P. M. Silin.

Oxidation-Reduction Reactions.

Oxidation-reduction reactions are those reactions in which the valence of certain elements changes (e.g., the transition from FeO to Fe₂O₃, from FeCl₃ to FeCl₂, from K₂Cr₂O₇ to CrCl₃, etc.).

According to views that have taken shape under the influence of electrochemistry, all atoms entering into the molecule of any compound may be regarded as charged with positive or negative electricity and as being held near one another owing to their opposite charges¹). The sum of all charges of a neutral molecule (not of an ion) is 0. The atom H in its compounds always has one and the same elementary positive charge (·). From the structure of the molecules H·—Cl′ and H·—O″—H· we conclude that Cl has one negative charge (′), and O two negative charges (almost always). From the structure of the more complex molecule H₂SO₄, or

\[ \begin{array}{c} \mathrm{H}\!\cdot\!-\mathrm{O}^{\prime\prime}\!\!\diagup\\[-2mm] \mathrm{H}\!\cdot\!-\mathrm{O}^{\prime\prime}\!\!\diagdown \end{array} \mathrm{S} \begin{array}{c} :\!=\mathrm{O}^{\prime\prime}\\[-1mm] :\!=\mathrm{O}^{\prime\prime} \end{array} \]

it is easy to understand that S has 6(·), etc. The chargedness may be calculated even without the structural formula (if the molecule contains only one atom of the element whose chargedness is being sought); e.g., in HMnO₄ the charge of H = +1, the charge of four O = −4·2, the charge of Mn = X (sought), but the sum \(+1 - 4\cdot 2 + X = 0\), whence \(X = +7\)²).

From the standpoint of the views set forth, valence is the number of elementary \(+\) or \(−\) charges (equal in absolute magnitude to the charge of H·) which an atom of an element has in the given compound. Thus, the valence of H· = +1, O″ = −2, S in H₂SO₄ = +6, etc. The valence of free elements may be regarded as = 0, e.g., O⁰, H⁰, Cl⁰, S⁰. A change of valence, i.e. oxidation or reduction of an atom, may be imagined as its recharging; moreover, taking into account that positive electricity is firmly bound to the atom and that only elementary negative charges—“electrons”—can pass from one atom to another, the following definition of oxidation-reduction reactions can be given.

Oxidation is the removal of electrons from an atom; here we consider that to remove an electron, i.e. a negative charge, is the same as to add a positive charge. Examples: Fe⁰—2(′) = Fe··; Fe··—(′) = Fe···; Cr···—3(′) = Crᵛᴵ; S″—2(′) = S⁰; Cl′—(′) = Cl⁰; S″—6(′) = S⁗.

Reduction is the addition of electrons. Examples: Fe·· + 2(′) = Fe⁰; Fe··· + (′) = Fe··; Crᵛᴵ + 3(′) = Cr···; S⁰ + 2(′) = S″; Cl⁰ + (′) = Cl′; S⁗ + 6(′) = S″.

Hence the conclusion: oxidation reactions always go together with reduction reactions, since an electron removed from some atom is, of course, attached to some other atom; i.e., if one atom is oxidized, then some other atom is necessarily reduced.

¹) See, e.g., Arrhenius, Theorien d. Chemie. Thomson, Corpuscular Theory of the Structure of Matter. Also Abegg’s views (Handbuch d. Anorg. Ch.). These ideas were first broadly developed and applied to the teaching of inorganic and organic chemistry by Prof. Ya. I. Mikhailenko (Tomsk, from 1908). The works of Berkenheim also belong here (electron chemistry of organic compounds, Zh. R. F. Kh. O., 1917).

²) Such calculations, in the case when the molecule contains more than one atom of the element, do not always give correct results. For example, for Na₂S₂O₃, by calculation we would find the valence of sulfur equal to +2. From the structural formula

\[ \mathrm{Na}\!\cdot\!\mathrm{O}^{\prime\prime} \backslash \mathrm{S}^{\prime\prime} - \mathrm{S} \begin{array}{c} =\mathrm{O}^{\prime\prime}\\ =\mathrm{O}^{\prime\prime} \end{array} \]

it is found that the valence of one of the sulfur atoms is +6, and of the other −2. The number found by calculation (+2) is the arithmetical mean of the valences of the two S atoms,

\[ \left(\frac{+6-2}{2}\right). \]

Thus, in order to oxidize any atom, i.e., to take electrons away from it, it is necessary to act upon it with an oxidizing agent—a substance capable of attaching electrons to itself. Conversely, in order to reduce, i.e., to attach electrons, a reducing agent is needed—a substance capable of giving up electrons.

In the following table (1) the valences and types of compounds are compared for those elements which can change their valence (only the more frequently encountered ones). Of course, in addition, each element may also have zero valence (i.e., be in the free state).

Table 1.

Elements with variable valence and types of their compounds.

The views set forth make it possible to deal easily with any, even the most complex, oxidation-reduction reaction, quickly to ascertain its essence, and to find the coefficients for the reaction equation. For example, it is known that $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ in acid-

medium (say, with $\mathrm{H_2SO_4}$) oxidizes $\mathrm{FeSO_4}$. In this process, of course, $\mathrm{K\cdot}$, $\mathrm{O'}$ and $\mathrm{S^{VI}}$ do not change valence. There remains $\mathrm{Cr^{VI}}$, which can pass into $\mathrm{Cr^{\cdots}}$ and, with the only acid present—$\mathrm{H_2SO_4}$—give the salt $\mathrm{Cr_2(SO_4)_3}$, while $\mathrm{Fe^{\cdot\cdot}}$ can pass into $\mathrm{Fe^{\cdots}}$ and give the salt $\mathrm{Fe_2(SO_4)_3}$. Since no chromic acid has been formed, $\mathrm{K\cdot}$ with sulfuric acid will give $\mathrm{K_2SO_4}$. In addition, $\mathrm{H\cdot}$ of sulfuric acid will give $\mathrm{H_2O}$. Thus:

\[ \mathrm{K_2|Cr_2O_7'' + Fe^{\cdot\cdot}|SO_4'' + H_2\cdot|SO_4'' \longrightarrow Cr_2^{\cdots}|(SO_4)_3'' + Fe_2^{\cdots}|(SO_4)_3'' + K_2|SO_4 + H_2O.} \]

