6D.4 Oxidation–Reduction Reactions

In a complexation reaction, a Lewis base donates a pair of electrons to a Lewis acid.

In an oxidation–reduction reaction, also known as a redox reaction,electrons are not shared, but are transferred from one reactant to another. As a result of this elec- tron transfer, some of the elements involved in the reaction undergo a change in ox- idation state. Those species experiencing an increase in their oxidation state are oxi- dized, while those experiencing a decrease in their oxidation state are reduced. For example, in the following redox reaction between Fe³⁺and oxalic acid, H2C2O4, iron is reduced since its oxidation state changes from +3 to +2.

2Fe³⁺(aq) + H₂C₂O₄(aq) + 2H₂O(l)

t

_2Fe²⁺_{(aq) + 2CO2(g) + 2H3O}⁺_{(aq) 6.22}

redox reaction

An electron-transfer reaction.

Oxalic acid, on the other hand, is oxidized since the oxidation state for carbon in- creases from +3 in H₂C₂O₄to +4 in CO₂.

Redox reactions, such as that shown in equation 6.22, can be divided into sepa- rate half-reactions that individually describe the oxidationand the reduction processes.

H₂C₂O₄(aq) + 2H₂O(l)→2CO₂(g) + 2H₃O⁺(aq) + 2e^– Fe³⁺(aq) +e^–→Fe²⁺(aq)

It is important to remember, however, that oxidation and reduction reactions al- ways occur in pairs.* This relationship is formalized by the convention of calling the species being oxidized a reducing agent,because it provides the electrons for the re- duction half-reaction. Conversely, the species being reduced is called an oxidizing agent.Thus, in reaction 6.22, Fe³⁺is the oxidizing agent and H₂C₂O₄is the reducing agent.

The products of a redox reaction also have redox properties. For example, the Fe²⁺in reaction 6.22 can be oxidized to Fe³⁺, while CO₂can be reduced to H₂C₂O₄. Borrowing some terminology from acid–base chemistry, we call Fe²⁺the conjugate reducing agent of the oxidizing agent Fe³⁺and CO₂the conjugate oxidizing agent of the reducing agent H₂C₂O_4.

Unlike the reactions that we have already considered, the equilibrium position of a redox reaction is rarely expressed by an equilibrium constant. Since redox reac- tions involve the transfer of electrons from a reducing agent to an oxidizing agent, it is convenient to consider the thermodynamics of the reaction in terms of the electron.

The free energy, ∆G_,associated with moving a charge, Q,under a potential, E, is given by

∆G=EQ

Charge is proportional to the number of electrons that must be moved. For a reac- tion in which one mole of reactant is oxidized or reduced, the charge, in coulombs, is

Q=nF

where nis the number of moles of electrons per mole of reactant, and Fis Faraday’s constant (96,485 C⋅mol^–1). The change in free energy (in joules per mole; J/mol) for a redox reaction, therefore, is

∆G= –nFE 6.23

where ∆Ghas units of joules per mole. The appearance of a minus sign in equation 6.23 is due to a difference in the conventions for assigning the favored direction for reactions. In thermodynamics, reactions are favored when ∆G is negative, and redox reactions are favored when Eis positive.

The relationship between electrochemical potential and the concentrations of reactants and products can be determined by substituting equation 6.23 into equation 6.3

–nFE= –nFE° +RTlnQ

where E°is the electrochemical potential under standard-state conditions. Dividing through by –nFleads to the well-known Nernst equation.

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Modern Analytical Chemistry

*Separating a redox reaction into its half-reactions is useful if you need to balance the reaction. One method for balancing redox reactions is reviewed in Appendix 4.

oxidation A loss of electrons.

reduction A gain of electrons.

reducing agent

A species that donates electrons to another species.

oxidizing agent

A species that accepts electrons from another species.

Nernst equation

An equation relating electrochemical potential to the concentrations of products and reactants.

Chapter 6 Equilibrium Chemistry

147

*ln(x) = 2.303 log(x)

Substituting appropriate values for R and F,assuming a temperature of 25 °C (298 K), and switching from ln to log* gives the potential in volts as

6.24 The standard-state electrochemical potential, E°, provides an alternative way of expressing the equilibrium constant for a redox reaction. Since a reaction at equilib- rium has a ∆Gof zero, the electrochemical potential, E,also must be zero. Substi- tuting into equation 6.24 and rearranging shows that

6.25 Standard-state potentials are generally not tabulated for chemical reactions, but are calculated using the standard-state potentials for the oxidation, E°ox, and reduction half-reactions, E°red. By convention, standard-state potentials are only listed for re- duction half-reactions, and E° for a reaction is calculated as

E°reac= E°red– E°ox

where both E°redand E°oxare standard-state reduction potentials.

Since the potential for a single half-reaction cannot be measured, a reference half- reaction is arbitrarily assigned a standard-state potential of zero. All other reduction potentials are reported relative to this reference. The standard half-reaction is

2H3O⁺(aq) + 2e^–

t

_{2H2O(l) + H2(g)}

Appendix 3D contains a listing of the standard-state reduction potentials for se- lected species. The more positive the standard-state reduction potential, the more favorable the reduction reaction will be under standard-state conditions. Thus, under standard-state conditions, the reduction of Cu²⁺to Cu (E° = +0.3419) is more favorable than the reduction of Zn²⁺to Zn (E° = –0.7618).

EXAMPLE

6.5

Calculate (a) the standard-state potential, (b) the equilibrium constant, and (c) the potential when [Ag⁺] = 0.020 M and [Cd²⁺] = 0.050 M, for the following reaction taking place at 25 °C.

Cd(s) + 2Ag⁺(aq)

t

_Cd²⁺(aq) + 2Ag(s) SOLUTION

(a) In this reaction Cd is undergoing oxidation, and Ag⁺ is undergoing reduction. The standard-state cell potential, therefore, is

(b) To calculate the equilibrium constant, we substitute the values for the standard-state potential and number of electrons into equation 6.25.

E^o= E^o_Ag⁺_/Ag –E^o_Cd²⁺_/Cd = 0 7996. V– (– .0 4030V)= 1 2026. V E RT

nF K

o = log

E = E

n Q

o – . 0 05916log E = E RT

nF Q

o – ln

1 2026 0 05916

. . 2

= logK

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Modern Analytical Chemistry

Solving for Kgives the equilibrium constant as logK= 40.6558 K= 4.527×10⁴⁰

(c) The potential when the [Ag⁺] is 0.020 M and the [Cd²⁺] is 0.050 M is calculated using equation 6.24 employing the appropriate relationship for the reaction quotient Q.