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7. Redox Equilibria

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This chapter focuses on equilibria involving electron transfers (redox reactions). A systematic approach to identify electron transfers and to balance redox reactions is introduced. Redox potential and the Nernst equation are linked to the Gibbs free energy.

7.1 Introduction

Redox reactions involve the transfer of electrons from an electron donor (reductant) to an electron acceptor (oxidant). The electron donor/reductant is then oxidized, while the electron acceptor (oxidant) is reduced. This ‘reductant is oxidized’ and ‘oxidant is reduced’ terminology could be seen as confusing, and for clarity we will avoid the use of the terms reductant and oxidant and limit the use of the terms oxidation and reduction. Electron transfers are linked to energy transfers and provide the basis for most life on Earth. Heterotrophs, including humans, utilize organic matter (an electron donor) and combine it with an electron acceptor (oxygen) to respire and release the energy for maintenance and growth. Some microbes (e.g. chemolitho(auto)trophs) use minerals such as pyrite (FeS2) or reduced gases such as methane (CH4) as an electron donor to generate energy upon electron transfer to an acceptor (oxygen, metal oxides or sulfate). The exposure of reduced substances (organic matter, iron sulfide) to atmospheric oxygen (oxidative weathering) causes not only transfer of electrons (oxidation), but also release of protons. An extreme example is acid mine drainage: the oxidation of metal sulfides causes proton release with acidic conditions as a result. Pyrite (FeS2) is a major constituent of mine waste tailings and reacts with oxygen:
$$ 4FeS_{2} + 15O_{2} + 10H_{2} O \to 4FeOOH + 8SO_{4}^{2 - } + 16H^{ + } $$
This reaction generates 4 protons per mol pyrite oxidized and leads to acidification.

7.2 Identifying Electron Transfer and Balancing Redox Reactions

Redox reactions change the oxidation state of substances. An oxidation state, or oxidation number, is a hypothetical charge of an atom, if all its bonds to other atoms were fully ionic. It quantifies the excess or shortage of electrons. Oxidation numbers are integers because only discrete electrons can be transferred. The rules for assigning oxidation numbers are as follows:
An atom in its elemental state has an oxidation number of 0: e.g., Na, O2, Ar
An atom in a monoatomic ion has an oxidation number equal to the charge of the ion: e.g., Ca2+, Fe3+, F
Oxygen has an oxidation state of −2, except when bonded to an oxygen or fluor atom.
Hydrogen has an oxidation state of + 1, except when bonded as a hydride (−1) to metals.
Alkaline and earth alkaline metals (columns 1 and 2 of periodic table) have an oxidation state of + 1 and + 2, respectively, and the halogens (F, Br, Cl, I) have an oxidation state of −1, except when bonded to oxygen or nitrogen.
The sum of oxidation numbers is equal to the net charge for a polyatomic ion.
For instance, in sulfuric acid, H2SO4, sulfur has an oxidation number of + 6, while nitrogen oxidation numbers in nitrate (\(NO_{3}^{ - } )\) and ammonium ( \(NH_{4}^{ + } )\) are + 5 and −3, respectively. Assigning oxidation states to sulfur is challenging as it may be −2, −1, 0, + 3, + 4 and + 6, depending on the molecular setting. It is important to recall that these oxidation numbers are hypothetical because one would infer that the charge of the two sulfur atoms in thiosulfate (\(S_{2} O_{3}^{2 - } )\) is + 2, while in fact one sulfur has a charge of −1 and the other + 5. Some common (Fe, Mn, Ti-bearing) minerals have elements with mixed oxidation number: e.g., in magnetite, Fe3O4, two of the iron atoms are + 3 and the other is + 2.
Balancing reactions involving electron transfers requires a systematic approach involving two steps, first the identification of the electron donor and acceptor, hence the electrons transferred, and then the balancing of the elements. There are two methods: the half-reaction, or ion–electron method, in which the oxidation and reduction processes are separated and each balanced before re-combining them again, and the oxidation state or whole-reaction method. The latter involves the following steps:
Assign oxidation state numbers to each of the atoms in the equation and write the numbers above the atom.
Identify the atoms that donate or accept electrons, i.e., that increase or decrease in oxidation number.
Identify the electron donating and electron accepting compounds on the left-hand side and quantify the number of electrons they donate/accept.
Balance the electron donating and accepting compounds on the left-hand side
Balance all elements, except O and H, on both sides of the equation
Balance oxygen on both sides of the equation through addition of H2O
Balance hydrogen on both sides of the equation through addition of H+
To illustrate this method, consider the following reaction that may occur in surface sediments or in groundwaters:
$$ FeS + NO_{3}^{ - } \to FeOOH + SO_{4}^{2 - } + N_{2} $$
First, we identify the oxidation numbers (oxygen and hydrogen are −2 and + 1, respectively)
$$ \begin{aligned} &+ 2 - 2\quad + 5\quad \quad \quad + 3 \quad \quad \quad+ 6\quad \quad \quad 0 \hfill \\ &FeS + \quad \quad NO_{3}^{ - } \to FeOOH + SO_{4}^{2 - } + N_{2} \hfill \\ \end{aligned} $$
and recognize that Fe donates one electron to go from + 2 to + 3, S donates 8 electrons to go from −2 to + 6 and N accepts 5 electrons to go from + 5 to 0. Accordingly, FeS donates 9 electrons (1 from Fe and 8 from S) and \(NO_{3}^{ - }\) accepts 5 electrons (N from + 5 to 0). Balancing the electron donor and acceptor on the left-hand side thus implies that 5 FeS react with 9 \(NO_{3}^{ - } .\)
$$ 5 FeS + 9 NO_{3}^{ - } \to FeOOH + SO_{4}^{2 - } + N_{2} $$
The rest of the balancing is simply a book-keeping, first, we balance Fe, S and N:
$$ 5 FeS + 9 NO_{3}^{ - } \to 5 FeOOH + 5 SO_{4}^{2 - } + 4.5 N_{2} $$
Next, we balance oxygen atoms by adding three molecules of H2O, in this case to the left-hand side:
$$ 5 FeS + 9 NO_{3}^{ - } + 3 H_{2} O \to 5 FeOOH + 5 SO_{4}^{2 - } + 4.5 N_{2} $$
And finally, we balance hydrogen atoms by adding one proton, in this case to the right-hand side:
$$ 5 FeS + 9 NO_{3}^{ - } + 3 H_{2} O \to 5 FeOOH + 5 SO_{4}^{2 - } + 4.5 N_{2} + H^{ + } $$

