO2(s)}\}\{\ce{H+}\}^4 \{e^-\}^2}
$$

Per chemical equilibrium "rules," activities of pure solids and liquids are equal to $1$. Therefore, $\{\ce{MnO2(s)}\} = \{\ce{H2O(l)}\} = 1$. The $K$ expression can be simplified and rearranged as 

$$\begin{align}
        K &= \frac{\{\ce{Mn^2+}\}}{\{\ce{H+}\}^4 \{e^-\}^2}\\
        \{e^-\}^2 &= \left[\frac{1}{K} \frac{\{\ce{Mn^2+}\}}{\{\ce{H+}\}^4} \right] \\
        \text{Apply}\ -\log_{10}\ \text{on both sides}\\
        pe &= \frac{1}{2} \left[ \log K - \log \{\ce{Mn^2+}\} -4p\ce{H} \right] 
\end{align}$$

If the unknown values (e.g., $\{\ce{Mn^2+}\}$, $p\ce{H}$, and $K$) are given in the above system, $pe$ can be determined.
```

##  $pe$ and $p\ce{H}$ relationships

Two components - protons ($\ce{H+}$) and electrons ($\ce{e-}$) - are omnipresent in the natural aqueous environments. In many half-reactions, both $\ce{H+}$ and $\ce{e-}$ participate to facilitate the redox transformation. Consider the reduction of hematite ($\ce{Fe2O3(s)}$) mineral into $\ce{Fe^2+(aq)}$ during weathering reactions. The balanced half-reaction can be shown as:

$$\ce{
Fe2O3 + 2e- + 6 H+ -> 2 Fe^2+ + 3 H2O \qquad $K$
}$$

The $pe$ expression for the above reaction can be derived from the equilibrium expression (rearrange the $\{\ce{e-}\}$ and apply $-\log_{10}$) to arrive at the following equation:

$$pe = \frac{1}{2} \left[ \log K - \log \frac{\{\ce{Fe2+}\}^2}{\{\ce{Fe2O3}\}} -6 p\ce{H} \right]$$

It can be seen from this expression that $pe$ is related to $p\ce{H}$.

## Redox Potential, $Eh$

Redox half-reactions explicitly involve $\ce{e-}$ and the law of mass action expressions show activities and/or concentrations of all participants in those reactions are determinable. However, concentrations or activities of the $\ce{e-}$ are not measurable directly. Like $\ce{H+}$ transfer in acid-base reactions is measured using a $p\ce{H}$-probe, the transfer of $\ce{e-}$ between two atoms can be quantified by measuring the electrical potential between two electrodes. This electrical potential that indirectly quantifies the $\ce{e-}$ activity is called redox potential, $Eh$, and reported in millivolts ($\pu{mV}$) or volts ($\pu{V}$). The transfer of electrons can be visualized using the operation of a galvanic cell, as shown in {numref}`galvanic-cell`.

```{figure} https://upload.wikimedia.org/wikipedia/commons/thumb/6/6d/Galvanic_cell_with_no_cation_flow.svg/1920px-Galvanic_cell_with_no_cation_flow.svg.png
---
name: galvanic-cell
figclass: margin-caption
---
A galvanic cell that shows the transfer of $\ce{e-}$ from $\ce{Zn}$ to $\ce{Cu}$.  The transfer of the $\ce{e-}$ is quantified by the voltage meter on the wire.  Image source: [Galvanic cell - Wikipedia](https://en.wikipedia.org/wiki/Galvanic_cell)
```

The $Eh$ is correlated to $pe$ as follows:

```{math}
:label: ehrt
Eh = 2.303 \frac{\text{R}T}{F} pe
```
Where, $R = \pu{8.314e-3 kJ mol-1 K-1}$, $T$ is temperature in $\pu{K}$, and $F$ is Faraday's constant and is equal to $\pu{96489 C mol−1}$ or $\pu{96.42 kJ g−1}$. At $\pu{25 ^\circ C}$, Eq. {eq}`ehrt`, is simplified to: 

```{math}
:label: ehpe
Eh = 0.059 pe
```


## $Eh-p\ce{H}$ or Pourbaix Diagrams

The environmental conditions, including $p\ce{H}$ and $Eh$, can have a strong influence on the predominance of any redox sensitive element. Relationships between redox species of different elements vary as a function of $p\ce{H}$ and $Eh$ conditions, a diagram that depicts these relationships can be useful in showing the likelihood of a particular form of a redox sensitive element under any given condition. The value of $Eh-p\ce{H}$ diagrams consists primarily in providing an aid for the interpretation of equilibrium data by permitting the simultaneous representation of many reactions.

Natural waters are often in a highly dynamic state with regard to redox status, rather than in or near equilibrium. Most redox reactions have a tendency to be much slower than acid-base reactions, especially in the absence of suitable biochemical conditions. However, these diagrams can greatly aid in understanding the possible redox patterns in natural waters systems. In {numref}`pourbaix-water`, different $p\ce{H}$ and $Eh$ environments are represented. More positive $Eh$ values represent oxidizing conditions, while more negative $Eh$ values represent reducing conditions. The possible limits of $Eh$ in natural environments are shown in {numref}`pourbaix-water`.