To find the coefficients and finally write the equation, let us isolate the essence of the reaction, writing out those atoms that change valence:

\[ \begin{array}{rcl|cc} \text{Oxidation} &—& \mathrm{Fe^{\cdot\cdot} - (') = Fe^{\cdots}} & 3 & 6\\ \text{Reduction} &—& \mathrm{Cr^{VI} + 3(') = Cr^{\cdots}} & 1 & 2 \end{array} \]

To $\mathrm{Cr^{VI}}$ there are attached $3(')$, while from $\mathrm{Fe^{\cdot\cdot}}$ only $1(')$ is removed; but since the number of removed $(')$ must be equal to the number of attached $(')$, it is clear that for 1 atom of $\mathrm{Cr^{VI}}$ one needs 3 $\mathrm{Fe^{\cdot\cdot}}$; consequently, for a molecule of $\mathrm{K_2Cr_2O_7}$ (in which there are 2 $\mathrm{Cr^{VI}}$) one needs 6 $\mathrm{Fe^{\cdot\cdot}}$. Thus, the coefficient at $\mathrm{K_2Cr_2O_7}$ is 1, and at $\mathrm{FeSO_4}$ is 6. To find the coefficient at $\mathrm{H_2SO_4}$, the following method is possible. When $\mathrm{Cr^{VI}}$ passes into $\mathrm{Cr^{\cdots}}$, the molecule $\mathrm{K_2Cr_2O_7}$ decomposes. There are liberated $7\mathrm{O'}$, which must be bound with $14\mathrm{H\cdot}$ into $7\mathrm{H_2O}$. Consequently, one needs $14\mathrm{H\cdot}$ or $7\mathrm{H_2SO_4}$ (often another method proves more convenient: counting and equalizing, in both parts of the equation, the number of acid anions, in the present case the groups $\mathrm{SO_4''}$). Knowing the coefficients of the left-hand side of the equation, we shall easily write the right-hand side as well:

\[ \mathrm{K_2|Cr_2O_7'' + 6Fe^{\cdot\cdot}|SO_4'' + 7H_2\cdot|SO_4'' = Cr_2^{\cdots}|(SO_4)_3'' + 3Fe_2^{\cdots}(SO_4)_3'' + K_2|SO_4'' + 7H_2O.} \]

In this way any reaction can be analyzed.¹

Examples:

\[ \mathrm{Zn^0 + Cl_2^0 = Zn^{\cdot\cdot}|Cl_2'} \qquad \mathrm{Zn^0 + 2H\cdot|Cl' = Zn^{\cdot\cdot}|Cl_2' + H_2^0} \]

\[ \begin{array}{rcl|c} \mathrm{Zn^0 - 2(')} &=& \mathrm{Zn^{\cdot\cdot}} & 1\\ \mathrm{Cl^0 + (')} &=& \mathrm{Cl'} & 2 \end{array} \qquad \begin{array}{rcl|c} \mathrm{Zn^0 - 2(')} &=& \mathrm{Zn^{\cdot\cdot}} & 1\\ \mathrm{H\cdot + (')} &=& \mathrm{H^0} & 2 \end{array} \]

\[ \mathrm{(Cl_2^0\text{—oxidizing agent}).} \qquad \mathrm{(H\cdot\text{—oxidizing agent}).} \]

\[ \mathrm{2Fe^{\cdots}|Cl_3' + H_2\cdot|S'' = 2Fe^{\cdot\cdot}|Cl_2' + 2H\cdot|Cl' + S^0} \]

\[ \begin{array}{rcl|c} \mathrm{S'' - 2(')} &=& \mathrm{S^0} & 1\\ \mathrm{Fe^{\cdots} + (')} &=& \mathrm{Fe^{\cdot\cdot}} & 2 \end{array} \]

\[ \mathrm{Na_3\cdot|CrO_3''' + 3Br^0 + 2NaOH = Na_2CrO_4 + 3NaBr + H_2O} \]

\[ \begin{array}{rcl|c} \mathrm{Cr^{\cdots} - 3(')} &=& \mathrm{Cr^{VI}} & 1\\ \mathrm{Br^0 + (')} &=& \mathrm{Br'} & 3 \end{array} \]

\[ \mathrm{Na_2Cr_2O_7 + 3H_2SO_3 + H_2SO_4 = Na_2SO_4 + Cr_2(SO_4)_3 + 4H_2O} \]

\[ \begin{array}{rcl|c} \mathrm{S^{IV} - 2(')} &=& \mathrm{S^{VI}} & 3\\ \mathrm{Cr^{VI} + 3(')} &=& \mathrm{Cr^{\cdots}} & 2 \end{array} \]

