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THE ELECTRONIC THEORY OF OXIDATION & REDUCTION

1. Oxidation & Reduction

Simple definitions of oxidation and reduction are based on the loss/gain of oxygen or the loss/gain of hydrogen. Oxidation is the gain of oxygen or the loss of hydrogen; reduction is the loss of oxygen or the gain of hydrogen. These definitions can only be used when a chemical reaction involves hydrogen and oxygen, and therefore their usefulness is limited.

A more basic and more useful definition of oxidation and reduction is based on the loss/gain of electrons.

OXIDATION IS LOSS OF ELECTRONS

REDUCTION IS GAIN OF ELECTRONS

In reactions involving simple ions, it is usually easy to tell whether electrons are lost or gained, but it is less easy to tell when complex ions or covalent molecules are involved. Oxidation number is a useful concept for helping to decide in these more awkward cases.

2. Oxidation Number

The oxidation number is used to express the oxidation state of an element, whether as the uncombined element or when combined in a compound; it consists of a + or – sign followed by a number, or it is zero.

Atoms of elements have no overall charge and are therefore given an oxidation number of zero. When two elements combine, the atoms or ions of the more electropositive element have a positive oxidation state, and those of the more electronegative element a negative oxidation state. Elements become more electronegative the higher their Group number and the lower their Period number; therefore, the most electronegative element is fluorine.

The oxidation number of an element in a compound is equal to the charge which a particle of the element would carry in the compound, assuming the compound is ionic. This is a purely theoretical idea, and it is does not matter whether the compound in question is really ionic or covalent.

TOPIC 13.20: ELECTRODE POTENTIALS 1

e.g. compound oxidation numbers

NaCl Na +1 Cl -1CCl4 C +4 Cl -1HBr H +1 Br -1H2S H +1 S -2

The following general rules are useful:

All free elements (i.e. those not combined with another element) have an oxidation number of 0.

e.g. Na, Mg, Br2

In simple ions, the charge on the ion is equal to the oxidation number.

e.g. ion Na+ Fe3+ Br- O2-

oxidation number +1 +3 -1 -2

Since fluorine is the most electronegative element, it always has an oxidation number in its compounds of –1.

Combined oxygen always has an oxidation number of –2, except when in combination with fluorine or in peroxides.


Fe2O3 Fe +3 O -2Mn2O7 Mn +7 O -2CrO3 Cr +3 O -2

BUT ….H2O2 H +1 O -1F2O O +2 F -1

Group I elements in their compounds always have an oxidation number of +1.

Group II elements in their compounds always have an oxidation number of +2.

Hydrogen in its compounds always has an oxidation number of +1, except when it has combined with a reactive metal.


H2O H +1 O -2HCl H +1 Cl -1CH4 H +1 C -4

BUT ….NaH Na +1 H -1

The sum of the oxidation numbers of all the atoms in an uncharged molecule is zero: in an ion, the sum is equal to the charge on the ion.


e.g. compound/ion oxidation numbers

NH3 H +1 N -3 (-3+1+1+1=0)NH4

+ H +1 N -3 (-3+1+1+1+1=+1)H2SO4 H +1 S +6 O -2 (+1+1+6-2-2-2-2=0)

3. Number of Oxidation StatesMany elements have several oxidation states:

e.g. sulphur H2S S SCl2 SO2 SO3+1 -2 0 +2 -1 +4 -2 +6 -2

chlorine HCl Cl2 HOCl ClF3 KClO3 KClO4+1-1 0 +1-2+1 +3-1 +1+5-2 +1+7-2

4. Redox ReactionsIf, during a chemical reaction, the oxidation number of an element increases (i.e. becomes more positive or less negative), then the element has lost electrons and has been OXIDISED.

Conversely, if the oxidation number of an element decreases (i.e. becomes less positive or more negative), then the element has gained electrons and has been REDUCED.

REDuction and OXidation occur together in what is called a REDOX reaction.

