Unit 6-53

Section 6.3 Redox Equilibrium ( 1 ) Redox reactions 1. Redox reactions in terms of electron transfer

The term ‘redox’ is used as an abbreviation for the processes of reduction and oxidation. These two processes usually occur simultaneously.

Redox reactions are electron transfer reactions. The separate equations showing which substance

gains electrons and which substance loses electrons are known as half-equations.

Oxidation is defined as the loss of electrons and reduction is defined as the gain of electrons. Oxidising agents are substances which accept electrons; reducing agents are substances which

donate electrons.

Example 1 When a piece of iron is immersed in copper (II) sulphate solution, it soon becomes coated with copper. The reaction may be represented by the ionic equation :

Fe(s) + Cu2+(aq) → Fe2+

(aq) + Cu(s) This shows that iron is the reducing agent and it is oxidized to Fe2+

(aq) : Oxidation half-equation : Fe(s) → Fe2+

(aq) + 2 e-

The oxidising agent is Cu2+ (aq) and it is reduced to copper : Reduction half-equation : Cu2+

(aq) + 2 e- → Cu(s)

Example 2

Chlorine can displace iodine from a solution of iodide ions : Cl2(aq) + 2 I-

(aq) → 2 Cl-(aq) + I2(aq)

Oxidation half-equation : 2 I-

(aq) → I2(aq) + 2 e_

Reduction half-equation : Cl2(aq) + 2 e- → 2 Cl-

(aq) In this redox reaction, chlorine is an oxidising agent and iodide ion is a reducing agent. Example 3

Some substances can act as either oxidising or reducing agents depending on reactions.

Hydrogen peroxide acts as an oxidising agent in the presence of a strong reducing agent: For example, it oxidized iodide ion which is a strong reducing agent : However, in the presence of a strong oxidising agent, it acts as a reducing agent: For example, it reduces manganate (VII) ions in the acidic medium :

Unit 6-54 Example 4 : Common oxidising agents

Oxidising agent Reduction half-reaction chlorine Cl2(aq) + 2 e- → 2 Cl-

(aq) bromine Br2(aq) + 2 e- → 2 Br-

(aq) oxygen

copper (II) ion Cu2+(aq) + 2 e- → 2 Cu(s)

iron (III) ion Fe3+(aq) + e- → Fe2+

(aq) hydrogen ion 2 H+

(aq) + 2 e- → H2(g) acidified manganate (VII) acidified dichromate (VI)

Test for oxidising agents: Oxidising agents change the colour of moist starch iodide paper from white to blue. The oxidising agent oxidizes the iodide ion to form iodine which reacts with starch to produce the blue colour. Example 5 : Common reducing agents

Reducing agent Oxidation half-reaction iodide ion 2 I-

(aq) → I2(aq) + 2 e_ bromide ion 2 Br-

(aq) → Br2(aq) + 2 e_ iron (II) ion Fe2+

(aq) → Fe3+(aq) + e_

sulphite ion thiosulphate ion

Test for reducing agents: Reducing agents decolorize acidified potassium manganate (VII) solution from purple to colourless. They also change the colour of acidified potassium dichromate (VI) solution from orange to green. Disproportionation is a redox reaction in which a particular chemical species is simultaneously oxidized and reduced. Example 6 : Iodine is oxidized to iodate (V) ion and reduced to iodide ion in alkaline medium.

Oxidation half-equation :

Reduction half-equation :

Overall reaction : Example 7 : Copper (I) ion disproportionates to copper (II) ion and copper in aqueous solution.

Write the redox reaction involved.

Unit 6-55 2. Oxidation states

There are some reactions which are regarded as redox processes in spite of the fact that they involve only molecules and do not have a complete transfer of electrons from one substance. For example, the reactions : 2 H2 + O2 → 2 H2O C + O2 → CO2

The concept of oxidation state (or oxidation number) provides a similar, but alternative, definition of redox to that involving electron transfer. 1. Oxidation number increases during oxidation. 2. Oxidation number decreases during reduction.

An oxidation number is a number assigned to an atom or an ion to describe relative state of

oxidation or reduction.. 1. The oxidation number of atoms in uncombined elements is 0. 2. In neutral molecules, the algebraic sum of the oxidation number is 0. 3. In ions, the algebraic sum of the oxidation number equals the charge on the ion. 4. In any substance, the more electronegative atom has the negative oxidation number, and the less electronegative atom has the positive oxidation number. 5. The oxidation number of hydrogen in all its compounds, except metal hydrides, is +1. 6. The oxidation number of oxygen in all its compounds, except in peroxides and in OF2 , is -2.

In assigning oxidation number in this way it is assumed that the electrons in each bond of the

molecule or ion belong to the more electronegative atom.