The method we have set forth makes it possible to clarify the essence of a reaction and quickly select the coefficients in its equation. But the fundamental question—whether the reaction will proceed or not—remains unanswered. This must be known in advance. The purpose of the present article is to draw attention precisely to this aspect—to the determination of the direction of chemical oxidation-reduction reactions. As for exchange reactions, a similar question is sufficiently clarified there from the point of view of the law of mass action and the theory of electrolytic dissociation. The question of the direction of oxidation-reduction reactions is inseparably connected with the question of the influence of

the reaction of the medium, since very often in an acidic medium the direction of a reaction is different from its direction in an alkaline medium. For example, in an acidic medium \(K_2Cr_2O_7\) oxidizes \(Br'\) to \(Br^0\), whereas in an alkaline medium, on the contrary, \(Br^0\) oxidizes \(Cr\ldots\) to \(Cr^{VI}\). Depending on the reaction of the medium, many compounds prove unstable: decomposition occurs, the essence of which is an oxidation–reduction reaction. For example, salts of manganous acid, such as \(K_2MnO_4\) (green in color), are stable in an alkaline medium, but upon acidification they liberate \(MnO_2\) and give raspberry-colored \(HMnO_4\) in solution, i.e. the reaction occurs:

\[ 3H_2MnO_4 = MnO_2 + 2HMnO_4 + 2H_2O \]

or:

\[ \left\{ \begin{array}{l} Mn^{VI} + 2(\ ) = Mn^{IV}\ \bigm| 1\\ Mn^{VI} - 1(\ ) = Mn^{VII}\ \bigm| 2 \end{array} \right., \]

i.e. part of \(Mn^{VI}\) is reduced to \(Mn^{IV}\), while another part is oxidized to \(Mn^{VII}\). What is the matter here?

Thus, it is desirable to clarify the direction of oxidation–reduction reactions, the influence on their course of the reaction of the medium, and also the conditions of stability and instability of individual compounds.

Of themselves, spontaneously, only those processes proceed in which work can be obtained (the second principle of thermodynamics). In oxidation–reduction reactions, indeed, work can also be obtained; it is especially convenient to obtain and measure it in the form of electrical energy in so-called oxidation–reduction cells.

Let us take two small glass beakers \(A\) and \(B\), filled with a solution of an electrolyte, for example \(H_2SO_4\). Through the bottom of each of them pass platinum wires, ending in electrodes \(a\) and \(b\). The beakers are connected with one another by siphon \(D\), filled with the same solution of \(H_2SO_4\). Electrodes \(a\) and \(b\) are connected to voltmeter \(C\). There is, of course, no current. But if on electrode \(a\) one places some oxidizing agent, for example a crystal of \(K_2Cr_2O_7\) (i.e. \(Cr^{VI}\)), and on electrode \(b\) a reducing agent, for example \(FeSO_4\) (i.e. \(Fe^{\bullet\bullet}\)), then a current will flow from \(a\) to \(b\), with \(Cr^{VI}\) being converted into \(Cr\ldots\), and \(Fe^{\bullet\bullet}\) into \(Fe\ldots\). The occurrence of the current may be imagined in this way: \(Cr^{VI}\), with a certain energy, strives to attach electrons to itself and pass into \(Cr\ldots\); by taking electrons from electrode \(a\), it charges it positively; conversely, \(Fe^{\bullet\bullet}\) readily gives off 1 \((\ )\), charges electrode \(b\) negatively, and is converted into \(Fe\ldots\). The intensity of the reduction reaction of \(Cr^{VI}\) and oxidation of \(Fe^{\bullet\bullet}\) is expressed by the difference of potentials of electrodes \(a\) and \(b\).

The action of “gas” cells may be represented in the same way, for example the oxygen–hydrogen cell, where near one of the platinum electrodes there is \(H_2\) gas, and near the other \(O_2\). Here \(H^0\), passing into \(H^\bullet\), gives up electrons. They go from the hydrogen electrode to the oxygen electrode and here charge negatively the atoms \(O^0\), which are converted into \(O''\). The electromotive force here expresses the sum of the intensities of the oxidation–reduction reactions. The ordinary Daniell cell \(Cu^0|Cu^{\bullet\bullet}|SO_4, Zn^{\bullet\bullet}|SO_4|Zn^0\) acts in the same way.

In order to give a numerical value of the potential for each separate electrode, it is necessary to adopt the potential of some electrode as fundamental. The electrode \(H_2^0|H^\bullet\)—hydrogen at a pressure of 1 atm. with respect to an acid solution with a concentration of \(H^\bullet\) ions equal to \(1N\), at a temperature of \(18^\circ\) (room temperature)—is taken as such. This potential is considered \(=0\). All the others are calculated with respect to it. Thus, they say that the potential \(Mn^{VII}|Mn^{IV}\) is \(+1.63\). This means that it is greater than the potential \(H_2|H^\bullet\), and in combination with it gives a cell with an electromotive force \(=+1.63\). In such a combination the electrode with \(Mn^{VII}\) is charged \(+\), i.e. \(Mn^{VII}\) will attach electrons, turning into \(Mn^{IV}\), i.e. being reduced, while \(H_2\) will be oxidized to \(H^\bullet\). The potential \(Zn^{\bullet\bullet}|Zn=-0.76\).

Here the phenomenon is the reverse: the electrode with Zn is negative with respect to the hydrogen electrode, i.e., here \(\mathrm{Zn}^{\bullet\bullet}\) cannot take electrons from \(\mathrm{H}_2\), but, on the contrary, \(\mathrm{Zn}^0\) gives up electrons, is transformed into \(\mathrm{Zn}^{\bullet\bullet}\); in so doing the zinc electrode is charged negatively, while the electrons flow through the external circuit to the hydrogen electrode and reduce \(\mathrm{H}^{\bullet}\) to \(\mathrm{H}_2\). Here \(\mathrm{H}^{\bullet}\) is the oxidizer with respect to \(\mathrm{Zn}^0\). The electromotive force expresses the strength of chemical affinity only in the case when all the free chemical energy is converted into electrical energy, when there is no “leakage”—conversion of part of this energy into heat. The guarantee is the “reversibility” of the process: the chemical reaction gives electrical energy—current; if there are no energy losses, then, when a current of the same magnitude is passed through the element in the reverse direction, i.e. when the energy obtained from the element is returned, the reaction must proceed backward and the initial starting substances must again be obtained from the reaction products. This condition is satisfied, for example, by the gas \(\mathrm{O}_2 \mid \mathrm{H}_2\) element, the Daniell element, and a number of others. But for many oxidizers it has not yet been possible to construct a reversible pair and reliably determine the e.m.f. (for example, for \(\mathrm{HNO}_3\), \(\mathrm{HNO}_2\), \(\mathrm{H}_2\mathrm{SO}_4\), etc.).¹

With temperature the e.m.f. changes, but, comparatively, weakly (a change in temperature affects the rate of chemical reactions much more strongly). Here we shall consider only the course of reactions at ordinary room temperature \((18^\circ\mathrm{C})\) or at a temperature differing little from it.