Example:

work out the oxidation numbers for all the elements in the reaction identify the oxidation and reduction steps work out the total number of electrons transferred in each step: they should be

equal oxidation: loss of 2e- x 1

Mg + 2HCl MgCl2 + H2

0 +1 -1 +2 -1 0

reduction: gain of 1e- x 2


Oxidation

-5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5

Reduction

check: 2e- x 1 = 1e- x 2

OXIDATION & REDUCTION

QUESTION SHEET 3

Work out the oxidation number of each element in the following chemical formulae

1. Ca …………………………………………………………………..

2. LiF ………………………………………………………………….

3. MgO ………………………………………………………………….

4. I2 ………………………………………………………………….

5. Cr3+ ………………………………………………………………….

6. Al2O3 ………………………………………………………………….

7. HF ………………………………………………………………….

8. Ni(NO3)2 ………………………………………………………………….

9. CrO42- ………………………………………………………………….

10. SrCO3 ………………………………………………………………….

11. NaClO4 ………………………………………………………………….

12. SO32- ………………………………………………………………….

13. NaIO3 ………………………………………………………………….

14. XeF4 ………………………………………………………………….

15. Pb(OH)2 ………………………………………………………………….

16. K2MnO4 ………………………………………………………………….

17. Al2(SO4)3 ………………………………………………………………….

18. NaVO3 ………………………………………………………………….

19. H3PO3 ………………………………………………………………….

20. NH4+ ………………………………………………………………….


OXIDATION NUMBERSTransition elements are able to form compounds in which they exhibit a variety of oxidation states. They form complex ions with oxygen and various ligands.

A ligand is a lone pair donor, which forms a coordinate bond with a central metal ion or atom. The ligand may be a neutral molecule or a negative ion. Examples of ligands are:

Ligand Formula Overall chargewater H2O 0

ammonia NH3 0

cyanide ion CN- -1

carbon monoxide CO 0

1,2-diaminoethane NH2CH2CH2NH2 or en 0

chloride ion Cl- -1

hydroxide ion OH- -1

thiosulphate ion S2O32- -2

ethanedioate ion C2O42- -2

bis[di(carboxymethyl)amino]ethane EDTA4- -4

Work out the oxidation number of the transition metal in each of the following complexes.

1. Cr2O72- …………………………………………………………………..

2. Ni(CO)4 ………………………………………………………………….

3. [CoCl4]2- ………………………………………………………………….

4. [Co(H2O)6]2+ ………………………………………………………………….

5. VO2+ ………………………………………………………………….

6. VO2+ ………………………………………………………………….

7. [Fe(CN)6]4- ………………………………………………………………….

8. [Ag(S2O3)2]3- ………………………………………………………………….

9. [Fe(H2O)4(OH)2]+ ………………………………………………………………….

10. [Cu(NH3)4(H2O)2]2+ ………………………………………………………………….

11. [Ag(NH3)2]+ ………………………………………………………………….


12. [CuCl2]- ………………………………………………………………….

13. Co(en)2Cl2 ………………………………………………………………….

14. [Fe(C2O4)3]3- ………………………………………………………………….

15. [Cu(EDTA)]2- ………………………………………………………………….

16. [Cr(OH)6]3- ………………………………………………………………….

17. [Cr(H2O)3(OH)3] ………………………………………………………………….

18. [Ag(CN)2]- ………………………………………………………………….

19. [Zn(OH)4]2- ………………………………………………………………….

20. [VCl2(H2O)4]+ ………………………………………………………………….


REDOX EQUATIONS

Constructing Half –Equations

The half-equation shows either the oxidation or the reduction step of a redox change. In a half-equation:

only one element changes its oxidation state

the atoms of each element must balance

the total charge on both sides of the equation must be the same

When constructing half-equations for reactions which take place in aqueous solution:

Water can be used as a source of oxygen atoms on the reactant side of the equation, the hydrogen appearing in the products as H+.

Any extra oxygen atoms on the reactant side of the equation can be converted into water by reaction with hydrogen ions from an acid.

The oxidation numbers of hydrogen and oxygen do not change.

Examples:


1. Deduce the half-equation for the reduction of VO3- to V2+ in acid

solution.

Vanadium is changing its oxidation state from +5 in VO3- to +2 in V2+ and

therefore needs to gain 3e-.VO3

- + 3e- V2+

The acid solution provides the six hydrogen ions needed to react with the three extra oxygen atoms in VO3

- to form water.

VO3- + 6H+ + 3e- V2+ + 3H2O

Finally, a check shows that the total charge on each side of the equation is the same.

l.h.s. -1 + 6 - 3 = +2 r.h.s. +2


2. Deduce the half-equation for the reduction of MnO4- to Mn2+ in acid solution.

Manganese is changing its oxidation state from +7 in MnO4- to +2 in Mn2+

and therefore needs to gain 5e-.MnO4

- + 5e- Mn2+

The acid solution provides the eight hydrogen ions needed to react with the four extra oxygen atoms in MnO4

- to form water.

