Chemistry: Further Redox: Standard Electrode Potential

Standard Electrode (Redox) Potential

Revision of Oxidation and ReductionThere are two very useful definitions of oxidation and reduction. One is in terms of loss or gain of electrons. The other is in terms of changes to the oxidation number of an element. In terms of electrons:1. Oxidation is the loss of electrons2. Reduction is the gain of electronsIn terms of changes in oxidation number:1. An element in a species is oxidised when its oxidation number increases2. An element in a species is reduced when its oxidation number decreasesThese definitions apply to all elements whether they are in the s-, p-, or d-block of the Periodic Table. How well you understand the topic of electrode potentials depends very heavily on your appreciation of redox reactions.

Standard Electrode (Redox) PotentialThe two terms "standard electrode potential" and "standard redox potential" mean the same. The reactions involved in the measurement of a standard electrode potential are redox reactions. Hence the term redox potential is also commonly used. In line with the Edexcel specification, we will use the term "standard electrode potential" throughout this chapter. There are many ways of teaching this topic, but very often a fundamental mistake is made at the outset. It is sometimes forgotten to emphasise that the reactions involved are equilibria. Too often you will find that the equations are written as if they were one-way reactions, rather than reversible. This very simple mistake can make the whole topic much more difficult to understand than is necessary. Throughout the topic we shall avoid this problem by always talking in terms of equilibria.

Background: The Different Tendencies of Metals to Release Electrons to form Positive IonsWhen a metal such as magnesium or copper is placed in water there is a very small tendency for the metal atoms to lose electrons and go into solution as positive ions:M(s) -> (M)n+(aq) + ne- (where M = Mg or Cu)The electrons will remain on the surface of the metal. In a very short time there will be a build-up of electrons on the surface of the metal, and the resulting negative charge attracts positive ions. In this way, a layer of positive ions is formed surrounding the metal. Some of the positive ions will regain their electrons from the surface of the metal and return to form part of the metal surface. (M)n+(aq) + ne- = M(s)Eventually, a dynamic equilibrium will be established in which the rate at which ions are leaving the surface of the metal to go into solution is the same as the rate at which they are joining it from solution. This equilibrium is represented by the equation:(M)n+(aq) + ne- -><- M(s)The equations for magnesium and copper are:Mg2+ +(aq) + 2e- -><- Mg(s)Cu2+ +(aq) + 2e- -><- Cu(s)The difference between magnesium and copper is that the equilibrium position will be further to the left-hand side for magnesium than for copper, because the tendency for magnesium to release electrons is greater than that of copper. So, with magnesium there will be a greater negative charge on the metal, and more positive ions in solution. In the case of both magnesium and copper, there is a potential difference is greater with magnesium than with copper. In each case, we would like to measure the potential difference between the metal and the solution. This is called the absolute potential difference. However, it is not possible to measure the absolute potential difference between a metal electrode and its solution. The reason is that although it is a simple matter to connect the metal electrode to one terminal of a voltmeter, the other terminal would have to be connected to the solution, and the only way of doing this is to dip another piece of metal into the solution. This second piece of metal would create its own potential difference. So, you would be measuring the potential difference between the two piece of metal, and not between the original metal and the solution. The way to solve this problem is to create a reference electrode and then measure the difference in potential between this reference electrode and the metal electrode.

The Standard Hydrogen ElectrodeThe reference of choice is the standard hydrogen electrode. This electrode consists of hydrogen gas at a pressure of 100kPa (1bar) bubbling over a piece of platinum foil dipped into a solution of hydrochloric acid (or sulphuric acid) with a hydrogen ion concentration of 1moldm-1, at a temperature of 298K. The surface of the platinum foil is covered in porous platinum. Porous platinum has a large surface area and allows an equilibrium between hydrogen ions in solution and hydrogen has to be established quickly. H+(aq) + e- -><- 1/2H2(g)It is this equilibrium that we are going to compare with all others.

The Importance of using Standard ConditionsYou will remember that the position of an equilibrium can be changed by altering the conditions. If we are going to make fair comparisons, it is therefore necessary to standardise the conditions used. The standard conditions are:1. Gas pressure, 100kPa (1 bar)2. Temperature, 298K3. Concentration of ions in solution, 1moldm-3. These standard conditions apply to all equilibria.

Measuring a Standard Electrode PotentialWe are now in a position to appreciate how to measure the standard electrode potential of a metal ion / metal system, such as that between magnesium ions and magnesium (Mg2+ / Mg).To do this, we connect the standard hydrogen electrode to the magnesium electrode via a circuit containing a high resistance voltmeter.

