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It has long been known that some metals are more "active" than others in the sense that a more active metal can "displace" a less active one from a solution of its salt. The classic example is the one we have already mentioned on the preceding page:
Zn(s) + Cu2+ → Zn2+ + Cu(s)
Here zinc is more active because it can displace (precipitate) copper from solution. If you immerse a piece of metallic zinc in a solution of copper sulfate, the surface of the zinc quickly becomes covered with a black coating of finely-divided copper, and the blue color of the hydrated copper(II) ion diminishes.
Similar comparisons of other metals made it possible to arrange them in the order of their increasing electron-donating (reducing) power. This sequence became known as the electromotive or activity series of the metals.
The most active (most strongly reducing) metals appear on top, and least active metals appear on the bottom. A more active metal (such as Zn) will donate electrons to the cation of a less active metal (Cu2+, for example.)
Notice the special role of hydrogen here; although H2 does not have the physical properties of a metal, it is capable of being "displaced" (a rather archaic term seldom used in modern chemistry) from H2O or H+-containing (acidic) solutions. Note that the "active" metals are all "attacked by acids"; what this really means is that they are capable of donating electrons to H+.
(This table was adapted from a now-disappeared one at General Chemistry Online .)
The activity series has long been used to predict the direction of oxidation-reduction reactions; see here for a nicely-done table with explanatory material. Consider, for example, the oxidation of Cu by metallic zinc that we have mentioned previously. The fact that zinc is near the top of the activity series means that this metal has a strong tendency to lose electrons. By the same token, the tendency of Zn to accept electrons is relatively small. Copper, on the other hand, is a poorer electron donor, and thus its oxidized form, Cu, is a fairly good electron acceptor. We would therefore expect the reaction
Zn(s) + Cu2+ → Zn2+ + Cu(s)
to proceed in the direction indicated, rather than in the reverse direction. An old-fashioned way of expressing this is to say that "zinc will displace copper from solution".
The above table is of limited practical use because it does not take into account the concentrations of the dissolved species. In order to treat these reactions quantitatively, it is convenient to consider the oxidation and reduction steps separately.
When a net reaction proceeds in an electrochemical cell, oxidation occurs at one electrode (the anode) and reduction takes place at the other electrode (the cathode.) We can think of the cell as consisting of two half-cells joined together by an external circuit through which electrons flow and an internal pathway that allows ions to migrate between them so as to preserve electroneutrality.
Each half-cell has associated with it an electrode-solution potential difference whose magnitude depends on the nature of the particular electrode reaction and on the concentrations of the dissolved electroactive species. The sign of this potential difference depends on the direction (oxidation or reduction) in which the electrode reaction proceeds. In order express them in a uniform way, we adopt the convention that half-cell potentials are always defined for the reduction direction. Thus the half-cell potential for the Zn/Zn2+ electrode (or couple as it is sometimes called) always refers to the reduction reaction
Zn2+ + 2e– → Zn(s)
In the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) the zinc appears on the left side, indicating that it is being oxidized, not reduced. For this reason, the potential difference contributed by the left half-cell has the opposite sign to its conventional half-cell potential. More generally, we can define the cell potential or cell EMF as
|Ecell = ΔV = Eright – Eleft||
in which "right" and "left" refer to the cell notation convention ("reduction on the right") and not, of course, to the physical orientation of a real cell in the laboratory. If we expand the above expression we see that the cell potential
Ecell = VCu – Vsolution + Vsolution – VZn
is just the difference between the two half-cell potentials Eright and Eleft.
The fact that individual half-cell potentials are not directly measurable does not prevent us from defining and working with them. Although we cannot determine the absolute value of a half-cell potential, we can still measure its value in relation to the potentials of other half cells. In particular, if we adopt a reference half-cell whose potential is arbitrarily defined as zero, and measure the potentials of various other electrode systems against this reference cell, we are in effect measuring the half-cell potentials on a scale that is relative to the potential of the reference cell.
The reference cell that has universally been adopted for this purpose is the hydrogen half-cell
Pt | ½ H2(g) | H+(aq) || ...
in which hydrogen gas is allowed to bubble over a platinum electrode having a specially treated surface which catalyzes the reaction
½ H2(g) → H+ + e–
When this electrode is operated under standard conditions of 1 atm H2 pressure, 25°C, and pH = 0, it becomes the standard hydrogen electrode, sometimes abbreviated SHE.