MnO4- + 8H+ + 5e- Mn2+ + 4H2O


l.h.s. -1 + 8 - 5 = +2 r.h.s. +2

3. Deduce the half-equation for the reduction of NO3- to NO in acid solution.

Nitrogen is changing its oxidation state from +5 in NO3- to +2 in NO and

therefore needs to gain 3e-.NO3

- + 3e- NO

The acid solution provides the four hydrogen ions needed to react with the two extra oxygen atoms in NO3

- to form water.

NO3- + 4H+ + 3e- NO + 2H2O


l.h.s. -1 + 4 - 3 = 0 r.h.s. 0

Combining Half –Equations

The overall equation for a redox change can be obtained by combining the half-equations for the oxidation and reduction steps in such a way that the number of electrons donated by the reducing agent is equal to the number accepted by the oxidising agent.

Any molecules or ions which appear on both sides of the overall equation, such as H+

and H2O, then need to be cancelled down.

Examples:


4. Manganate(VII) ions (MnO4-) oxidise iron(II) ions (Fe2+) in acid solution,

forming iron(III) ions (Fe3+), and are reduced to manganese(II) ions (Mn2+). Deduce the two half-equations for this reaction, and hence derive an overall equation.

Manganese is changing its oxidation state from +7 in MnO4- to +2 in Mn2+ and

therefore needs to gain 5e-.MnO4

- + 5e- Mn2+

The acid solution provides the eight hydrogen ions needed to react with the four extra oxygen atoms in MnO4

- to form water.

MnO4- + 8H+ + 5e- Mn2+ + 4H2O

Iron is changing its oxidation state from +2 in Fe2+ to +3 in Fe3+ and therefore needs to lose 1e-.

Fe2+ Fe3+ + e-

To balance the electrons, the first half-equation needs to be multiplied x1:

MnO4- + 8H+ + 5e- Mn2+ + 4H2O

and the second x5:5Fe2+ 5Fe3+ + 5e-

The two half-equations are now added together:

5Fe2+ + MnO4- + 8H+ + 5e- Mn2+ + 4H2O + 5Fe3+ + 5e-

The electrons cancel out to give the overall equation:

5Fe2+ + MnO4- + 8H+ Mn2+ + 4H2O + 5Fe3+


l.h.s. +10 -1 + 8 = +17 r.h.s. +2 +15 = +17


5. Dichromate(VI) ions (Cr2O72-) oxidise sulphate(IV) ions (SO32-) in acid solution, forming sulphate(VI) ions (SO42-), and are reduced to chromium(III) ions (Cr3+). Deduce the two half-equations for this reaction, and hence derive an overall equation.

Chromium is changing its oxidation state from +6 in Cr2O72- to +3 in Cr3+ and

therefore needs to gain 6e-.

Cr2O72- + 6e- 2Cr3+

The acid solution provides the fourteen hydrogen ions needed to react with the seven extra oxygen atoms in Cr2O7

2- to form water.

Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O

Sulphur is changing its oxidation state from +4 in SO32- to +6 in SO4

2- and therefore needs to lose 2e-.

SO32- SO4

2- + 2e-

One water molecule is needed to provide the extra oxygen atom:

SO32- + H2O SO4

2- + 2H+ + 2e-


Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O

and the second x3:

3SO32- + 3H2O 3SO4

2- + 6H+ + 6e-


3SO32-+ 3H2O + Cr2O7

2- + 14H+ + 6e- 2Cr3+ + 7H2O + 3SO42- + 6H+ + 6e-

The electrons cancel out, and the numbers of H+ ions and H2O molecules are simplified to give the overall equation:

3SO32- + Cr2O7

2- + 8H+ 2Cr3+ + 4H2O + 3SO42-


l.h.s. -6 -2 + 8 = 0 r.h.s. +6 -6 = 0

Sometimes the same species oxidises and reduces itself simultaneously. This process is called disproportionation.

ELECTRODE POTENTIALSIf a metal is placed in water or a solution of one of its salts, the metal tends to throw off positive ions into the liquid. For example:

Zn(s) + aq. Zn2+(aq) + 2e-


6. Under alkaline conditions, chlorine disproportionates to form chloride ions (Cl-) and chlorate(V) ions (ClO3

-). Deduce the two half-equations for this reaction, and hence derive an overall equation.