The two components of the apparatus as known as half-cells. The two half-cells combine to make a complete cell. The salt bridge is needed to complete the electrical circuit. It usually contains a concentrated solution of potassium nitrate in the form of liquid or a gel. It functions by allowing the movement of ions. Theoretically, the salt bridge can contain any ionic salt, but neither of the ions present should interfere with the components of the half-cells. For example, if the right-hand half-cell contains a solution of silver nitrate, then potassium chloride would be unsuitable for the salt bridge because the chloride ions would interact with the silver ions to form a precipitate of silver chloride. There is a reason why a high resistance voltmeter is used. Ideally, the voltmeter should have infinite resistance so that there is no flow of electrons (i.e. no current flowing) around the external circuit. If this were possible, the reading on the voltmeter would represent the difference in potential between the two half-cells when both reactions are in equilibrium. The two reactions concerned are:H+(aq) + e- -><- 1/2H2(g)Mg2+ +(aq) + 2e- -><- Mg(s)With the magnesium half-cell connected to the standard hydrogen electrode, the reading on the voltmeter is 2.37V. If the magnesium half-cell is replaced by a copper half-cell (i.e. copper metal dipped into a solution of copper sulphate with [Cu2+(aq)] = 1moldm-1), then the reading on the voltmeter would be 0.34V. There is, however, a further issue to be considered. In the case of magnesium, if the voltmeter was taken out of the circuit, electrons would flow from the magnesium electrode to the hydrogen electrode. This means that magnesium is the negative electrode of the cell. With copper, electrons would flow from the hydrogen electrode to the copper electrode, making the copper the positive electrode of the cell. This difference in behaviour is recognised by using a sign convention. A negative sign (-) indicates that the metal electrode is negative with respect to the hydrogen electrode, whilst a positive sign indicates the reverse. The difference in potential measured, together with the appropriate sign, is called the standard electrode potential of the metal. Standard electrode potential is given the symbol E. Standard electrode potentials are quoted together with the relevant half-cell reaction. For example, for magnesium and copper they are:Mg2+ +(aq) + 2e- -><- Mg(s) E = -2.37VCu2+ +(aq) + 2e- -><- Cu(s) E =+0.34VBy convention, the standard electrode potential of the standard hydrogen electrode is zero. H+(aq) + e- -><- 1/2H2(g) E = 0.00V

What do E values tell us?It is important to remember that the standard electrode potential of a metal ion / metal half-cell is the potential difference measured when the half-cell is connected to a standard hydrogen electrode. The potential difference provides a comparison between the position of the metal ion / metal equilibrium and the position of equilibrium in the hydrogen electrode. To illustrate this point, let us consider the following two standard electrode potentials:Mg2+ +(aq) + 2e- -><- Mg(s) E = -2.37VH+(aq) + e- -><- 1/2H2(g) E = 0.00VCu2+ +(aq) + 2e- -><- Cu(s) E =+0.34VThe negative sign for Mg2+(aq) / Mg(s) indicates that the equilibrium position of this reaction is further to the left than the equilibrium position of the reaction in the hydrogen electrode. The positive sign for Cu2+(aq) / Cu(s) indicates that the equilibrium position of this reaction is further to the right than the equilibrium position of the reaction in the hydrogen electrode. This is another way of saying that magnesium releases electrons more readily than hydrogen (and also than copper), but that copper releases electrons less readily than hydrogen. Magnesium is, therefore, a better reducing agent than copper. This point will be developed later in the topic.

SummaryE values provide a method of comparing the positions of equilibria when metal atoms lose electrons to form ions in solution. 1. The more negative the E value, the further the equilibrium lies to the left, i.e. the more readily the metal loses electrons to form ions.2. The more positive (or less negative) the E value, the further the equilibrium lies to the right, i.e. the less readily the metal loses electrons to form ions. Later in this topic we will extend this to include both molecules and ions losing electrons.

Electromotive ForceAs already mentioned, when we are determining a standard electrode potential, the potential difference is measured when no electrons are flowing (i.e. no current is flowing) through the external circuit. The potential difference measured under these conditions is called the electromotive force (emf) of the cell. The standard emf of a cell is given the symbol Ecell and is sometimes called the standard cell potential.Emf values can be positive or negative, depending on the reference point against which the potential difference is measured. When measuring standard electrode potentials, the reference point is the potential of the standard hydrogen electrode, which is set at zero. The emf of a cell in which the hydrogen electrode is the positive electrode will have a negative value. Conversely, the emf of a cell in which the hydrogen electrode is the negative electrode will have a positive value. This now allows us to define standard electrode potential as follows:The standard electrode potential of a half-cell is the emf of a cell containing the half-cell connected to the standard hydrogen electrode. Standard conditions of 298K, 100kPa pressure of gases and solution concentrations of 1moldm-3 apply.