Example 1 : What is the oxidation number of each element in the following ? Mg Cl2 Br- Fe3+ H2O HCl PH3 H2S CO2 NaH Na2O Na2O2 OF2 Example 2 : What is the oxidation number of each element in the following ? MgCl2 NaOH CO SO2 SO3

2- SO42-

MnO4

- Cr2O72- CrO4

2- Example 3 : The oxidation number concept is often important in naming compounds. Recommended name HClO chloric(I) acid HClO3 KMnO4 potassium manganate(VII) FeCl3 iron(III) chloride NaIO3 Example 4: Nitric(III) acid disproportionates to form nitric(V) acid and nitrogen monoxide : 3 HNO2(aq) → HNO3(aq) + 2 NO(g) + H2O(l)

Unit 6-56 3. Balancing redox equations Half-equations method: 1. Work out the two relevant half-equations. 2. Balance for atoms other than oxygen and hydrogen first. 3. Balance for oxygen by adding the appropriate number of water molecules. 4. Acidic or neutral medium : Balance for hydrogen by adding H+

(aq) ions. Alkaline medium : Balance for H+

(aq) by adding OH-(aq) ions.

5. Balance the overall charge on each side of the half-equation by adding the appropriate number of electrons. 6. Balance the half-equations so that the number of electrons produced by the one equals the number used by the other. 7. Add the two half-equations together and cancel out any particles that appear on both sides of the equation. Example 1

Potassium manganate (VII) oxidizes ethanedioic acid to carbon dioxide under acidic conditions at 60oC. 2 x MnO4

- + 8 H+ + 5 e- → Mn2+ + 4 H2O(l)

5 x (COOH)2 → 2 CO2 + 2 H+ + 2 e- 2 MnO4

-(aq) + 5 (COOH)2(aq) → 2 Mn2+

(aq) + 10 CO2(g) + 8 H2O(l) Example 2 Balance the following redox reactions in acidic medium : (a) MnO4

-(aq) + 5 Fe2+

(aq) + 8 H+(aq) → Mn2+

(aq) + 5 Fe3+(aq) + 4 H2O(l)

MnO4

- + 8 H+ + 5 e- → Mn2+ + 4 H2O(l)

5 x Fe2+ → Fe3+ + e- (b) Cr2O7

2-(aq) + SO3

2-(aq) + H+

(aq) → Cr3+ (aq) + SO4

2- (aq) + H2O(l)

Example 3

Balance the following disproportionation reactions : (a) MnO4

2-(aq) + H+

(aq) → MnO4-

(aq) + MnO(s) + H2O(l) (b) Br2(aq) + OH-

(aq) → Br-(aq) + BrO3

-(aq) + H2O(l)

Unit 6-57 Example 4

CrO42-

(aq) ions react with S2O32-

(aq) ions to form Cr(OH)4-(aq) and SO4

2-(aq) ions in basic medium.

(a) Write balanced equations for each half reaction and for the overall reaction. (b) What volume of 0.50 M Na2CrO4 solution is needed to react completely with 40.0 cm3 of 0.20 M

Na2S2O3 solution in basic medium ? Oxidation states method : 1. Work out the increase in oxidation state of the reducing agent and the decrease in oxidation state of the oxidising agent. 2. Find the coefficients for the reactants that will make the total increase in oxidation state balance the total decrease. 3. The rest of the equation is then balanced by inspection. Example 5 SO2(aq) + Br2 (aq) + H2O(l) → H+

(aq) + SO42-

(aq) + Br-(aq)

Example 6 Cu(s) + NO3

- (aq) + H+ (aq) → Cu2+

(aq) + NO(g) + H2O(l) Example 7 Sn2+

(aq) + H2O2 (aq) + H+ (aq) → Sn4+

(aq) + H2O(l)

Unit 6-58 ( 2 ) Electrochemical cell 1. E.m.f. measurement of electrochemical cells of metal-metal ion systems

If a piece of zinc metal is placed in a solution of copper (II) sulphate, the blue colour of the copper

sulphate slowly fades, the zinc dissolves and red-brown copper metal takes its place. The ionic equation for the overall reaction taking place is :

Zn (s) + Cu2+

(aq) → Zn2+ (aq) + Cu(s)

This reaction produces energy, which is lost as heat if we simply carry out the process in a single

reaction vessel. However, if we separate the reaction in two half-cells, we can harness the energy through the flow of electrons taking place between the cells, as figure shows. This combination of half-cells is an electrochemical cell (galvanic cells or voltaic cells).

Electrochemical cell : An electrochemical cell is a device which produces an electromotive force (e.m.f.) as a result of

chemical reactions taking place at the electrodes. An electrochemical cell is thus a device which converts chemical energy into electrical energy.

Each cell consists of two half-cells. In one half-cell an oxidation half-reaction takes place. In the

other a reduction half-reaction takes place.

Electrode signs of electrochemical cells : Anode Cathode Half-reaction Electron flow Sign

Measurement of electromotive force (e.m.f.) : By using a high resistance voltmeter, the current in the external circuit is virtually zero and the cell

registers its maximum potential difference. This maximum p.d. is called the e.m.f. This e.m.f. gives a quantitative measure of the likelihood of the redox reaction taking pace in the cell. A digital multimeter is a cheaper and more handy alternative to the high impedance voltmeter in measuring e.m.f. of cells.