Chlorine is changing its oxidation state from 0 in Cl2 to -1 in Cl- and therefore needs to gain 1e-.

½ Cl2 + e- Cl-

Chlorine is changing its oxidation state from 0 in Cl2 to +5 in ClO3- and therefore

needs to lose 5e-.1/2 Cl2 ClO3

- + 5e-

Six hydroxide ions in the alkaline solution provide the three extra oxygen atoms needed to form ClO3

-, and also form water.

1/2 Cl2 + 6OH- ClO3- + 3H2O + 5e-


21/2 Cl2 + 5e- 5Cl-

and the second x1:

1/2 Cl2 + 6OH- ClO3- + 3H2O + 5e-


3Cl2 + 6OH- + 5e- 5Cl- + ClO3- + 3H2O + 5e-

The electrons cancel out to give the overall equation:

3Cl2 + 6OH- 5Cl- + ClO3- + 3H2O


l.h.s. -6 r.h.s. -1 -5 = -6

When positive ions are given off into the liquid, an excess of electrons is left on the surface of the metal, which therefore acquires a negative charge. Owing to the attraction of opposite charges, the hydrated cations remain close to the metal and an electric double layer is formed.

There is also a tendency for ions in the solution to be deposited on the metal:

Zn2+(aq) + 2e- Zn(s)

so that eventually an equilibrium is established; the rate of formation of the ions is equal to the rate of deposition:


The formation of the electric double layer causes a potential difference between the surface of the metal and the liquid; this is called the electrode potential.

The further to the left the above equilibrium lies, the greater the potential difference will be.

For a given metal, the position of the equilibrium depends on the concentration of metal ions already present in the solution; the more dilute the solution, the higher the electrode potential will be.

For different metals placed in solutions containing equal concentrations of their ions at the same temperature, the position of the equilibrium, and therefore the electrode potential, is governed by the overall energy change associated with the formation of hydrated ions from the metal. This can be broken down into a series of energy changes:


-----

-----

-----

-----

Zn2+

Zn2+

Zn2+

Zn2+

Zn2+

Zn2+

Zn2+

Zn2+

Zn2+

Zn2+

Zn

Energy change Na Mg Zn CuHsub +109 + 150 + 130 + 339

HI +494 + 736+1450

+ 908+1730

+ 745+1960

Hhyd -406 -1920 -2050 -2100H +197 + 416 + 718 + 944

As a result of these energy changes, from Na to Cu, the dissociation into ions will be less extensive. Therefore, the electrode potential will decrease from Na to Cu.

ELECTRODESThere are three types of electrodes:

Metal ElectrodeA metal electrode consists of a metal dipping into a solution of its ions in which the following type of equilibrium is established:


Metal electrodes are represented by convention as, for example

Zn2+(aq) Zn(s)

where the vertical line represents a phase boundary.

The redox couple (e.g. Zn2+/Zn), which is always written with the reduced species on the right, is a convenient shorthand way of representing the half reaction.

Gas ElectrodeA gas electrode consists of a gas in equilibrium with a solution of its ions in the presence of an inert metal, usually platinum. An example of a gas electrode is the hydrogen electrode (H+/H2).

Pt(s) H2(g) H+(aq)

By convention, the reduced species is written nearer to the Pt.Redox ElectrodeTOPIC 13.20: ELECTRODE POTENTIALS 13

M(s) + aq.

M(g) + aq.

Mn+(g) + ne- + aq.

Mn+(aq) + ne-

HHsub

HI

Hhyd

Energy

A redox electrode consists of an element in two different oxidation states in the presence of an inert metal, usually platinum. An example of a redox electrode is Fe3+/Fe2+.

Pt(s) Fe2+(aq), Fe3+(aq)

By convention, the reduced species is written nearer to the Pt. The two different ions of iron are in the same phase and are separated by a comma.

Half-Equation Convention The reaction occurring in an electrode is represented by a half-equation. By IUPAC convention, redox half-equations are written as reductions, for example:


Electrochemical Cells An electrode is half a cell. If two different electrodes are connected together, an electrochemical cell is formed. Usually, two electrode compartments are joined by a salt bridge. A salt bridge allows ions to flow, thus completing the electrical circuit, but prevents the ions in the two electrode compartments from mixing to any great extent. A salt bridge consists usually of a saturated solution of KNO3 which may or may not be thickened with gelatine or agar. A salt bridge is written in a cell representation as a double vertical line.