Measuring Standard Electrode Potentials of More Complicated Redox SystemsSystems Involving GasesSuppose we wanted to measure the standard electrode potential of the following redox system:1/2Cl2(g) + e- -><- Cl-(aq)We can do this by setting up a half-cell in which chlorine gas is bubbled into a solution containing chloride ions. However, in order to establish an equilibrium between the chlorine molecules and the chloride ions, and also to provide an electrical connection to the external circuit, we would need to place a piece of platinum into the solution. This half-cell is then connected to a standard hydrogen electrode and the emf is measured in the usual way. The measured emf is +1.36V. Hence:1/2Cl2(g) + e- -><- Cl-(aq) E = +1.36VOther Systems1. Non-metal elements and their ions in solutionTo determine the E of the redox system 1/2Br2 / Br-, a half-cell containing a solution of Br2 and Br- ions, each of concentration 1moldm-3 is connected to a standard hydrogen electrode. A similar set up is used to determine the E values of the redox system 1/2I2 / I-.1/2Br2(g) + e- -><- Br-(aq) E = +1.09V1/2I2(g) + e- -><- I-(aq) E = +0.54V2. Ions of the same element with different oxidation numbersA half-cell containing a solution of Fe3+ and Fe2+ is set up, each with a concentration of 1moldm-3 and connected to a standard hydrogen electrode. To provide connection to the external circuit, a piece of platinum is placed in the solution.1/2Fe3+(aq) + e- -><- Fe2+(aq)

The Electrochemical SeriesArranging Redox Equilibria in Order of their E ValuesThe electrochemical series is built up by arranging various redox equilibria in order of their standard electrode (redox) potentials. The most negative E values are placed at the top of the series, and the most positive at the bottom.

Reducing Agents and Oxidising AgentsThe species on the right-hand side of the half-cell reactions are all capable of behaving as reducing agents because they can lose electrons. The most powerful reducing agent in the table above is lithium because its redox system has the most negative E value; the equilibrium position of its half-cell reaction is furthest to the left. The least powerful reducing agent in the list is hydrogen fluoride, HF(aq) because its redox system has the least negative (most positive) E value. The position of equilibrium of its half-cell reaction is furthest to the right. The species on the left-hand side of the half-cell reactions are all capable of acting as oxidising agents because they can gain electrons. The most powerful oxidising agent is fluorine and the least powerful is the lithium ions, Li+(aq).

SummaryThe more negative the E value, the more the equilibrium position lies towards the left - and the more readily the species on the right loses electrons. The more negative the E value , the more powerful the reducing agent. The more positive the E value, the more the equilibrium position lies toward the right - and the more readily the species on the left loses electrons. The more positive the E value, the more powerful the oxidising agent.

Electrochemical Cells

Electrochemical CellsAn electrochemical cell is a device for producing an electric current from chemical reaction. It is constructed from two half-cells. The diagram below shows the apparatus used to construct an electrochemical cell from a Zn2+ / Zn half-cell and a Cu2+ / Cu half-cell under standard conditions.

The relevant standard electrode potentials for the redox systems involved are:Zn2+(aq) + 2e- -<>- Zn(s) E = -0.76VCu2+(aq) + 2e- -<>- Cu(s) E = +0.34VThe E value for the Zn2+ / Zn half-cell is the more negative, so the zinc electrode will be the negative electrode of the cell. When the cell is in operation, i.e. it is generating an electric current, electrons will flow through the external circuit from the zinc electrode to the copper electrode. The reactions taking place under these conditions are shown below:At the negative electrode: Zn(s) -> Zn2+ + 2e-At the positive electrode: Cu2+(aq) + 2e- -> Cu(s)The overall cell reaction is:Zn(s) + Cu2+(aq) 0> Zn2+(aq) + Cu(s)