```{figure} https://upload.wikimedia.org/wikipedia/commons/1/1d/PourbaixWater.png
---
name: pourbaix-water
figclass: margin-caption
---
$Eh$ and $p\ce{H}$ conditions for natural waters on Earth's surface at $\pu{25 ^\circ C}$.  The upper and lower limits represent oxidation or reduction of water to $\ce{O2(g)}$ or $\ce{H2(g)}$, respectively.  Image source: [Pourbaix diagram - Wikipedia](https://en.wikipedia.org/wiki/Pourbaix_diagram)
```

The upper and lower limits of the $Eh-p\ce{H}$ diagram are represented by the oxidation and reduction of water to $\ce{O2(g)}$ and $\ce{H2(g)}$, respectively, as shown in the table below:

| Redox  | Reaction | $K$ values |
|-----------|----------------------------------------|-----------------:|
| Oxidation | $\ce{O2(g) + 4 H+(aq) +  4 e- -> 2 H2O (l)}$ | $10^{83.1}$ |
| Reduction | $\ce{2 H+(aq) + 2 e- -> H2(g)}$              | $1$           |


```{dropdown} Example: Pourbaix diagram for $\ce{As}$ 

The arsenic problem in the Bengal basin is cause by redox transformation of multiple $\ce{As}$ species in soil and groundwater environments. The most common reactions are listed below:

| Reaction                                         | $K$ values       |
|--------------------------------------------------|------------------:|
| $\ce{H2AsO4- + 3 H+ + 2 e- -> As(OH)3 + H2O }  $   | $10^{21.68}$ |
| $\ce{HAsO4^2- + 4 H+ + 2 e- -> As(OH)3 + 2 H2O }$  | $10^{28.44}$ |
| $\ce{H3AsO4 + 2 H+ + 2 e- -> As(OH)3 + H2O } $     | $10^{19.44}$ |
| $\ce{HAsO4^2- + 3 H+ + 2 e- -> As(OH)4- + 2 H2O }$ | $10^{19.22}$ |

The resulting Pourbaix diagram is shown in the figure below. If we know the redox conditions prevalent in the soil and groundwater environments, we can easily read from the diagram what As form is likely to predominate under those conditions.

![Arsenic Pourbaix diagram](https://chem.libretexts.org/@api/deki/files/240373/clipboard_e3ceaf5d6995b5f745fcfa8d73981aa9f.png?revision=1)

Image source: [Introduction to Inorganic Chemistry/Redox Stability and Redox Reactions - Wikibooks, open books for an open world](https://en.wikibooks.org/wiki/Introduction_to_Inorganic_Chemistry/Redox_Stability_and_Redox_Reactions)
```


## Microorganisms and Redox Reactions

Redox reactions can be spontaneous as written, however, these reactions occur at a slow rate. Often, microorganisms play an important role, as catalysts, in these reactions. These microorganisms can manufacture or decompose organic matter, produce or consume oxygen, and oxidize or reduce sulfur, iron, nitrogen, and a variety of other substances. They may also influence the rate of a geochemical reaction in which they are producers or consumers of the reactants or products. It is likely that microorganisms contributed significantly to the formation of many types of mineral deposits found in Earth's near surface environments.

Consider the following reaction in which $\ce{N}$ is being oxidized:

$$ \ce{
NH4+ + 2 O2 -> NO3- + H2O + 2 H+
} $$

$\Delta _{rxn}G^\circ = \pu{-266.5 kJ mol-1}$ for this reaction, therefore, this reaction is spontaneous and favored to occur as written. However, $\ce{NH4+}$ is very stable in the environment and the above reaction occurs slowly. This reaction can proceed at a faster rate if the aqueous system contained a certain type of bacteria, *Nitrosomonas* sp. and *Nitrobacter* sp. These bacteria use $\ce{NH4+}$ as nutrient during their metabolism, while facilitating the redox reaction. There are many such reactions that occur in nature.

Note that microorganisms act as catalysts and do not actually participate in the reactions.