The diagram below shows an electrochemical cell constructed from two metal electrodes.

One electrode will form the negative terminal of the cell, the other the positive terminal. When the cell is operating, electrons flow in the external circuit from the negative terminal to the positive terminal. Therefore an oxidation reaction occurs at the negative terminal and a reduction reaction at the positive terminal. The reactions occurring in this particular cell are:

Cu2+(aq) + 2e- Cu(s)

Zn(s) Zn2+(aq) + 2e-

Measurement of Electrode PotentialsThe electrode potential cannot be measured absolutely. Measurement of the potential difference between, for example, a metal and the solution of its ions necessitates making an electrical connection between the metal and its solution. The connection to the solution could only be made by immersing some other metal in the solution, which would simply introduce another electrode potential.


copper

copper sulphate

zinc

zinc sulphate

salt bridge

Only potential difference can be measured, and it is therefore necessary to define a standard electrode, relative to which all other potentials are measured. The electrode used for this purpose is the STANDARD HYDROGEN ELECTRODE (S.H.E.).

Pt(s) H2(g,1bar) H+(aq, 1.0 mol.dm-3) E = 0V at 298K

In this gas electrode, hydrogen at a pressure of 1 bar is bubbled over a strip of platinum coated in platinum black and immersed in an acid solution containing a hydrogen ion concentration of 1.0 mol.dm-3. Temperature = 298K

The following equilibrium is established:

H+(aq) + e- 1/2H2(g)

To measure the standard electrode potential of a particular electrode, it is connected to a S.H.E. to form an electrochemical cell. The cell potential is then measured. If the electrode under test forms the negative pole of the cell, its standard electrode potential is negative; if it forms the positive pole, the standard electrode potential is positive. The following standard conditions are required:

Zero current. This is achieved by measuring the cell potential with a high resistance (solid-state) voltmeter.

Ion concentrations of 1.0 mol.dm-3

Temperature 298K Pressure 100 kPa (1 bar)

The S.H.E. is made the left hand electrode and the unknown the right hand electrode. If the voltmeter is connected as indicated, the electrode potential and its sign can be read directly from the voltmeter. In this case, the reading is -0.76V. Therefore the standard electrode potential of Zn2+/Zn is -0.76V.


Zn

[Zn2+] = 1.0 mol.dm-3[H+] = 1.0 mol.dm-3

H2 at 100kPa

Pt coated in Pt black

salt bridge – KNO3 soln.

T = 298K

V +-

high resistance voltmeter

If a current is allowed to flow in the above cell, electrons will flow from the negative electrode (Zn2+/Zn) to the positive electrode (H+/H2). Therefore zinc is oxidised and hydrogen reduced; the reactions occurring in the cell are:

Zn(s) Zn2+(aq) + 2e-

2H+(aq) + 2e- H2(g)

The overall cell reaction is: Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

Secondary Standard ElectrodesThe hydrogen electrode is difficult to set up and use. Therefore, it is often more convenient to use a secondary standard electrode, whose potential with respect to the S.H.E. is accurately known. The electrode most frequently used as a secondary standard electrode is the calomel electrode:

Pt(s) Hg(l) Hg2Cl2(s), KCl(aq) Eo = +0.27V

Cell Representations The more negative electrode is drawn on the left Phase boundaries are shown as single vertical lines A salt bridge is shown as a double vertical line Substances in the same phase are separated by a comma In gas electrodes and redox electrodes, where an inert metal is present, the

component in the reduced state is written nearest to the inert metal


Zn2+/Zn SHE calomel Cu2+/Cu

-0.76 0 +0.27 +0.34 VEcell = +0.76V

Ecell = +1.03V

Ecell = +0.07V

Ecell = +0.34V

Example 2: A cell comprises the electrodes Fe3+/Fe2+ and H+/H2. Write the representation of this cell.

The hydrogen electrode is more negative and is shown on the left of the cell.

Pt(s) H2(g) H+(aq) Fe3+(aq), Fe2+(aq) Pt(s)

Example 1: A cell comprises the electrodes Mg2+/Mg and Sn4+Sn2+. Write the representation of this cell.

The magnesium electrode is more negative and is shown on the left of the cell.