7.3 Redox Potentials and the Nernst Equation

Electron transfers are associated with energy transfer. Electrochemical cells (e.g., a battery) convert Gibbs free energy of a reaction into electrical energy and vice versa. Galvanic cells make use of spontaneous chemical reactions to generate electricity, while electrolysis requires electrical energy to drive a non-spontaneous reaction.
Consider a strip of zinc metal in an aqueous solution of copper sulfate, the strip will become darker because of copper precipitates and the blue color solution will fade because of the spontaneous reaction consuming the colored Cu2+
$$ Zn \left( s \right) + Cu^{2 + } \left( {aq} \right) \to Zn^{2 + } \left( {aq} \right) + Cu\left( s \right) $$
Zn donates two electrons which are accepted by Cu2+ and the energy released goes into heat. If the electron donating (oxidation) and accepting (reduction) part of the reaction are physically separated but connected via a wire and salt bridge (an ionic solution allowing migration of cations and anions) to maintain charge balance, an electrical current will flow through the wire because of a difference in potential (E, expressed in V(olt)). Electrochemists split redox reactions in two half-reactions, or half-cells, one for the oxidation and one for the reduction process. For reaction 7.1, these half-reactions are:
$$ Zn\left( s \right) \to Zn^{2 + } \left( {aq} \right) + 2e^{ - } {\text{for the oxidation }}\left( {\text{electrons donating reaction}} \right) $$
$$ Cu^{2 + } \left( {aq} \right) + 2e^{ - } \to Cu \left( s \right){\text{for the reduction }}\left( {\text{electrons accepting reaction}} \right). $$
The potentials for these half reactions have been measured relative to that of protons/hydrogen gas conversions which has been set at 0 V for both reduction and oxidation:
$$ 2H^{ + } \left( {aq} \right) + 2e^{ - } \to H_{2 } \left( g \right),{\text{ reduction}},{\text{ electrons accepting reaction}} $$
$$ H_{2 } \left( g \right){ } \to 2H^{ + } \left( {aq} \right) + 2e^{ - } ,{\text{ oxidation}},{\text{ electrons donating reaction}} $$
The standard reduction potentials \(E^{0}\) (with electrons on the left-hand side and at SATP) have been tabulated and can be used to calculate the redox potential of reactions (Table 7.1). Oxidation potentials are the reverse of reduction potentials. For instance, the standard reduction potential for Zn and Cu are:
$$ \begin{gathered} Zn^{2 + } + 2e^{ - } \to Zn\quad {\text{E}}^{0} = - 0.{76}\,{\text{V}} \hfill \\ \hfill \\ \end{gathered} $$
$$ Cu^{2 + } + 2e^{ - } \to Cu\quad {\text{E}}^{0} = + 0.{34}\,{\text{V}} $$
Table 7.1
Standard reduction potentials at 25 °C
Reduction Half-Reaction
Eo (V)
Co3+ (aq) + e
\(\to\) Co2+ (aq)
H2O2 (aq) + 2 H+ (aq) + 2 e
\(\to\) 2 H2O (l)
Ce4+ (aq) + e
\(\to\) Ce3+ (aq)
Mn3+ (aq) + e
\(\to\) Mn2+ (aq)
Cr2O72− (aq) + 14 H+ (aq) + 6 e
\(\to\) 2 Cr3+ (aq) + 7 H2O (l)
O2 (g) + 4 H+ (aq) + 4 e
\(\to\) 2 H2O (l)
Ag+ (aq) + e
\(\to\) Ag (s)
Fe3+ (aq) + e
\(\to\) Fe2+ (aq)
Cu2+ (aq) + 2 e
\(\to\) Cu (s)
Cu2+ (aq) + e
\(\to\) Cu+ (aq)
2H+(aq) + 2 e
\(\to\) H2 (g)
Pb2+ (aq) + 2 e
\(\to\) Pb (s)
Sn2+ (aq) + 2 e
\(\to\) Sn (s)
Co2+ (aq) + 2 e