The influence of concentration is clarified by the following considerations. Suppose that a substance \(A\) (higher degree of oxidation) is transformed into \(B\) (lower degree of oxidation): \(A \longrightarrow B\). At normal \((1N)\) concentrations of \(A\) and \(B\), this chemical process gives the “normal” potential \(E_0\). If the concentration of the oxidizer \(A\) is greater, for example \(C_A\), then in passing from concentration \(C_A\) to concentration \(1N\) we could obtain additional work of the osmotic forces (analogous to the work of an expanding gas). Consequently, the potential will be greater than for concentration \(1N\). Conversely, if the concentration of the reaction product (of the lower degree of oxidation)—\(B\)—is greater, for example \(C_B\), then, in comparison with the process at normal concentrations, the potential will be smaller, since here the reaction product must be brought from the normal concentration to the higher one—\(C_B\), which is analogous to compression of a gas and, consequently, requires expenditure of energy.²

\[ > EnF = E_0 nF + P > \]

or

\[ > E = E_0 + \frac{1}{nF}\,P. > \]

The work of the osmotic forces in the transition of 1 gram-molecule of the higher oxidation state \(A\) from concentration \(C_A\) to concentration \(=1\) (analogously to the work of expansion of a gas \(= RT\ln \frac{P_1}{P_2}\)) is equal to: \(RT\ln C_A\). For the lower oxidation state—for the work of transition from concentration \(=1\) to concentration \(=C_B\)—we obtain the same expression, but with the opposite sign, \(-RT\ln C_B\). Assuming that \(m_A\) molecules of the higher state participate in the reaction and that \(m_B\) moles of the lower state are obtained, we find that the whole work of the osmotic forces is

\[ > P = RT\,(m_A \ln C_A - m_B \ln C_B). > \]

Substituting into the basic equation:

\[ > E = E_0 + \frac{RT}{nF}\,(m_A \ln C_A - m_B \ln C_B). > \]

But \(R = 8.317\) joules, \(T = 290^\circ\) (for room temperature \(t = 17^\circ\mathrm{C}\)), \(F = 96540\); moreover, if one passes from natural logarithms to common logarithms, a factor \(2.303\) appears and the equation takes the form:

\[ > E = E_0 + \frac{8.317 \cdot 290 \cdot 2.303}{n \cdot 96540} > \times (m_A \lg C_A - m_B \lg C_B) > \]

or

\[ > E = E_0 + \frac{0.058}{n}\,(m_A \lg C_A - m_B \lg C_B). > \]

With ordinary changes in concentration—no more than by a factor of 10 (i.e., instead of 1 N, 10 N or 0.1 N)—this is reflected only weakly in the magnitude of the electromotive force (in hundredths of a volt). But a considerable deviation from the normal electromotive force may result if the initial substance or the reaction product proves to be “insoluble” (e.g., \(AgCl\)) or complex (e.g., \(Ag(NH_3)_2Cl'\)), i.e., when the concentration of our ion is extremely small (in the presence of a precipitate of \(AgCl\), a little \(Ag^+\) is found in solution, but its concentration is very small; likewise, together with the complex \(Ag(NH_3)_2\) there are always very few \(Ag^+\) ions). The following table shows how significantly the electromotive forces change in this case. However, all these deviations are intelligible and explicable from the standpoint set forth above concerning the influence of concentrations.

Let us first consider the influence of the reaction of the medium on the electromotive force for the simplest electrode—the hydrogen electrode. In an acid medium its electromotive force is taken as \(=0.00\). In an alkaline medium, the electromotive force in the conversion of \(2H^+\) into \(H_2\) is equal to \(-0.82\)—a negative value and, consequently, smaller than in an acid medium. This is understandable, since in an acid medium the concentration of the oxidizer \(H^+\) is sufficiently great (1 N), whereas in an alkaline medium these \(H^+\) ions are obtained only through the electrolytic dissociation

\[ H_2O \rightleftarrows H^+ + OH', \]

and in general in a very insignificant amount, and here this is further suppressed by the presence of \(OH'\) ions (alkaline medium). Thus, one may imagine that, for an alkaline medium as compared with an acid one, excess energy is spent on obtaining \(H^+\) ions from \(H_2O\) (more precisely—on bringing the \(H^+\) ions from the negligible concentration corresponding to the dissociation \(H \mid OH'\) in the presence of \(OH'\) up to concentration \(=1\) N). Consequently, the work which the reaction can furnish, and hence also the electromotive force for the alkaline medium, will be less¹.

The same takes place in all oxidation reactions for which \(H^+\) ions are required; and this usually occurs with oxygen-containing oxidizers, if they decompose while liberating \(O''\), which reacts with \(2H^+\) and gives \(H_2O\); consequently, it is precisely \(H^+\) ions that are required here.