Unit 6-59 The Daniell Cell :

This is a classic example of a simple electrochemical cell. One half-cell consists of a zinc rod dipping into zinc sulphate solution. Oxidation takes place in this half-cell :

Zn(s) → Zn2+

(aq) + 2 e-

The zinc is the anode since oxidation occurs at this electrode.

The other half-cell consists of a copper rod dipping into copper (II) sulphate solution. Reduction takes place in this half-cell :

Cu2+

(aq) + 2 e- → Cu(s)

The copper is the cathode since reduction occurs at this electrode.

Salt bridge : Simple electrochemical cells used in a laboratory often consist of two half-cells separate by a salt

bridge. It may consist of an inverted U-tube containing salt solution. It is plugged at both ends by cotton wool. Alternatively it may consist of strip of filter paper soaked in salt solution. Salts used for this purpose include ammonium nitrate, potassium nitrate and potassium chloride. The salt used is chosen so that it does not react with the ions in either half-cell.

Salt bridge thus serves two main functions :

1. It completes the circuit by allowing ions carrying charge to move towards one half-cell from the other.

2. It provides cations and anions to replace those consumed at the electrodes. Electromotive force (e.m.f.) :

Ecell is the e.m.f. of the cell in volts. Changes that occur spontaneously have positive e.m.f., while changes that do not occur spontaneously have negative e.m.f.

Standard conditions :

At 25o C, 1 atm and unit concentration, the measured e.m.f. is the standard e.m.f. of the cell, E0cell.

Thus : Zn(s) + Cu2+

(aq) → Zn2+(aq) + Cu(s) E0

cell = + 1.10 V

Cu(s) + Zn2+(aq) → Cu2+

(aq) + Zn(s) E0cell =

Unit 6-60 2. E.m.f. values to compare the relative tendencies of half-cells to release or gain electrons

We can set up electrochemical cells with the Cu (s) | Cu2+(aq) half-cell as the oxidation half-cell and

other metal-metal ion half-cells as the reduction half-cell. E.m.f. values of the cells are then measured and shown in table :

Cu(s) | Cu2+(aq) M Mn+

(aq) | M (s)

Oxidation half cell Reduction half-cell ECell (Cell e.m.f.) Cu(s) | Cu2+

(aq) Ag+(aq) | Ag (s) +0.46

Cu(s) | Cu2+(aq) Cu2+

(aq) | Cu (s) 0.00 Cu(s) | Cu2+

(aq) Ni2+(aq) | Ni (s) -0.59

Cu(s) | Cu2+(aq) Zn2+

(aq) | Zn (s) -1.10 Cu(s) | Cu2+

(aq) Mn2+(aq) | Mn (s) -1.53

The cells in table represent the following reactions : Electrode potential

Cu(s) + 2Ag+(aq) → Cu2+

(aq) + 2Ag(s) Ag+(aq) | Ag (s) E = + 0.46 V

Cu(s) + Cu2+

(aq) → Cu2+(aq) + Cu(s) Cu2+

(aq) | Cu (s) E = 0.00 V

Cu(s) + Ni2+(aq) → Cu2+

(aq) + Ni(s) Ni2+(aq) | Ni (s) E = - 0.59 V

Cu(s) + Zn2+(aq) → Cu2+

(aq) + Zn(s) Zn2+(aq) | Zn (s) E = - 1.10 V

Cu (s) + Mn2+

(aq) → Cu2+ (aq) + Mn(s) Mn2+ (aq) | Mn (s) E = -1.53 V

Ag+

(aq) ion is a stronger oxidising agent than Cu2+(aq) ion, oxidising copper metal to copper(II). The

first reaction therefore happens spontaneously in the direction written, with a decrease in the energy of the system. The last three reactions are different. Mn2+

(aq) , Zn2+(aq) and Ni2+

(aq) ions are weaker oxidising agents than Cu2+

(aq) ions, so the reaction written here do not happen spontaneously. Instead, the reverse reactions are the energetically favourable ones, in which Cu2+

(aq) ions oxidize manganese metal to Mn2+(aq)

ions, nickel metal to Ni2+(aq) ions and zinc metal to Zn2+

(aq) ions. When we measure the e.m.f. of a cell, we can imagine that the electron flow arises from the

competition for electrons between two half-cells. Each half-cell reaction will have its own tendency to attract electrons, a tendency measured by the electrode potential of the half-cell. This is the e.m.f. measured when that half-cell forms an electrochemical cell with a reference half-cell.