\(\to\) Co (s)
Cd2+ (aq) + 2 e
\(\to\) Cd (s)
Fe2+ (aq) + 2 e
\(\to\) Fe (s)
Zn2+ (aq) + 2 e
\(\to\) Zn (s)
Mn2+ (aq) + 2 e
\(\to\) Mn (s)
Mg2+ (aq) + 2 e
\(\to\) Mg(s)
The overall potential for the complete reaction 7.1 is then – (− 0.76) + 0.34 = 1.1 V, resulting in a positive voltage as it should be for a spontaneous reaction. Redox potentials are intensive properties, i.e., independent of the amount, and additive.
The standard redox potential \(E^{0}\) is related to the standard Gibbs free energy via the Nernst expression:
$$ \Delta_{r} G^{o} = - n FE^{0} $$
where \(\Delta_{r} G^{o}\) is the standard Gibbs free energy of a reaction, n is the number of electrons transferred and F is the Faraday constant (the electrical charge of 1 mol of electrons: 96,500 C (C) per mol e). Recall from physics class that 1 J = 1 V·C. The minus sign in Eq. (7.5) is needed because spontaneous reactions have a negative Gibbs free energy and a positive redox potential. Accordingly, we can calculate the Gibbs free energy of reaction 7.1:
$$ \Delta_{r} G^{o} = - n FE^{0} = - 2*96500*1.1 = - 212 kJ $$
The standard redox potential is also related to the equilibrium constant because \(\Delta_{r} G^{o}\) is linked to K: (Eq. 5.10):
$$K = \exp \left( {\frac{{ - \Delta_{r} G^{o} }}{RT}} \right) = \exp \left( {\frac{{n FE^{0} }}{RT}} \right)$$
Redox potentials, like Gibbs free energy changes, depend on temperature and the composition of the reaction mixture. For Gibbs free energy, we have (Eq. 5.9):
$$ \Delta_{r} G = \Delta_{r} G^{o} + RT\ln Q $$
where Q is the reaction quotient. Since \(\Delta_{r} G^{o} = - n FE^{0}\) and \(\Delta_{r} G = - n FE\), we arrive at
$$ - n FE = - n FE^{0} + RTlnQ $$
Dividing by −nF, this results in the Nernst equation
$$ E = E^{0} - \frac{RT}{{nF}}lnQ = E^{0} - \frac{2.303RT}{{nF}}logQ $$
Recall from math class that \(\ln 10 = 2.303\log 10 = 2.303\).
At 25 °C, the term \(\frac{2.303RT}{F}\) has a value of 0.0592 V and thus
$$ E = E^{0} - \frac{0.0592}{n}logQ. $$
Returning to our initial reaction (7.1), we consider the case that \(\left[ {Zn^{2 + } } \right]\) has a concentration of 0.5 M and \(\left[ {Cu^{2 + } } \right]\) = 0.01 M. The potential would then be
$$ E = E^{0} - \frac{0.0592}{n}logQ = 1.1 - \frac{0.0592}{2}log\frac{{\left[ {Zn^{2 + } } \right]{ }}}{{\left[ {Cu^{2 + } } \right]{ }}} = 1.05{\text{ V}}. $$
The redox reaction is favorable.
We conclude with articulating the internal consistency and interchangeability of equilibrium constants between different branches of chemistry. Equilibrium constants can be derived from:
thermochemistry, i.e., \(\Delta G_{f}^{o} \) data obtained on pure phases (Eq. 5.9): \(K = exp\left( {\frac{{ - \Delta G_{r}^{0} }}{RT}} \right)\)
the composition of solutions at equilibrium (Eq. 5.4) \(K = \frac{{C_{eq}^{\gamma } D_{eq}^{\delta } }}{{A_{eq}^{\alpha } B_{eq}^{\beta } }}\)
and redox potentials (Eq. 6): \(K = exp\left( {\frac{{n FE^{0} }}{RT}} \right)\).
In the absence of uncertainties, the equilibrium constants should be identical.
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Redox Equilibria
Jack J. Middelburg
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