Geochemistry. William M. White

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CH₂O refers to carbohydrate, the basic product of photosynthesis.

and called the hydrogen scale potential, or E_H, of this reaction. Thus, the E_H for the reduction of Zn⁺² to Zn⁰ is −0.763 V. The hydrogen scale potentials of a few half-cell reactions are listed in Table 3.3. The sign convention for E_H is that the sign of the potential is positive when the reaction proceeds from left to right (i.e., from reactants to products). Thus, if a reaction has positive E_H, the metal ion will be reduced by hydrogen gas to the metal. If a reaction has negative E_H, the metal will be oxidized to the ion and H⁺ reduced. The standard state potentials (298 K, 0.1 MPa) of more complex reactions can be predicted by algebraic combinations of the reactions and potentials in Table 3.3 (see Example 3.11).

      The half-cell reactions in Table 3.3 are arranged in order of increasing . Thus, a species on the product (right) side of a given reaction will reduce (give up electrons to) the species on the reactant side in all reactions listed below it. Thus, in the Daniell cell reaction in Figure 3.18, Zn metal will reduce Cu²⁺ in solution. Zn may thus be said to be a stronger reducing agent than Cu.

      Electrochemical energy is another form of free energy and can be related to the Gibbs free energy of reaction as:

(3.106)

      and

(3.107)

      where z is the number of electrons per mole exchanged (e.g., 2 in the reduction of zinc) and is the Faraday constant ( = 96,485 coulombs; 1 joule = 1 volt-coulomb). The free energy of formation of a pure element is 0 (by convention). Thus, the ΔG in a reaction that is opposite one such as 3.105, such as:

      is the free energy of formation of the ion from the pure element. From eqn. 3.106 we can calculate the ΔG for the reduction of zinc as 147.24 kJ/mol. The free energy of formation of Zn²⁺ would be −147.24 kJ/mol. Given the free energy of formation of an ion, we can also use eqn. 3.105 to calculate the hydrogen scale potential. Since

(3.108)

      we can substitute eqns. 3.106 and 3.107 into 3.108 and also write

(3.109)

Equation 3.109 is known as the Nernst equation.† At 298 K and 0.1 MPa it reduces to:

      (3.110)

      We can deduce the meaning of this relationship from the relationship between ΔG and E in eqn. 3.106. At equilibrium ΔG is zero. Thus, in eqn. 3.108, activities will adjust themselves such that E is 0.

Example 3.11 Calculating the EH of net reactions

      We can calculate E_H values for reactions not listed in Table 3.3 by algebraic combinations of the reactions and potentials that are listed. There is, however, a catch. Let's see how this works.

      Calculate the E_H for the reaction:

      Answer: This reaction is the algebraic sum of two reactions listed in Table 3.3:

Since the reactions sum, we might assume that we can simply sum the E_H values to obtain the E_H of the net reaction. Doing so, we obtain an E_H of 0.33 V. However, the true E_H of this reaction is −0.037 V. What have we done wrong?

      We have neglected to take into consideration the number of electrons exchanged. In the algebraic combination of E_H values, we need to multiply the E_H for each component reaction by the number of electrons exchanged. We then divide the sum of these values by number of electrons exchanged in the net reaction to obtain the E_H of the net reaction, i.e.,

(3.111)

      where the sum is over the component reactions i. Looking at eqn. 3.106, we can see why this is the case. By Hess's law, the ΔG of the net reaction must be the simple sum of the component reaction ΔGs, but E_H values are obtained by multiplying ΔG by z. Equation 3.111 is derived by combining eqn. 3.106 and Hess's law. Using eqn. 3.111, we obtain the correct E_H of −0.037 V.

       3.11.1.2 Alternative representation of redox state: pε

      Consider again the reaction:

      (3.102)

      If we were to express the equilibrium constant for this reaction, we would write:

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