The more positive a half-cell’s electrode potential, the greater its tendency to attract electrons. Ag+(aq) ion is the strongest oxidising agent here as it is the strongest competitor for electrons. Mn2

(aq) ion is the weakest oxidising agent, as it is the weakest species to attract electrons. Using this idea and the data in figure, calculate the e.m.f. of the cell : Ni(s) | Ni2+

(aq) M Ag+(aq) | Ag(s) Ecell = + 0.46 - ( - 0.59) = + 1.05 V

The e.m.f. of this cell is the difference between the electrode potentials of the two half-cells. The

positive value confirms that the reaction proceeds spontaneously in the forward direction : Ni(s) + 2 Ag+

(aq) → Ni2+

(aq) + 2 Ag(s)

Unit 6-61 3. Types of half-cells Metal - Metal ion half-cell When a metal electrode is dipped into a solution containing ions of the same metal, atoms from the metal lattice pass into solution and form hydrated metal ions. At the same time hydrated metal ions gain electrons at the electrode and form metal atoms. Eventually an equilibrium is established : Mn+

(aq) + n e- M(s)

The position of equilibrium depends on a number of factors including the nature of the metal, the concentration of ions and the temperature. If the equilibrium lies to the right, reduction predominates and the electrode acquires a positive charge. If the equilibrium lies to the left, oxidation predominates and thus the electrode acquires a negative charge. Non-metal - Ion half-cell Example 1

Chlorine gas is allowed to bubble through a solution containing chloride ions. The electrode consists of platinum foil coated with Pt black. The electrode is inert although it allows the chlorine to adsorb on its surface.

The following equilibrium is established : Cl2(g) + 2 e- 2 Cl-

(aq)

This half-cell is represented as : Example 2

A Pt electrode is dipping into a solution containing I2 (aq) and iodide ions. The following equilibrium is established :

I2(aq) + 2 e- 2I-(aq)

This half-cell is represented as :

Ion - Ion half-cell This type of half-cell consists of an inert electrode such as Pt electrode dipping into a solution containing ions of the same metal in two different oxidation states.

A typical example is the iron (II) - iron (III) half-cell. Fe3+

(aq) + e- Fe2+(aq)

This half-cell is represented as :

Metal - Metal salt half-cell

This type of half-cell consists of an inert electrode such as Pt electrode coated with the metal and the same metal salt. A typical example is :

which represents the following half-reaction :

Unit 6-62 ( 3 ) Electrode Potential 1. The standard hydrogen electrode

It is impossible to obtain the electrode potential for a single half-cell. E.m.f. values can only be

measured for a complete circuit with two electrodes. This can be done by arbitrarily assigning an electrode potential of zero to one particular half-cell. Other half-cell can then be compared with this standard.

The standard chosen for electrode potentials is standard hydrogen electrode. This consists of H2

gas at one atmosphere pressure and 25o C bubbling around a platinised platinum electrode. The electrode is immersed in a 1.0 moldm-3 solution of H+ ions. Hydrogen is adsorbed on the platinum and an equilibrium is established between the adsorbed layer of H2 and H+ ions in the solution :

The platinised platinum electrode has three functions : 1. It acts as an inert metal connection to the H2/H+ system. There is no tendency for Pt to form ions itself. 2. It allows H2 gas to be adsorbed onto its surface. 3. It is covered by a loosely deposited layer of finely divided platinum. This increases its surface area so that it can established an equilibrium between H2 (g) and H+ (aq) as rapidly as possible. The standard electrode potential E0 for this reference half-cell is taken as zero. i.e. Cell diagram for standard hydrogen electrode :

Unit 6-63 2. Standard reduction potentials and the electrochemical series

Since electrode potential depend on temperature, concentration and also pressure, it is necessary to standardize them if they are to be compared.

The standard conditions chosen are similar to those for thermochemical measurements : 1. All solutions have a concentration of 1 moldm-3. 2. Any gas involved has a pressure of 1 atmosphere. 3. The temperature is 298 K (25o C). 4. Platinum is used as the electrode when the half-cell system does not include a metal. The standard electrode potential of an electrode ( half-cell ) is thus defined as the e.m.f. of a cell in

which the electrode on the left is a standard hydrogen electrode and that on the right is the standard electrode in question.

Standard electrode potentials are sometimes called standard reduction potentials because they

related to the reduction of the more oxidized species. Oxidized species + n e- Reduced species A standard reduction potential is a measured of the tendency for reduction to occur. It is also a

measure of the strength of the oxidising agent to accept electrons. Example 1: When a standard zinc half-cell is connected to a standard hydrogen half-cell, the e.m.f.

measured is - 0.76 volts. In this case the zinc electrode is negative.

Thus, the standard electrode potential for the Zn2+

(aq) / Zn(s) half-cell is - 0.76 volts, i.e. Zn2+

(aq) + 2 e- Zn(s) E0 = -0.76 V The negative sign indicates that Zn2+ ions have a lower tendency to accept electrons than H+ ions. The redox reaction occurring at the electrochemical cell is :

Unit 6-64 Example 2 : When a standard chlorine half-cell is connected to a standard hydrogen half-cell, the e.m.f.

measured is + 1.36 volts. In this case the chlorine electrode is positive.