In order to measure the relative potential of some other electrode couple M2+/M, we can set up a cell
Pt | H2(g) | H+ || M2+ (aq) | M(s)
whose net reaction is
H2(g) + M2+(aq) → 2H+ + M(s)
the potential difference between the platinum and M electrodes will be
Ecell = VM – Vsolution + Vsolution – V Pt
but since the difference Vsolution – V Pt is by definition zero for the hydrogen half-cell, the cell potential we measure corresponds to
Ecell = VM – Vsolution
which is just the potential (relative to that of the SHE) of the half-cell whose reaction is
M2+ + 2e– → M(s)
Measurement of a standard reduction potential.
The M2+/M half-cell is on the left, and the standard hydrogen cell is on the right. The two half-cells are joined through the salt bridge in the middle. The more "active" the metal M (the greater its tendency to donate electrons to H+), the more negative will be Ecell = ΔV = Eright – Eleft
Standard [reduction] potentials are commonly denoted by the symbol E°. E° values for hundreds of electrodes have been determined (mostly in the period 1925-45, during which time they were referred to as "oxidation potentials") and are usually tabulated in order of increasing tendency to accept electrons (increasing oxidizing power.)
Table 2: some standard reduction potentials
A much more extensive table can be found here. Note particularly that
Given the E° values for two half reactions, you can easily predict the potential difference of the corresponding cell: simply add the reduction potential of the reduction half-cell to the negative of the reduction potential (that is, to the oxidation potential) of the oxidation reaction.
Find the standard potential of the cell
Cu(s) | Cu2+ || Cl– | AgCl(s) | Ag(s)
and predict the direction of electron flow when the two electrodes are connected.
Solution: The above notation represents a cell in which metallic copper undergoes oxidation, delivering electrons to a reactant on the right, which gets reduced. But what species actually receives the electron? It cannot be Ag, because metallic elements do not form negative ions. The only reducible species on the right is the Ag+ contained within the insoluble salt AgCl. Reduction of this ion to metallic Ag is accompanied by a release of Cl– ions into the solution. The two half-reactions and their standard potentials are
|Cu (s) → Cu2+ (aq) + 2 e–||–(+0.337) v|
|AgCl (s) + e– → Ag (s) + Cl– (aq)||+0.222 v|
Note that because Cu is being oxidized (rather than reduced), we take the negative of the standard reduction potential.
Multiplying the second equation (but not its reduction potential) by 2 in order to balance electrons, we have the net reaction
2 AgCl (s) + Cu (s) → 2 Ag (s) + 2 Cl– (aq) + Cu2+ (aq)
Combining the two half-cell potentials shown above, the net cell potential is
Ecell = (–0.337 + 0.222) v = –0.115 v
Because this potential is negative, we know that the net cell reaction (and, of course, the two half reactions) proceed in the reverse directions to those depicted above. Thus when the two electrode are connected, electrons pass from the silver electrode through the external circuit to the copper electrode. Note that this corresponds to the rule in noted in Table 2 above: AgCl acts as an oxidizing agent (electron sink) to metallic copper which appears below it in the reductant column.
If you are wondering wny we did not multiply E° the for the Cu2+/Cu couple by two, the reason for this will be explained in the next section.
From the above, it should be apparent that the potential difference between the electrodes of a cell is a measure of the tendency for the cell reaction to take place: the more positive the cell potential, the greater the tendency for the reaction to proceed to the right. But we already know that the standard free energy change expresses the tendency for any kind of process to occur under the conditions of constant temperature and pressure. Thus ΔG° and E° measure the same thing, and are related in a simple way:
ΔG° = –nFE°
... or in more detail (see below for explanations of the units given for voltage)
A few remarks are in order about this very fundamental and important relation:
The negative sign on the right indicates that a positive cell potential (according to the sign convention discussed previously) implies a negative free energy change, and thus that the cell reaction will spontaneously proceed to the right.
Electrical work is done when an electric charge q moves through a potential difference ΔV. The right side of Eq. 2 refers to the movement of n moles of charge across the cell potential E°, and thus has the dimensions of work.