Cell DiagramsIt is not always convenient to draw a diagram of the full apparatus for a cell. To simplify matters, chemists use a shorthand notation to represent half-cells. The half-cell made from zinc ions and zinc metal is written as:Zn2+(aq) / Zn(s) E = -0.76VThe solid vertical line indicates a phase boundary, in this case between an aqueous phase, Zn2=(aq), and a solid phase, Zn(s). When the electrode, i.e. the electrical connection between the solution and the external circuit, is a piece of platinum foil, the following convention is used:Pt(s) / Fe3+(aq) , Fe2+(aq) E = +0.77VNotice that because there is no phase boundary between Fe3+(aq) and Fe2+(aq), a comma is used to separate them, not a solid vertical line. The standard hydrogen electrode is represented as follows:H+(aq) / 1/2H2(g) / Pt(s)These shorthand notations can now be used to represent a cell comprising of two half-cells. Convention dictates two things:1. The two reduced forms of the species are shown on the outside of the cell diagram. 2. The positive electrode is shown on the right-hand side of the cell diagram. Applying these conventions produces the following cell diagram for a cell formed by combining the Zn2+(aq) / Zn(s) and the Cu2+(aq) / Cu(s) half-cells:Zn(s) / Zn2+(aq) ¦¦ Cu2+(aq) / Cu(s)The double vertical lines (¦¦) represent the salt bridge. The emf, E cell, of this cell is simply the difference between two standard electrode potentials of the two half-cells.Zn(s) / Zn2+(aq) ¦¦ Cu2+(aq) / Cu(s)E = -0.76 V E = +0.34VThe difference between these two numbers of 1.10, so the emf of the cell is 1.10V. To indicate that the right-hand electrode, i.e. the copper, is the positive electrode of the cell, the emf is given a positive (+) sign.So, the complete cell diagram is:Zn(s) / Zn2+(aq) ¦¦ Cu2+(aq) / Cu(s) E cell = +1.10V

What is meant by the "difference" between two standard electrode potentials?The easiest way to explain this is to represent the two values on a number scale. To move from one number to the other on the number scale involves a change of 1.10 units. Breaking the cell conventionChemists always reserve the right to break a convention if it suits their purpose. This is exactly what we do when we draw cell diagrams to represent the measurement of a standard electrode potential. In this case, the standard hydrogen electrode is always written on the left-hand side. The cell diagrams for the set-up used when measuring the standard electrode potential of the Zn2+(aq) / Zn(s) and the Cu2+(aq) / Cu(s) half-cells are therefore:Pt(s) / 1/2H2(g) / H+(aq) ¦¦ Zn2+(aq) / Zn(s) E cell = -0.76VPt(s) / 1/2H2(g) / H+(aq) ¦¦ Cu2+(aq) / Cu(s) E cell= +0.34VAs before, the sign of the E cell indicates the polarity of the right-hand electrode. Zinc is the negative electrode of the cell formed in combination with the standard hydrogen electrode. Copper is the positive electrode of the cell formed in combination with the standard hydrogen electrode.

Standard Electrode Potentials and Thermodynamic Feasibility

Using standard electrode potentials is one way of measuring how easily a species loses electron. This method provides information on how far to the left an equilibrium is relative to the equilibrium in the standard hydrogen electrode. For example, for the following two equilibria, the equilibrium of the top reaction lies further to the left than that of the hydrogen electrode, and the equilibrium for the bottom reaction lies further to the right than that of the hydrogen electrode.Zn2+(aq) + 2e- -<>- Zn(s) E = -0.76VCu2+(aq) + 2e- -<>- Cu(s) E = +0.34VIt follows, therefore, that the position of equilibrium of the Zn2+(aq) / Zn(s) reaction lies further to the left than that of the Cu2+(aq) / Cu(s) reaction. If these two equilibria are linked by combining the two half-cells to make an electrochemical cell then electrons will flow from the zinc electrode to the copper electrode. Therefore, the equilibria will be disturbed and the following reactions will take place in each half-cell:Zn(s) -> Zn2+(aq) + 2e-Cu2+(aq) + 2e- -> Cu(s)Electrons flow from the half-cell with the more negative E value to the half-cell with the less negative (i.e. more positive) E value.

Linking this to a Test Tube ReactionIs the following reaction thermodynamically feasible?Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)That is, will zinc displace copper from an aqueous solution containing copper(II) ions?To answer this questions, we will look at the E values for the relevant half-cell reactions:Zn2+(aq) + 2e- -<>- Zn(s) E = -0.76V equilibrium 1Cu2+(aq) + 2e- -<>- Cu(s) E = +0.34V equilibrium 2The E value for equilibrium 1 is more negative than that for equilibrium 2. So, the position of equilibrium 1 will shift to the left, releasing electrons, and the position of equilibrium 2 will shift to the right, accepting electrons. This means that the reaction between zinc and copper(II0 ions is thermodynamically feasible.