Thus, the standard electrode potential for the Cl2(g) / Cl-(aq) half-cell is + 1.36 volts ,

i.e. Cl2(g) + 2 e- 2 Cl-(aq) E0 = + 1.36 V

The positive sign indicates that Cl2 have a higher tendency to accept electrons than H+ ions. The redox reaction occurring at the electrochemical cell is :

3. The electrochemical series

When redox systems are arranged in order of their standard electrode potentials, the electrochemical series is obtained. The following table shows standard electrode potentials of some common redox systems :

Unit 6-65 4. Uses of the standard electrode potential values To compare the strength of oxidising / reducing agents :

The standard electrode potential of a half-cell is a measure of the oxidising or reducing power of the species in it, i.e. their ability to compete electrons.

In general the stronger an oxidising agent, the more positive its electrode potential. Example 1 : Ozone gas is a more powerful oxidising agent than chlorine gas : O3(g) + 2 H+

(aq) + 2 e- O2(g) + H2O(l) EO = + 2.08 V Cl2(g) + 2 e- 2 Cl-

(aq) EO = + 1.36 V A strong reducing agent has a large negative electrode potential. Example 2 : Calcium metal is a more powerful reducing agent than lead metal : Pb2+

(aq) + 2 e- Pb(s) EO = - 0.13 V Ca2+

(aq) + 2 e- Ca(s) EO = - 2.87 V ∴ Pb(s) Pb2+

(aq) + 2 e- EO = + 0.13 V Ca(s) Ca2+

(aq) + 2 e- EO = + 2.87 V To calculate the e.m.f. of cells : 1. The half-cell with the more positive electrode potential is the half-cell which reduction occurs, E1. 2. The half-cell with the less positive electrode potential is the half-cell which oxidation occurs, E2. 3. The e.m.f. of an electrochemical cell is given by :

E cell = E1 - E2

Example 1 From the following data determine :

(a) the standard cell e.m.f., (b) the cell reaction,

Data : Zn2+(aq) + 2 e- Zn(s) EO = - 0.76 V

Ni2+(aq) + 2 e- Ni(s) E0 = - 0.25 V

Example 2

A cell based on the reaction : 6 Fe2+

(aq) + Cr2O72-

(aq) + 14 H+(aq) → 6 Fe3+

(aq) + 2 Cr3+(aq) + 7 H2O(l)

consists of one platinum electrode dipped into a beaker containing an acidified solution K2Cr2O7(aq), and another platinum electrode dipped into a separate beaker containing FeSO4(aq), with the two solutions being connected by a salt bridge. With reference to this cell : (a) write equations for the reactions occurring at the cathode and the anode; (b) state the direction in which electrons move when the electrodes are connected externally; (c) calculate the e.m.f. of the cell, given that the standard reduction potentials of Fe3+ / Fe2+ and

acidified Cr2O72- / Cr3+ are respectively +0.77 V and +1.33 V.

Unit 6-66 5. Prediction of the feasibility of redox reactions

The standard electrode potentials enable us to predict whether a redox reaction is spontaneously or not. In general, redox reactions with an overall positive standard cell potential ( cell e.m.f. ) are energetically feasible whereas those with an overall negative value are not so. If E0 cell > 0 the reaction will proceed from left to right as written. If E0 cell = 0 the reaction is at chemical equilibrium. If E0 cell < 0 the reaction will not proceed in the direction written, but it will take place in the opposite direction. Example 1

When copper and silver metals are in contact with their ions, what redox reaction will occur ? Data : Cu2+

(aq) + 2 e- Cu(s) E0 = + 0.34 V Ag+

(aq) + e- Ag(s) E0 = + 0.80 V Note : E0 relates to the probability of a reaction occurring not to the quantity of materials reacting.

Thus, if E0 for Ag+(aq) + e- Ag(s) is + 0.80 V,

E0 for 2Ag+(aq) + e- 2Ag(s) is also + 0.80 V , not +1.60 V.

Example 2

Given that : Br2(aq) + 2 e- 2 Br-(aq) E0 = + 1.09 V

O2(g) + 2 H2O(l) + 4 e- 4 OH-(aq) E0 = +0.40 V

(a) Show how you would predict whether Br2 or O2 is a stronger oxidising agent. Br2 is a stronger oxidising agent than O2 as standard reduction potential of Br2/Br- > O2/OH-. (b) Show how you would experimentally demonstrate your prediction. Add aqueous solution of NaOH(aq) to bromine water. The reddish brown colour of Br2(aq) would

disappear : 2 Br2(aq) + 4 OH-

(aq) → 2 Br-(aq) + O2(g) + 2 H2O(l) E0

cell = + 0.69 V > 0 Example 3

Given : Ag+(aq) + e- Ag(s) E0 = + 0.80 V

AgCl(s) + e- Ag(s) + Cl-(aq) E0 = + 0.22 V

O2(g) + 2 H2O(l) + 4 e- 4 OH-(aq) E0 = +0.40 V

Show that silver cannot under oxidation in the presence of only oxygen and pure water but that it can do so if chloride ions are also present.