Mg(s) Mg2+(aq) Sn4+(aq), Sn2+(aq) Pt(s)

Calculating Cell e.m.f.To calculate the e.m.f. of an electrochemical cell made from any two electrodes, the cell is first arranged so that the electrode with the more positive potential is on the right hand side. Then

where ER and EL are the reduction potentials of the right and left electrodes.


Eocell = Eo

R - EoL

Example: Calculate the standard e.m.f. of a cell with copper (Cu2+/Cu; Eo = +0.34V) and zinc (Zn2+/Zn; Eo = -0.76V) electrodes. Write the representation of the standard cell.

The more positive electrode (Cu2+/Cu) is made the right hand electrode, then

Eocell = Eo

R - EoL

Eocell = +0.34 - (-0.76) V

Eocell = +1.10V

Cell representation:

The zinc electrode is more negative and is shown on the left of the cell.

Zn(s) Zn2+(aq, 1.0mol.dm-3) Cu2+(aq, 1.0mol.dm-3) Cu(s)

Electrochemical Series

The list is presented either in descending order, with the most positive potentials first, or in ascending order, with the most negative potentials first. Reactions are written as reductions. Part of the electrochemical series is given below.

Reduction half-equation Eo /V

F2(g) + 2e- 2F-(aq) +2.85MnO4

-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) +1.51Cl2(g) + 2e- 2Cl-(aq) +1.36Cr2O7

2-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + 7H2O(l) +1.33MnO2(s) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) +1.23Br2(g) + 2e- 2Br-(aq) +1.09Cu2+(aq) + I-(aq) + e- CuIs) +0.86Ag+(aq) + e- Ag(s) +0.80Fe3+(aq) + e- Fe2+(aq) +0.77I2(g) + 2e- 2I-(aq) +0.54Cu+(aq) + e- Cu(s) +0.52Cu2+(aq) + 2e- Cu(s) +0.34Hg2Cl2(aq) + 2e- 2Hg(l) + 2Cl-(aq) +0.27AgCl(s) + e- Ag(s) + Cl-(aq) +0.22Cu2+(aq) + e- Cu+(aq) +0.15Sn4+(aq) + 2e- Sn2+(aq) +0.152H+(aq) + 2e- H2(g) -0.00Pb2+(aq) + 2e- Pb(s) -0.13Sn2+(aq) + 2e- Sn(s) -0.14Ni2+(aq) + 2e- Ni(s) -0.25V3+(aq) + e- V2+(aq) -0.26Cr3+(aq) + e- Cr2+(aq) -0.41Fe2+(aq) + 2e- Fe(s) -0.44Zn2+(aq) + 2e- Zn(s) -0.76 Al3+(aq) + 3e- Al(s) -1.66Mg2+(aq) + 2e- Mg(s) -2.36Na+(aq) + e- Na(s) -2.71Ca2+(aq) + 2e- Ca(s) -2.87K+(aq) + e- K(s) -2.92Li+(aq) + e- Li(s) -3.05


The electrochemical series is a list of standard electrodepotentials in order of their voltages.

better oxidising

agent

better reducing

agent

Spontaneous ReactionsThe feasibility of a reaction is governed by the sign of the free energy change (G) for the reaction; a spontaneous change will occur only if G is negative. G is related to the e.m.f. of a cell (E):

G = -nEFF and n are both positive constants, therefore a cell reaction is spontaneous (G is negative) only if Ecell is positive.

Ecell = ER - EL

Ecell will only be positive if ER is more positive than EL. In other words, the right-hand electrode must be the site of reduction and the left-hand electrode the site of oxidation.

Deducing Cell Reactions & Cell e.m.f.Cell reactions and the cell e.m.f. can be deduced from the electrochemical series.

Example: Deduce the cell reactions and the cell e.m.f. for a cell comprising the electrodes Cl2/Cl- and V3+/V2+

Zn2+(aq) + 2e

- Zn(s)

V3+(aq) + e

- V2+(aq)

Fe3+(aq) + e

- Fe2+(aq)

Cl2 (g) + 2e

- 2Cl -(aq)

MnO

4 -(aq) + 8H+(aq) + 5e

- Mn

2+(aq) + 4H2 O

(l)

-0.76

-0.26

+0.77

+1.36

+1.51

2V2+(aq) + Cl2(g) 2V3+(aq) + 2Cl-(aq)

Ecell = ER - EL

Ecell = +1.36 - (-0.26)

= 1.62V Predicting the Feasibility of a ReactionTOPIC 13.20: ELECTRODE POTENTIALS 19

The electrochemical series is first rotated so that the more positive potentials are to the r.h.s.