The value of ΔG° expresses the maximum useful work that a system can do on the surroundings. "Useful" work is that which can be extracted from the cell by electrical means to operate a lamp or some other external device. This excludes any P-V work that is simply a consequence of volume change (which could conceivably be put to some use!) and which would be performed in any case, even if the reactants were combined directly. This quantity of work –ΔG° can only be extracted from the system under the limiting conditions of a thermodynamically reversible change, which for an electrochemical cell implies zero current. The more rapidly the cell operates, the less electrical work it can supply.
1 J = 1 watt-sec = 1 (amp-sec) × volts
If Eq. 2 is solved for E°, we have
This states explicitly that the cell potential is a measure of the free energy change per mole of electrons transferred, which is a brief re-statement of the principle explained immediately above.
To see this more clearly, consider the cell
Cu(s) | Cu2+ || Cl– | AgCl(s) | Ag(s)
for which we list the standard reduction potentials and ΔG°s of the half-reactions:
cathode: 2 × [AgCl(s) + e– → Ag(s) + Cl–]
anode: Cu(s) → Cu2+ + 2 e–
net: 2 Ag(s) + 2 Cl–(aq) + Cu2+(aq) → AgCl(s) + Cu(s)
cell: Cu(s) | Cu2+(aq) || AgCl(s) | Cl–(aq) | Ag(s)
Here we multiply the cathodic reaction by two in order to balance the charge. Because the anodic reaction is written as an oxidation, we reverse the sign of its E° and obtain Ecell = Eright – Eleft = –.115 volt for the cell potential. The negative cell potential tells us that this reaction will not proceed spontaneously.
Note, however, that if we are combining two half reactions to obtain a third half reaction, the E° values are not additive, since this third half-reaction is not accompanied by another half reaction that causes the charges to cancel. Free energies are always additive, so we combine them, and use ΔG° = –nFE° to find the cell potential.
Calculate E° for the electrode Fe3+/Fe(s) from the standard potential of the couples Fe3+/Fe2+ and Fe2+/Fe(s)
Solution: Tabulate the values and calculate the ΔG°s as follows:
|(i) Fe3+ + e– → Fe2+||E°1 = .771 v , ΔG°1 = –.771 F|
|(ii) Fe2+ + 2 e– → Fe(s)||E°2= –.440 v , ΔG°2 = +.880 F|
|(iii) Fe3+ + 3 e– → Fe(s)||E°3 = ? , ΔG°3 = +.109 F|
A table of standard half-cell potentials summarizes a large amount of chemistry, for it expresses the relative powers of various substances to donate and accept electrons by listing reduction half-reactions in order of increasing E° values, and thus of increasing spontaneity. The greater the value of E°, the greater the tendency of the substance on the left to acquire electrons, and thus the stronger this substance is as an oxidizing agent.
If you have studied elementary chemical thermodynamics, you will have learned about the role that a quantity called the Gibbs free energy, usually referred to as simply the "free energy", plays in determining the direction of any chemical change. The rule is that all spontaneous change (that is, all reactions that proceed to the "right") is associated with a fall in the free energy, and the greater the degree of that fall (ΔG°), the greater will be the tendency for the reaction to take place.
If you are not familiar with the concept of free energy, just think of it as something like potential energy, which similarly decreases when spontaneous mechanical events occur, such as the dropping of a weight.
Since oxidation-reduction processes involve the transfer of an electron from a donor to an acceptor, it makes sense to focus on the electron and to consider that it falls from a higher-free energy environment (the reductant, or "source") to a lower-free energy one (the oxidant, or "sink".)
As can be seen from the diagram below, this model makes it far easier to predict what will happen when two or more oxidants and reducants are combined; the electron "falls" as far as it can, filling up oxidizing agents (sinks) from the bottom up, very much in the same way as electrons fill atomic orbitals as we build up larger atoms.
Electron-free energy diagram of redox couples
This chart is essentially an abbreviated form of a table of standard potentials in which the various couples are displayed on a vertical scale corresponding to
At this point, it might be worth calling your attention to the similar way of depicting acid-base reactions as representing the "fall of the proton" as shown below and described much more thoroughly here.