Unit 6-67 Limitation of prediction :

Theoretically, redox reactions having a positive cell e.m.f. are feasible reactions and will take place spontaneously. However, since electrode potentials are affected by changes in concentration and temperature, a prediction based on standard electrode potential values may not be valid under conditions which are not standard. As a rule of thumb, if the magnitude of the cell e.m.f. calculated from standard electrode potential values is greater than 0.4 V, it is quite safe to say that the prediction is valid. A cell e.m.f. only gives information about the feasibility of a redox reaction from an energetic point of view. It cannot tell how fast a feasible reaction is likely to proceed. Therefore, some feasible reactions may not appear to take place just because they are too slow. Example 1

The positive cell e.m.f. value shows that the oxidation of iron(II) salts in acidic solution is energetically feasible : Fe3+

(aq) + e- Fe2+(aq) E0 = + 0.77 v

O2 (g) + 4 H+(aq) + 4 e- 2 H2O(l) E0 = +1.23 V

Overall : However, the oxidation of Fe2+ under acidic conditions is extremely slow because kinetic factors increase the stability of Fe2+ in acidic solutions. Example 2 The standard electrode potential for the half-reaction Fe3+

(aq) + e- Fe2+(aq) E0 = + 0.77 V

What happens to the electrode potential when the concentration of •Fe3+

(aq) rises •Fe3+

(aq) falls •Fe2+

(aq) rises •Fe2+

(aq) falls •both Fe3+

(aq) and Fe2+(aq) are doubled ?

Example 3 Given the following standard reduction potentials, E0 : E0 / V [Cr2O7

2-(aq) + 14 H+

(aq)], [2 Cr3+(aq) + 7 H2O(l)] Pt(s) + 1.33

Cr3+(aq) , Cr2+

(aq) Pt(s) - 0.41 Zn2+

(aq) Zn(s) - 0.76 (a) Predict the products of the reaction of zinc with dichromate(VI) solution. Explain your

prediction. (b) (i) Predict the effect of an increase in the concentration of Zn2+

(aq) on the reduction potential of Zn2+

(aq) Zn(s) . Explain your prediction. (ii) Give one other factor which can affect the reaction potential.

Unit 6-68 6. The Nernst equation The processes that occur at an electrode surface are reversible and are governed by the equilibrium law. If the concentration of a reactant or product is altered, the position of equilibrium also alters and therefore the electrode potential changes : oxidised species + n e- reduced species

e.g. Fe3+(aq) + e- Fe2+

(aq) The exact relationship between the electrode potential of a half-cell and the concentrations of the ionic species involved is given by the Nernst equation :

E = EO + ][][ln

nFRT

redox

where E = electode potential EO = standard electrode potential R = gas constant T = absolute temperature n = number of electrons transfered F = Faraday’s constant (the charge of one mole of electrons) An increase in temperature results in the electrode potential becoming more positive compared with the standard hydrogen electrode.

At 298 K, the Nernst equation can be simplified as an empirical relationship of the variation of

electrode potential with concentration of ionic species involved :

E = EO + ][][log

n0.059

redox

The concentration of a pure element and water are taken as unity.

Example 1 If a strip of aluminium dips into a solution of aluminium ions at a concentration of 0.100 mol dm-3, calculate the electrode potential at 298 K. Given : The standard electrode potential for the half-reaction Al3+

(aq) + 3 e- Al(s) E0 = - 1.66 V Example 2 For the dichromate redox system : Cr2O7

2-(aq) + 14 H+

(aq) + 6 e- 2 Cr3+(aq) + 7 H2O(l) E0 = + 1.33 V

If dichromate is used in neutral solution instead of acid solution, calculate the electrode potential at 298 K. Assume : [H+

(aq)] = 1 x 10-7 mol dm-3 and [Cr2O72-

(aq)] = [Cr3+(aq)] = 1.0 M

In neutral solution, the electrode potential is reduced from 1.33 V to 0.37 V.

Unit 6-69 Example 3 : Concentration cell (a) If a silver electrode dips into a solution of Ag+ ions at a concentration of 0.001 mol dm-3, calculate

the electrode potential at 298 K. Given : The standard electrode potential for the half-reaction Ag+

(aq) + e- Ag(s) E0 = + 0.80 V (b) A cell can be set up by using the half-cell in (a) and the standard Ag+

(aq) / Ag(s) half-cell. It is called a concentration cell because the e.m.f. of the cell depends on the difference in

concentration in the two half-cells : Calculate the e.m.f. of the above concentration cell.

In the above cell, the electrode dipping into the more dilute solution is negative compared with the standard half-cell. Electrons then flow from left to right through the wire. Silver dissolves in the left-hand beaker and is deposited on the electrode of the right-hand beaker. The process continues until the concentration of silver ions in the two beakers becomes equal.