The more positive potential (Cl2/Cl-) forms the positive pole of the cell; reduction takes place here.

The more negative potential (V3+/V2+) forms the negative pole of the cell; oxidation takes place here.

At the +ve (right) electrode:Cl2(g) + 2e- 2Cl-(aq)

At the -ve (left) electrode:V2+(aq) V3+(aq) + e-

The reduction equation given in the table needs to be reversed.

To balance the electrons, the second equation must be multiplied by 2:

2V2+(aq) 2V3+(aq) + 2e-

The two equations are then added together:

+-

Electrode potentials can be used to predict the feasibility of redox changes. The change can be broken down into two half-reactions, one of which is a reduction, the other an oxidation. These two electrodes can be used to construct a cell, the reduction step occurring at the right-hand electrode. Ecell is then calculated. If Ecell is positive, the change is feasible.


Example 1: Is the disproportionation of copper(I) to copper(II) and copper a spontaneous reaction?

The reaction is: 2Cu+(aq) Cu2+(aq) + Cu(s)

This redox change can be broken down into two half-reactions:

1. Cu+(aq) Cu2+(aq) + e- Eo = +0.15V

2. Cu+(aq) + e- Cu(s) Eo = +0.52V

Reaction 1. is an oxidation and must be written as the left-hand electrode.Reaction 2. is a reduction and must be written as the right-hand electrode.

Ecell = ER - EL

Ecell = +0.52 - (+0.15)

Ecell = +0.37V

Since the cell e.m.f. is positive, the reaction will go spontaneously.

Example 2: Will copper react spontaneously with an aqueous acid to form hydrogen?

The reaction is: Cu(s) + 2H+(aq) Cu2+(aq) + H2(g)


1. Cu(s) Cu2+(aq) + 2e- Eo = +0.34V

2. 2H+(aq) + 2e- H2(g) Eo = +0.00V

Reaction 1. is an oxidation and must be written as the left-hand electrode.Reaction 2. is a reduction and must be written as the right-hand electrode.

Ecell = ER - EL

Ecell = +0.00 - (+0.34)

Ecell = -0.34V

Since the cell e.m.f. is negative, the reaction will not go spontaneously.

Example 3: Will dichromate(VI) ions in acid solution oxidise chloride ions spontaneously under standard conditions?

The reaction is:6Cl-(aq) + Cr2O7

2-(aq) + 14H+(aq) 2Cr3+(aq) + 7H2O(l) + 3Cl2(g)


1. Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + 7H2O(l) Eo = +1.33V

2. 6Cl-(aq) 3Cl2(g) + 6e- Eo = +1.36V

Reaction 1. is a reduction and must be written as the right-hand electrode.Reaction 2. is an oxidation and must be written as the left-hand electrode.

Ecell = ER - EL

Ecell = +1.33 - (+1.36)

Ecell = -0.03V

Since the cell e.m.f. is negative, the reaction will not go spontaneously.

Example 4: Will dichromate(VI) ions in acid solution oxidise bromide ions spontaneously under standard conditions?

The reaction is:6Br-(aq) + Cr2O7

2-(aq) + 14H+(aq) 2Cr3+(aq) + 7H2O(l) + 3Br2(g)


1. Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + 7H2O(l) Eo = +1.33V

2. 6Br-(aq) 3Br2(g) + 6e- Eo = +1.09V

Reaction 1. is a reduction and must be written as the right-hand electrode.Reaction 2. is an oxidation and must be written as the left-hand electrode.

Ecell = ER - EL

Ecell = +1.33 - (+1.09)

Ecell = +0.24V

Since the cell e.m.f. is positive, the reaction will go spontaneously.

Commercial Cells

Batteries and cells

There are three main types of commercial electrochemical cell:


• Primary cells are not rechargeable and are thrown away after they run down. These types of cell are commonly used in torches or remote controls.

• Secondary cells can be recharged after they run down. These are found in mobile phones and mp3 players.

• Fuel cells produce electricity from gaseous or liquid fuels.