Proton-free energy diagram of acid-base systems
Acids are proton sources (donors), bases are proton sinks. Protons "fall" (in free energy) whenever a base is present that presents proton-empty free energy levels. The red arrows show what happens when acetic acid is titrated with a strong base; the results are acetate ion and water. Note here again the crucial role of water, both as a proton acceptor (forming hydronium ion) and as a proton donor (forming hydroxide ion.) Note also that the pH of a solution is a direct measure of the average free energy of protons in the solution (relative to H3O+.)
An important difference between proton transfer and electron transfer reactions is that the latter can vary greatly in speed, from almost instantaneous to so slow as to be unobservable. Acid-base reactions are among the fastest known.
Considerable insight into the chemistry of a single element can be had by comparing the standard electrode potentials (and thus the relative free energies) of the various oxidation states of the element. The most convenient means of doing this is the Latimer diagram. As an example, consider the Latimer diagram for iron:
The formulas of the species that represent each oxidation state of the element are written from left to right in order of decreasing oxidation number, and the standard potential for the reduction of each species to the next on the right is written in between the formulas. Potentials for reactions involving hydrogen ions will be pH dependent, so separate diagrams are usually provided for acidic and alkaline solutions (effective hydrogen ion concentrations of 1M and 10–14 M, respectively).
The more positive the reduction potential, the greater will be the tendency of the species on the left to be reduced to the one on the right. To see how Latimer diagrams are used, look first at the one for iron in acid solution. The line connecting Fe3+ and Fe2+ represents the reaction
Fe3+ + e– → Fe2+
whose positive E° (.440 v) indicates that metallic iron will dissolve in acidic solution to form Fe2+. Because the oxidation of this species to the +3 state has a negative potential (-.771v; moving to the left on the diagram reverses the sign), the +2 state will be the stable oxidation state of iron under these conditions.
This Latimer diagram for chlorine illustrates an important principle:
When the potential on the left of a species is less positive than that on the right. This indicates that the species can oxidize and reduce itself, a process known as disproportionation. As an example, consider Cl2 in alkaline solution. The potential for its reduction to Cl– is sufficiently positive (+1.36 v) to supply the free energy necessary for the oxidation of one atom of chlorine to hypochlorite. Thus elemental chlorine is thermodynamically unstable with respect to disproportionation in alkaline solution, and the same it true of the oxidation product, ClO– (hypochlorite ion).
Behavior of chlorine in water
Cl2 can oxidize water (green arrows, top) and also undergo disproportionation (purple arrows, bottom). In the latter process, one Cl2 molecule donates electrons to another.
Bear in mind that many oxidation-reduction reactions, unlike most acid-base reactions, tend to be very slow, so the fact that a species is thermodynamically unstable does not always mean that it will quickly decompose. Thus the two reactions shown in the figure are normally very slow.
The free energy change for a process represents the maximum amount of non-PV work that can be extracted from it. In the case of an electrochemical cell, this work is due to the flow of electrons through the potential difference between the two electrodes. Note, however, that as the rate of electron flow (i.e., the current) increases, the potential difference must decrease; if we short-circuit the cell by connecting the two electrodes with a conductor having negligible resistance, the potential difference is zero and no work will be done. The full amount of work can be realized only if the cell operates at an infinitesimal rate; that is, reversibly.
You should recall that this is exactly analogous to the expansion of an ideal gas. The full amount of work w = PdV is extracted only under the special condition that the external pressure P opposing expansion is only infinitesimally smaller than the pressure of the gas itself. If the gas is allowed to expand into a vacuum (P = 0), no work will be done.
The total amount of energy a reaction can supply under standard conditions at constant pressure and temperature is given by ΔH°. If the reaction takes place by combining the reactants directly (no cell) or in a short-circuited cell, no work is done and the heat released is ΔH. If the reaction takes place in a cell that performs electrical work, then the heat released is diminished by the amount of electrical work done. In the limit of reversible operation, the heat released becomes
ΔH = ΔG° + T ΔS
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
Pt | H2(g) | H+ || M2+ (aq) | M(s)
whose left half consists of a standard hydrogen electrode (SHE) and whose net reaction is
H2(g) + M2+(aq) → 2H+ + M(s)