Example 4 : Determination of solubility product The solubility product of silver chloride can be measured by making up a saturated solution of silver chloride at 298 K. By dipping of silver electrode into this solution, a concentration cell is made up with a standard Ag+

(aq) / Ag(s) half-cell as the other half-cell. The measured e.m.f. is 0.29 V at 298 K.

Calculate the Ksp of silver chloride at this temperature. Let E be the electrode potential of the Ag+

(aq) / Ag(s) half-cell with saturated solution of silver chloride.

Unit 6-70 Example 5 : Measurement of pH (a) For the hydrogen electrode 2 H+

(aq) + 2 e- H2(g) E0 = 0.00 V

E = EO + ][][log

20.059

)(2

2)(

g

aq

HH +

Calculate the electrode potential of the following solutions : (i) [H+

(aq)] = 0.01 mol dm-3 (pH = 2) (ii) [H+

(aq)] = 0.0001 mol dm-3 (pH = 4) (iii) [H+

(aq)] = 0.025 mol dm-3 (b) The pH of unknown solution A hydrogen-platinum black electrode is dipped into the solution whose [H+

(aq)] is to be measured. A concentration cell is set up by connecting the half-cell to a standard hydrogen half-cell. The e.m.f. of the cell is measured.

If the measured e.m.f. is 0.18 V at 298 K, what is the pH of the solution ? Let E be the electrode potential of the hydrogen half-cell with unknown pH.

Unit 6-71 ( 4 ) Secondary Cell Batteries :

A battery is a portable source of electricity. Batteries store energy in the form of chemicals, the energy being released when a conductor is connected between the terminals of the battery. Battery consists of one or more electrochemical cells.

Primary and Secondary cell :

A cell which does not regenerate reactants is called a primary cell. A secondary cell is one which can be recharged by regenerating the cell reactants.

1. The Lead-acid Accumulator Structure :

The lead-acid accumulator is a secondary cell--the half-reactions at the electrodes are readily reversible. It consists of a lead anode and a grid of lead packed with lead (IV) oxide as the cathode. The electrolyte is sulphuric acid. Electrochemical processes :

When discharging, the half-reactions at the electrodes are at the anode ( - ) : at the cathode ( + ) : The overall reaction is

The current which can be obtained from the cell is increased by constructing the cathode of a number of plates which alternate with several anode plates. One such cell provides about 2 V. The battery is recharged by applying a current from an external source. This reverses the electrode reactions.

When charging, anode is connected to the negative terminal of an external transformer and cathode

is connected to the positive terminal of an external transformer. Overall reaction is : Uses: Car batteries consist of six lead accumulator cells joined in series to give an e.m.f. of 12 V. When the car is in motion, it drives a generator which charges the battery. When the lead storage cell is fully charge, the sulphuric acid has a relative density of about 1.275. Upon discharging, lead (II) sulphate forms at both electrodes, thus reducing the concentration and relative density of the sulphuric acid. The relative density thus indicates the state of charge in the battery. If there is too much stopping and starting, the battery loses its charge and becomes ‘flat’, until it is recharged by the passage of a direct current from a transformer. It is important that discharge does not continue for too long, otherwise the precipitate of fine lead (II) sulphate changes to a coarser, inactive and non-reversible form. If this happens the accumulator becomes much less efficient.

Unit 6-72 2. The Lithium-ion battery

Lithium ion batteries (or Li-ion) have become very common and dropped in price recently. They provide one of the best energy-per-weight ratios of rechargeable batteries at present. They have succeeded nickel metal hydride and nickel-cadmium batteries in consumer electronics such as mobile phones, digital photo/video cameras, and notebook computers.

Lithium-ion batteries have a nominal voltage of 3.6 V and a typical charging voltage of 4.2 V. Lithium-ion batteries cannot be fast-charged and typically need at least four hours to fully charge. Li-ion batteries do not suffer from the memory effect, but they are not as durable as NiMH or NiCd designs and can be extremely dangerous if mistreated.

At a typical 100% charge level (notebook battery, full most of the time) at 25 degrees Celsius, Li-ion batteries irreversibly lose approximately 20% capacity per year from the time they are manufactured, even when unused. (6% at 0 °C, 20% at 25 °C, 35% at 40 °C. When stored at 40% charge level, these figures are reduced to 2%, 4%, 15% at 0, 25 and 40 degrees Celsius respectively.) Every (deep) discharge cycle decreases their capacity. The degradation is sloped such that 100 cycles leave the battery with about 75% to 85% of the original. When used in notebook computers or cellular phones, this rate of deterioration means that after three to five years the battery will have capacities too low to be still usable.

One great advantage of Li-Ion batteries is their low self-discharge rate of only approximately 5% per month, compared with over 30% per month and 20% per month in nickel metal hydride batteries and nickel cadmium batteries respectively.