Examples: 1) \(O_2 \mid 2O''\)—the oxygen electrode; in an acid medium, the \(2O''\) formed with \(4H^+\) give \(2H_2O\):

\[ 2O'' + 4H^+ = 2H_2O, \]

whereas in an alkaline medium energy must be expended in order to obtain the \(4H^+\) needed for this reaction from \(4H \mid OH'\). The electromotive force in an acid medium is \(+1.23\), while in an alkaline one it is \(+0.41\)—smaller. 2) \(Mn^{VII} \mid Mn^{VI}\), or, more explicitly,

\[ (MnO_4)' + 3(\cdot) = MnO_2 + 2O''; \]

then

\[ 2O'' + 4H^+ = 2H_2O; \]

in an acid medium there are ready \(H^+\) (with concentration 1 N), whereas in an alkaline medium, in comparison with the acid medium, additional work is required to obtain \(4H^+\) from \(4H \mid OH'\) (to bring \(H^+\) from the low concentration in an alkaline reaction to concentration \(=1\) N). The electromotive force in an acid medium is \(=+1.63\), and in an alkaline one \(=+0.52\)².

\[ \underset{\text{alk. med.}}{4F \cdot E_2} = \underset{\text{acid med.}}{4F \cdot E_1} -4F \cdot 0.82; \]

it is known that \(E_1=+1.23\); consequently, \(E_2=+0.41\), which corresponds to reality.

Thus all oxygen-containing oxidizing agents oxidize more vigorously in an acid medium, and oxygen-free oxidizing agents prove to be much less vigorous; for them \( \mathrm{H}^{\cdot} \) is of no significance, and therefore the electromotive force is the same in acid as in alkaline medium. Such are the haloids \((\mathrm{Cl}_2, \mathrm{Br}_2, \mathrm{J}_2)\); \(\mathrm{FeCy}_6^{\prime\prime\prime}\), passing into \(\mathrm{FeCy}_6^{\prime\prime\prime\prime}\) (it is often said that they oxidize more vigorously in alkaline medium; in reality their oxidizing ability is the same, but they seem stronger in comparison with oxygen-containing oxidizing agents, whose vigor is diminished for alkaline medium); the electromotive forces of the electrodes \(\mathrm{Na}^{\cdot}|\mathrm{Na}\), \(\mathrm{K}^{\cdot}|\mathrm{K}\), etc., likewise do not change as a function of the reaction of the medium.

The influence of the reaction of the medium is so considerable that it is necessary to compare and study the strength of oxidizing agents and oxidation reactions separately for acid and for alkaline medium. In Table II\(^1\) oxidizing agents are given with their “normal” (for \(18^\circ\mathrm{C}\), conc. \(1N\), and with respect to \(\mathrm{H}_2|\mathrm{H}^{\cdot}\)) electromotive forces, both for acid and (in a special column) for alkaline medium (found experimentally or calculated). In the first column stand the oxidizing agents themselves; in the second—what the oxidizing agents are converted into in oxidation reactions, by attaching electrons (lower oxidation states—“reducing agents”); in the third column—the corresponding electromotive forces. The oxidizing agents are arranged according to their electromotive forces in decreasing order. Thus, the first column represents a list of oxidizing agents arranged in decreasing order according to their strength. The second column may be regarded as a list of reducing agents arranged, conversely, in increasing order according to their strength. At the top, therefore, stand the strongest oxidizing agents (column I) and the weakest reducing agents (column II), and at the bottom—the weakest oxidizing agents (column I) and the strongest reducing agents (column II).

The table embraces and predicts all oxidation-reduction reactions possible with the oxidizing agents given in it. Let \(A\) and \(A_1\) be any two oxidizing agents (from column I), with \(A\) standing higher (and therefore being stronger) than \(A_1\); \(B\) and \(B_1\) the corresponding lower oxidation states. The following systems are possible: 1) \(A + A_1\)—a stable system; no reaction can be expected, since these are two oxidizing agents: both can take away electrons, but there is no element that could

For the electrode \(\mathrm{Mn}^{\mathrm{VII}}|\mathrm{Mn}^{\mathrm{IV}}\), where gr.-atom Mn in passing from 7-valent to 4-valent requires 3 F and where 4 \(\mathrm{H}^{\cdot}\) are needed:

\[ \begin{array}{c} \text{alkaline medium} \qquad \text{acid medium} \\ 3F \cdot E_2 = 3F \cdot E_1 - 4F \cdot 0.82; \qquad E_1 = +1.63 \quad (\text{known}); \end{array} \]

therefore:

\[ E_2 = +0.53 \quad (\text{in reality } +0.52), \text{ etc.} \]

However, such calculations are possible only in the case when in alkaline medium no side reactions occur that change the concentration (precipitation of precipitates, formation of complexes).

Of course, with a change in the degree of acidity or alkalinity (concentration of \(\mathrm{H}^{\cdot}\) or \(\mathrm{OH}^{\prime}\)) the electromotive force also changes. With an increase in the concentration of \(\mathrm{H}^{\cdot}\) \((C_{\mathrm{H}})\) it will be greater; with an increase in the concentration of \(\mathrm{OH}^{\prime}\) \((C_{\mathrm{OH}})\), which causes a decrease in the number of \(\mathrm{H}^{\cdot}\) in \(\mathrm{H}|\mathrm{OH}^{\prime}\), the electromotive force will be smaller. Mathematically, this dependence is easily derived in the same way as on p. 5 (influence of concentration). Thus, for acid medium, depending on the concentration of \(\mathrm{H}^{\cdot}\), there will appear the correction term

\[ +\frac{m}{n}\,0.058 \lg C_{\mathrm{H}^{\cdot}}, \]

and for alkaline medium

\[ -\frac{m}{n}\,0.058 \lg C_{\mathrm{OH}^{\prime}}, \]

where \(m\) is the number of ions \(\mathrm{H}^{\cdot}\) required for the reaction, and \(n\) is the number of electrons transferred. When the concentration of \(\mathrm{H}^{\cdot}\) or \(\mathrm{OH}^{\prime}\) is changed in one direction or the other by a factor of 10, the variations of the electromotive forces will be comparatively small, in hundredths of a volt. Thus it turns out that for each oxidizing agent two characteristic electromotive forces are obtained: one for acid and another for alkaline medium; when the concentration of the oxidizing agent is changed, as well as the degree of alkalinity or acidity, they change, but comparatively only a little. A neutral medium is not maintained, since as a result of the oxidation-reduction reaction either \(\mathrm{H}^{\cdot}\) or \(\mathrm{OH}^{\prime}\) accumulates. (It can be created artificially, for example, in the presence of \(\mathrm{Zn(OH)}_2\). It is also possible, by acidifying with a weak acid, e.g. acetic acid, in the presence of an excess of a salt of this same acid, e.g. \(\mathrm{NaCH}_3\mathrm{COO}\), to create an extremely small concentration of \(\mathrm{H}^{\cdot}\). Under these special conditions the electromotive forces will, of course, be intermediate between the forces for acid and for alkaline medium.)