The Reaction Between Manganese(IV) Oxide and Hydrochloric AcidOne method of preparaing chlorine in the laboratory is to react manganese(IV) oxide with hydrochloric acid:MnO2(s) + 4HCl(aq) -> Mn2+(aq) + 2Cl-(aq) + 2H2O(l) + Cl2(g)However, the standard electrode potentials suggest that this reaction is not feasible:MnO2(s) + 2H+(aq) + 2e- -><- Mn2+(aq) + 2H2O(l) E = +1.23V equilibrium 1Cl2(g) + 2e- -><- 2Cl-(aq) E = +1.36V equilibrium 2The E value for equilibrium 2 is more positive than that for equilibrium 1. So, chloride ions cannot release electrons to MnO2, and this is required for the reaction shown above to take place. The reaction is not thermodynamically feasible under standard conditions, in which the concentration of the hydrochloric acid, and therefore the concentration of both the hydrogen ions and chloride ions, is 1moldm-3. The key to making the reaction take place is to increase the concentration of hydrochloric acid. If concentrated hydrochloric acid (-10moldm-3) is used, then the concentration of both the hydrogen ions and the chloride ions increases. The effect is to shift the position of equilibrium 1 to the right, and the position of equilibrium 2 to the left. As a result, the electrode potential for equilibrium 1 becomes more positive because the redox tem is now a better electron acceptor. The electrode potential for equilibrium 2 becomes less positive because the redox system is now a better electron releaser.The net effect is that the electrode potential of equilibrium 2 becomes less positive than that of equilibrium 1, so the chloride ions can now release electrons to manganese(IV) oxide. The reaction has become thermodynamically feasible under these non-standard conditions.

Summary1. The thermodynamic feasibility of a chemical reaction can be predicted using standard electrode potentials. 2. Although the standard electrode potentials indicate that a reaction is thermodynamically stable, it may not take place for two reasons:a. the reactants may kinetically stable because the activation energy for the reaction is very large, andb. the reaction may not be taking place under standard conditions.3. A reaction that is not thermodynamically feasible under standard conditions may become feasible when the conditions are altered. 4. Changing the concentrations may alter the electrode potential, E, of a half-cell because the position of equilibrium of the half-cell reaction may change.

Disproportionation ReactionsPreviously, we learnt that a disproportionation reaction is one in which an element in a species is simultaneously oxidised and reduced in the same reaction. A typical disproportionation reaction is the conversion of copper(I) ions in aqueous solution into copper(II) ions and copper atoms. Cu+(aq) + Cu+(aq) -> Cu2+(aq) + Cu(s)This reaction can be explained in terms of standard electrode potentials.Cu2+(aq) + e- -> Cu+(aq) E = +0.15V equilibrium 1Cu+(aq) + e- -> Cu(s) E = +0.52V equilibrium 2The E value for equilibrium 1 is more negative than that for equilibrium 2, so the position of equilibrium 1 will move to the left, and the equilibrium position of equilibrium 2 will move to the right. Therefore, Cu+ ions will both release and accept electrons to one another to form Cu2+ ions and Cu atoms. That is, Cu+ ions disproportionate into Cu2+ ions and Cu atoms.

Relationship Between Total Entropy and E cellThe standard Gibbs energy change, ΔG, taking place in a cell and the emf of the cell, E cell, are related by the following expression:ΔG = -nFEcell where n is the number of moles of electrons involved in the cell reaction and F is the Faraday constant. ΔG = ΔH - TΔSsystembut, ΔSsurroundings = -ΔH/Ttherefore, ΔH = -TΔSsurroundingsand hence:ΔG = -TΔSsurroundings - TΔSsystem = -T(ΔSsurroundings + ΔSsurroundings) = -TΔStotalAs n is a constant for a given cell reaction, and F is also a constant, it follows that, at a given temperature, the total entropy change of the cell reaction is proportional to the emf of the cell. If Ecell is positive, the reaction as written as written from left to right in the cell diagram is thermodynamically feasible because ΔStotal will be positive.

Relationship Between the Equilibrium Constant and E cellThe standard Gibbs energy change for a reaction and the equilibrium constant, K, are related by the following equation:ΔG = -RTlnKHowever:ΔG = -nFE cellTherefore:-RTlnK = -nFE cellor: lnK = nFE cell / RTAs n, F, and R are constants, it follows that at a given temperature, T, lnK is proportional to E cell.E cell can therefore be used to calculate the thermodynamic equilibrium constant, K, for a cell reaction.