Lithium ion internal design is as follows. The anode is made from carbon, the cathode is a metal

oxide, and the electrolyte is a lithium salt in an organic solvent. Since the lithium metal which might be produced under irregular charging conditions is very reactive and might cause explosion, Li-ion cells usually have built-in protective electronics and/or fuses to prevent polarity reversal, over voltage and over-heating.

The Li-ion battery required nearly 20 years of development before it was safe enough to be used on a

mass market level. A unique drawback that we can see to the Li-ion battery is that its life cycle is dependent upon aging from time of manufacturing (shelf life) regardless if it was charged or not and not on the number of charge/discharge cycles. A more advanced lithium-ion battery design is the lithium polymer cell. Example : The nickel-cadmium rechargeable battery The nickel-cadmium rechargeable battery has electrodes of solid cadmium and solid nickel(IV) oxide, NiO2, coated on a conductor. When the battery discharges, solid cadmium hydroxide, Cd(OH)2, and solid nickel(II) hydroxide, Ni(OH)2, are formed. The electrolyte is an aqueous solution of potassium hydroxide. The standard reduction potential for Cd(OH)2 going to Cd is – 0.76V and that for NiO2 going to Ni(OH)2 is + 0.49V. (a) Write the half equation occurring at each electrode when the battery discharges. (b) What is the standard e.m.f. of the cell ?

Unit 6-73 ( 5 ) Fuel Cell

A fuel cell is an electrochemical device similar to a battery, but differing from the latter in that it is designed for continuous replenishment of the reactants consumed; i.e. it produces electricity from an external fuel supply as opposed to the limited internal energy storage capacity of a battery.

Typical reactants used in a fuel cell are hydrogen on the anode side and oxygen on the cathode side (a hydrogen-oxygen cell). In contrast, conventional batteries consume solid reactants and, once these reactants are depleted, must be discarded, recharged with electricity by running the chemical reaction backwards, or, at least in theory, by having their electrodes replaced. Typically in fuel cells, reactants flow in and reaction products flow out, and continuous long-term operation is feasible virtually as long as these flows are maintained.

Fuel cells are very attractive in modern applications for their high efficiency and ideally emission-free advantage, where the only by-product of a hydrogen fuel cell is water vapor. However, using more-available fuels such as methane or natural gas will still generate some carbon dioxide. The other concern is the energy-consuming process of manufacturing hydrogen, which may generate pollution.

Hydrogen-oxygen fuel cell

A hydrogen-oxygen fuel cell is a primary cell which directly converts the energy supplied by the oxidation of hydrogen (the fuel) into electrical energy. A simplified diagram of the hydrogen-oxygen fuel cell is shown below :

In the fuel cell, the fuel (hydrogen) and the oxidising agent (oxygen) are respectively supplied in a

continuous flow into the anode and cathode compartments. In these compartments, the porous nickel electrodes not only act as the electrical conductors connecting the fuel cell to the external circuit, but also act as the catalyst for the reactions which occur in the cell. The product and the operation procedures of the hydrogen-oxygen fuel cell are non-polluting. However, the cost is high and thus its use is limited in spacecrafts. The water produced from the fuel cell is usually removed and consumed by the astronauts.

The half-reactions and the overall reaction occurring in the cell are : Anode : Cathode : Net reaction :

Unit 6-74 Direct-methanol fuel cells

Direct-methanol fuel cell or DMFC is a type of proton-exchange fuel cell.

The fuel methanol is fed directly to the fuel cell. Storage of methanol is much easier than that of hydrogen because it does not need to be done at high pressures, as methanol is a liquid. The energy density of methanol (the amount of hydrogen in a given volume) is orders of magnitude greater than even highly compressed hydrogen.

The DMFC relies upon the oxidation of methanol on a catalyst layer to form carbon dioxide. Water is consumed at the anode and is produced at the cathode. Protons (H+) are transported across the proton exchange membrane to the cathode where they react with oxygen to produce water. Electrons are transported via an external circuit from anode to cathode providing power to external devices.

The half reactions are:

Anode:

Cathode:

Net reaction:

Because water is consumed at the anode in the reaction, pure methanol cannot be used without provision of water via either passive transport such as back diffusion (osmosis), or active transport such as pumping. The need for water limits the energy density of the fuel.

Achievements

As of 2005, the record for the smallest commercially available fuel cell is held by Toshiba, at 22 x 56 x 4.5 millimeters. This device outputs 100 milliwatts at 10 hours per milliliter of fuel, and takes advantage of new technology allowing the use of undiluted (99.5%) methanol.

Uses of fuel cells

Fuel cells are very efficient, with 70% or more of the chemical energy being converted into electricity. By comparison, the most modern power plants can only achieve a conversion of chemical energy to electricity of about 45%. Moreover, the product and the operation procedures are usually non-polluting. Therefore, fuel cell seems to be a desirable energy converter. However, the cost is high and thus its use is limited. More research and development are still needed before fuel cells can be used for other purposes. Some applications that have been suggested include

• emergency backup power,

• portable electronics,

• electrically-powered vehicles, and

• mobile phone power.