\(^1\) The figures are chiefly from Abegg, Luther and Auerbach, Messungen d. elek.-motor. Kräfte, 1911. Part of the electromotive forces has been recalculated from acid medium to alkaline. Figures printed in italics are either the result of calculations from thermochemical data or in general are less reliably determined. For some oxidizing agents there are no corresponding electromotive forces, since they have not been determined. The position of such oxidizing agents in the table is only approximately indicated according to their reactions.

Electromotive force of oxidation-reduction electrodes (“normal”: relative to \(H_2 \mid H^\bullet\) at \(18^\circ\mathrm{C}\) and conc. \(1\,N\)).

Table II.

give them up. 2) \(B + B_1\)—also a stable system; there is no reaction, since these are two reducing agents—both can only give up electrons; there is no element that could accept them. 3) \(A_1 + B\)—a stable system; there is no reaction; although \(A_1\) can accept electrons and \(B\) can give them up, this will not occur, since the energy liberated in the transition of \(A_1\) into \(B_1\) (the addition of electrons to \(A_1\)) is less than that which must be expended in removing electrons from \(B\) (and converting it into \(A\)), and which is expressed by the magnitude of the e.m.f. corresponding to \(A\). Thus, oxidizing agents cannot oxidize the lower oxidation states corresponding to stronger oxidizing agents, or, more simply, an oxidizing agent does not react with reducing agents standing above it in the table. 4) \(A + B_1\)—the only case of an unstable system. \(A\), being the more energetic oxidizing agent, will, by accepting electrons, be converted into \(B\). The energy thereby liberated is quite sufficient (with an excess) to take electrons from \(B_1\) and convert it into \(A_1\) (a comparatively weak oxidizing agent). Thus, the oxidizing agents standing higher in the table react with the reducing agents standing lower (the direction is indicated by the arrow).

Any element appearing in this table will be converted from its lower oxidation state to a higher one (will be oxidized) by all the stronger oxidizing agents standing above it. Conversely, its higher oxidation state will itself be an oxidizing agent with respect to all the weaker oxidizing elements standing below it, which with respect to it will already prove to be reducing agents. Hence the complete relativity of the concepts of “oxidizing agents” and “reducing agents” is evident: an oxidizing agent cannot always oxidize, nor a reducing agent always reduce.

The order of arrangement of oxidizing agents in the table for an alkaline medium is different from that for an acid medium; therefore it is clear that the direction of reactions, as determined by the tabular order, is also different. Thus, in an acid medium, $\mathrm{Cr}^{\cdots}$ will be converted into $\mathrm{Cr}^{\mathrm{VI}}$ by means of $\mathrm{Mn}^{\mathrm{VII}}$ (i.e. $\mathrm{H}\cdot|\mathrm{MnO}_4'$), $\mathrm{Pb}^{\mathrm{IV}}(\mathrm{PbO}_2)$, $\mathrm{Mn}^{\mathrm{IV}}(\mathrm{MnO}_2)$, but $\mathrm{Br}'$, $\mathrm{J}$, $\mathrm{FeCy}_6^{\prime\prime}$ will not oxidize $\mathrm{Cr}^{\cdots}$; on the contrary, $\mathrm{Cr}^{\mathrm{VI}}$, passing into $\mathrm{Cr}^{\cdots}$, will convert (oxidize) $\mathrm{Br}'$ into $\mathrm{Br}^0$, $\mathrm{J}'$ into $\mathrm{J}^0$, $\mathrm{FeCy}_6^{\prime\prime\prime\prime}$ into $\mathrm{FeCy}_6^{\prime\prime\prime}$. For an alkaline medium the order proves to be different. In an acid medium the strongest oxidizing agents are oxygen-containing ones, such as $\mathrm{Mn}^{\mathrm{VII}}(\mathrm{H}\cdot|\mathrm{MnO}_4')$, $\mathrm{Pb}^{\mathrm{IV}}(\mathrm{PbO}_2)$, $\mathrm{Mn}^{\mathrm{IV}}(\mathrm{MnO}_2)$, $\mathrm{Cr}^{\mathrm{VI}}(\mathrm{H}_2|\mathrm{Cr}_2\mathrm{O}_7^{\prime\prime})$. In an alkaline medium their potential is much lower, and the strongest oxidizing agents prove to be the halides $\mathrm{Cl}$, $\mathrm{Br}$, $\mathrm{J}$, only then $\mathrm{Mn}^{\mathrm{VII}}$, while $\mathrm{Cr}^{\mathrm{VI}}$ stands quite low. Thus in an alkaline medium it is easy to oxidize $\mathrm{Cr}^{\cdots}$ to $\mathrm{Cr}^{\mathrm{VI}}$. This can be done by means of $\mathrm{Cl}$, $\mathrm{Br}$, $\mathrm{J}$, $\mathrm{Mn}^{\mathrm{VII}}$, $\mathrm{FeCy}_6^{\prime\prime\prime}$, $\mathrm{Pb}^{\mathrm{IV}}$. Here $\mathrm{Cr}^{\mathrm{VI}}$ is, comparatively, a weak oxidizing agent; nevertheless it will oxidize $\mathrm{Fe}^{\cdots}$, $\mathrm{As}^{\cdots}$, $\mathrm{Sn}^{\cdot\cdot}$, $\mathrm{S}^{\prime\prime}$, $\mathrm{S}^{\mathrm{IV}}$.

The table also explains and predicts a number of interesting reactions. For example, it is known that $\mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{S} = 2\mathrm{H}_2\mathrm{O} + 3\mathrm{S}$; this is understandable, since in an acid medium $\mathrm{S}^{\mathrm{IV}}$ with respect to $\mathrm{S}^{\prime\prime}$ is an oxidizing agent ($\mathrm{S}^{\mathrm{IV}}|\mathrm{S}^0$ stands higher than $\mathrm{S}^0|\mathrm{S}^{\prime\prime}$); oxidation of $\mathrm{S}^{\prime\prime}$ to $\mathrm{S}^0$ and reduction of $\mathrm{S}^{\mathrm{IV}}$ likewise to $\mathrm{S}^0$ take place. Conversely, in an alkaline medium the electromotive force of $\mathrm{S}^{\mathrm{IV}}|\mathrm{S}^0$ is lowered (an oxygen-containing oxidizing agent) and proves to be lower than the electromotive force of $\mathrm{S}^0|\mathrm{S}^{\prime\prime}$; consequently, here $\mathrm{S}^0$, passing into $\mathrm{S}^{\prime\prime}$, will oxidize another atom of $\mathrm{S}^0$ to $\mathrm{S}^{\mathrm{IV}}$, i.e. a reaction will occur that may be called a reaction of self-oxidation:

$$ 3\mathrm{S}^0 + 6\mathrm{KOH} = 2\mathrm{K}_2\mathrm{S} + \mathrm{K}_2\mathrm{SO}_3 + 3\mathrm{H}_2\mathrm{O} $$

(in addition, the reaction will go still farther: $\mathrm{K}_2\mathrm{SO}_3 + \mathrm{S} = \mathrm{K}_2\mathrm{S}_2\mathrm{O}_3$; this is also explicable: in $\mathrm{K}_2\mathrm{S}_2\mathrm{O}_3$ the atoms of $\mathrm{S}$ have different valencies—one is $\mathrm{S}^{\prime\prime}$, the other $\mathrm{S}^{\mathrm{VI}}$; consequently, the substance reacts

$$ \begin{vmatrix} \mathrm{S}^0 + 2() = \mathrm{S}^{\prime\prime} \\ \mathrm{S}^{\mathrm{IV}} - 2() = \mathrm{S}^{\mathrm{VI}} \end{vmatrix}; $$

and this is understandable, since the electromotive force of $\mathrm{S}^0|\mathrm{S}^{\prime\prime}$ in an alkaline medium is greater than the electromotive force of $\mathrm{S}^{\mathrm{VI}}|\mathrm{S}^{\mathrm{IV}}$). The reaction of self-oxidation that occurs when $\mathrm{Cl}_2$ is passed into alkali is entirely analogous and understandable:

$$ 2\mathrm{K}'|\mathrm{OH}' + \mathrm{Cl}_2 = \mathrm{K}'|\mathrm{Cl}' + \mathrm{K}'|\mathrm{OCl}' + \mathrm{H}_2\mathrm{O}; $$

this reaction is possible because in an alkaline medium the electromotive force of $\mathrm{Cl}^0|\mathrm{Cl}'$ is greater than the electromotive force of $\mathrm{Cl}'|\mathrm{Cl}^0$. In an acid medium the reverse reaction occurs:

$$ \mathrm{H}\cdot|\mathrm{Cl}' + \mathrm{H}\cdot|\mathrm{OCl} = \mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O}, $$

since here, conversely, the electromotive force of $\mathrm{Cl}'|\mathrm{Cl}^0$ is greater than the electromotive force of $\mathrm{Cl}^0|\mathrm{Cl}'$.

The reactions of $\mathrm{S}^0$ in an alkaline medium and of $\mathrm{Cl}^0$ in an alkaline medium can be generalized. Here we are dealing with three oxidation states: for sulfur—$\mathrm{S}^{\mathrm{IV}}$, $\mathrm{S}^0$, and $\mathrm{S}^{\prime\prime}$, and for chlorine—$\mathrm{Cl}'$, $\mathrm{Cl}$, and $\mathrm{Cl}'$. If the potential in passing from the first state to the second ($\mathrm{S}^{\mathrm{IV}}|\mathrm{S}^0$ and $\mathrm{Cl}'|\mathrm{Cl}^0$) is greater than in passing from the second state to the third ($\mathrm{S}^0|\mathrm{S}^{\prime\prime}$ and $\mathrm{Cl}\cdot|\mathrm{Cl}'$), then the intermediate state ($\mathrm{S}^0$ and $\mathrm{Cl}^0$) is stable; the mixture of the first and third oxidation states is unstable (it is converted into the second state). Such is the situation for $\mathrm{S}^0$ and $\mathrm{Cl}^0$ in an acid medium (in general form the scheme is given in the accompanying table, where the oxidation states are denoted by Roman numerals: I—the highest, II—the middle, and III—the lowest).

$$ \left| \begin{array}{c} \mathrm{I}—\mathrm{II} \\ \searrow \\ \mathrm{II}—\mathrm{III} \end{array} \right| $$

If, conversely, the potential $\mathrm{II}|\mathrm{III}$ is higher than the potential $\mathrm{I}|\mathrm{II}$ (see the second scheme), then the intermediate oxidation state (II) will be unstable: it decomposes into the higher (I) and lower (III), and self-oxidation–self-reduction will occur (such is the relation of $\mathrm{S}^0$ in an alkaline medium and $\mathrm{Cl}^0$ in an alkaline medium).

$$ \left| \begin{array}{c} \mathrm{II}—\mathrm{III} \\ \searrow \\ \mathrm{I}—\mathrm{II} \end{array} \right| $$

From this point of view the instability of \(H_2MnO_4\) in an acidic medium is understandable \((3H_2MnO_4 = 2HMnO_4 + MnO_2 + 2H_2O)\), as is that of \(HNO_2\) in an acidic medium \((3HNO_2 = HNO_3 + 2NO + H_2O)\), and of \(KClO\), both in acidic and in alkaline media \((3KClO = 2KCl + KClO_3)\). In all these cases there are likewise three degrees of oxidation of one and the same element. Although the electromotive forces needed by us here have not all been determined \(^{1}\) (there are none for \(Mn^{VI}\) and \(N\)), we may assert that the electromotive forces for \(Mn^{VII}\,|\,Mn^{VI}\), \(N^{\cdots}\,|\,N^{\cdot}\), \(Cl^{\cdot}\,|\,Cl'\) for an acidic medium (and, for the last of these, also for an alkaline medium) are respectively greater than the electromotive forces \(Mn^{VII}\,|\,Mn^{VI}\), \(N^V\,|\,N^{\cdots}\), and \(Cl^V\,|\,Cl^{\cdot}\). Chemical reactions agree with such a statement: for example, \(HNO_2\) often proves to be a stronger oxidizing agent than \(HNO_3\) (for example, the liberation of \(J^0\) and \(Br^0\) from \(HJ\) and \(HBr\)). The stability of \(Mn^{VI}\) \((K_2MnO_4)\) and \(N^{\cdots}\) \((KNO_2)\) in an alkaline medium is also understandable, since by calculation (analogously to p. 6) one can be convinced that the potentials \(Mn^{VI}\,|\,Mn^{IV}\) and \(N^{\cdots}\,|\,N^{\cdot}\) decrease in an alkaline medium, as compared with an acidic one, more than the potentials \(Mn^{VII}\,|\,Mn^{VI}\) and \(N^V\,|\,N^{\cdots}\) decrease, and probably, as a result, become less than the latter, which is the condition for the stability of the intermediate degrees of oxidation \((Mn^{VI}\) and \(N^{\cdots})\). For the pairs \(Cl^{\cdot}\,|\,Cl'\) and \(Cl^V\,|\,Cl^{\cdot}\), on passing to an alkaline medium, the lowering of the electromotive forces, and consequently also the stability of the intermediate degree of oxidation \((Cl^{\cdot})\), are the same in acidic and in alkaline media; this is indeed observed experimentally.

Thus, by using the table, one can determine the direction of oxidation-reduction reactions, can take into account the influence of the reaction medium, and can clarify for oneself—and sometimes even predict—the stability or instability of certain compounds. In studying the reactions of any new oxidizing agent, it is necessary only, either from its electromotive force or from some of its reactions, to find its place in the table. Then all its further relations become clear.

In using the table, however, the following remarks must be borne in mind.

1) It is suitable only for temperatures close to room temperature \((18^\circ C)\).

2) If the potentials are close to one another (for example, \(Mn^{VII}\,|\,Mn^{\cdot\cdot} — +1.52\) and \(Pb^{IV}\,|\,Pb^{\cdot\cdot} — +1.44\)), then a chemical equilibrium results; depending on the ratio of concentrations, the reaction may proceed in one direction or the other.

3) For many oxidizing agents the electromotive forces have not yet been determined, and their positions in the table are indicated only approximately \((N^V,\ Sn^{IV},\ S^{IV},\ S^{VI})\).

4) Table II indicates only the possibility of reactions, but it says nothing about their rate, which, according to Nernst, is

\[ = V = \frac{E}{R} = \]

the difference of the chemical potential divided by some resistance \(R\), which has so far been completely unstudied. Sometimes \(R\) is very large; the reaction proceeds infinitely slowly and can be carried out only if a corresponding catalyst is found (this applies, it is true, chiefly only to reactions in which the gases \(O_2\) and \(H_2\) participate, and also to almost all oxidation-reduction reactions in organic chemistry). Even in this case Table II may sometimes be useful: it indicates whether the given reaction is conceivable and whether it makes sense to look for a catalyst for carrying it out.

\(^{1}\) The same, properly speaking, must also be said with regard to \(S^{IV}\) (see above), but its position in the table could be indicated approximately, on the basis of chemical relations.

1. Work of electricity \(=\) its quantity \(\times\) e.m.f. Let us denote by \(F\) the quantity of electricity corresponding to the charge of 1 gram-ion \(H^+\) (\(F=96400\) coul.). The work for an acid medium (per 1 gram-molecule \(H_2\)) is equal to \(2FE_1 = 2F \cdot 0.00\); for an alkaline medium the work is \(=2FE_2 = 2F(-0.82)\); but the work for an alkaline medium equals the work for an acid medium minus the work of transferring \(H^+\) from the low concentration in the alkaline medium to concentration 1 N (acid medium); let us denote this work for \(1H^+\) by \(A\). Then: \(-2F \cdot 0.82 = 2F \cdot 0.00 - 2A\); consequently, \(A=+F \cdot 0.82\). ↩↩↩

2. Knowing the e.m.f. for an acid medium, one may calculate it also for an alkaline one, using the equation: work for the alkaline medium \(=\) work for the acid medium minus the work expended on transferring \(H\) from the concentration in an alkaline medium to the concentration in an acid one (i.e., \(A=+F \cdot 0.82\)—for normal concentrations of alkali and acid, as was already explained in the preceding note). For example, for the oxygen electrode, where for 1 gram-molecule \(O_2\) 4 \(F\) of electricity and 4 \(H^+\) are required, we obtain: ↩↩