1) Primary cellsEarly designs of primary cells were wet cells. The Daniell cell was invented by John Daniell in 1836. It consists of a piece of zinc dipped into aqueous zinc sulphate. The zinc acts as the anode. Electrons are lost and the zinc is oxidized:

Zn(s) → Zn2+ (aq) + 2e-

The copper acts as the cathode. Electrons are gained and copper(II)

ions

are reduced:

Cu2+ (aq) + 2e-→ Cu(s)

The overall cell reaction is

Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s)

with an Ecell of 1.1 V. This is however not very portable because of the liquid electrolytes it contains. Most modern primary cells are dry cells instead like the zinc-carbon cells you are most familiar with.

Zinc–carbon cells

Zinc–carbon cells are cheap to make but easily run down under heavy use. They produce a potential difference of about 1.5 V, which gradually reduces to around 0.8 V with use.


Anode reactionZinc reacts with ammonia formed from the ammonium chloride electrolyte.

Zn(s) + 2NH3(aq) → [Zn(NH3)2]2+(aq) + 2e-

Cathode reactionsThe reaction involves hydrogen ions released from the ammonium ions:

2MnO2(s) + 2H+(aq) + 2e- → Mn2O3(s) + H2O(l)

The overall cell reaction is:

Zn(s) + 2MnO2(s) + 2NH4+(aq) → [Zn(NH3)2]2+ (aq) + Mn2O3(s) + H2O(l)

Alkaline dry cells

Alkaline cells produce the same potential difference as zinc–carbon cells but they last longer. They contain potassium hydroxide as the electrolyte rather than ammonium chloride.

Anode reactionsAt the anode, hydroxide ions react with the zinc ions:

Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2e−

Cathode reactions

2MnO2(s) + H2O(l) + 2e− → Mn2O3(s) + 2OH-(aq)

This is the overall cell reaction:TOPIC 13.20: ELECTRODE POTENTIALS 23

Zn(s) + 2MnO2(s) → ZnO(s) + 2 Mn2O3 (s)

2) Rechargeable cells

Rechargeable cells are secondary cells. They must be charged before use by connecting them to the electricity supply. The lead–acid battery, the type used in cars, is the oldest design.

The lead–acid battery

A typical car battery comprises six cells in series, each producing 2V, giving a total voltage of 12V. The electrolyte is 6M H2SO4.

Anode reactionsThis is the reaction that happens at the anode when the battery discharges:

Pb(s) + SO42-(aq) → PbSO4(s) + 2e-

The reaction is reversed during charging:

Cathode reactionsThis is the reaction that happens at the cathode when the battery discharges:PbO2(s) + 4H+(aq) + SO4

2-(aq) + 2e- → PbSO4(s) + 2H2O(l)


Nicads

The most common rechargeable cell in everyday use is the nickel–cadmium cell, often just called a nicad. The anode is made from


cadmium and the cathode from nickel(III) hydroxide. Potassium hydroxide is the electrolyte.

Anode reactionsThis is the reaction that happens at the anode when the battery discharges:

Cd(s) + 2OH-(aq) → Cd(OH)2(s) + 2e-


Cathode reactionsThis is the reaction that happens at the cathode when the battery discharges:

NiO(OH)(s) + e- + H2O → Ni(OH)2(s) + OH- (aq)


Write the overall reaction in the space below

……………………………………………………………………………………

……………………………………………………………………………………

Lithium-ion batteries

Lithium-ion batteries are used in portable devices such as mobile phones and laptop computers. They are more complex than the other secondary TOPIC 13.20: ELECTRODE POTENTIALS 25

cells described here, and include a computer chip to control charging and discharging. They produce a much higher potential difference, too, typically 3.7 V. The anode is made of graphite and the cathode is madeof lithium cobalt oxide, LiCoO2.

Anode reactions

This is an example of the reaction that happens at the anode when the battery discharges:

Li+ + CoO2 + + e- → Li+(CoO2)-


Cathode reactionsThis is an example of the reaction that happens at the cathode when the battery discharges:

Li → Li+ + e-


3) Fuel Cells The alkaline hydrogen–oxygen fuel cell

Fuel cells transform the chemical energy in a fuel such as hydrogen or methanol directly into electrical energy. The fuel is oxidized by oxygen from TOPIC 13.20: ELECTRODE POTENTIALS 26

the air using electrochemical reactions in the fuel cell. A typical alkaline hydrogen–oxygen fuel cell comprises two flat electrodes, each coated on one side by a thin layer of platinum catalyst. A semi-permeable membrane is sandwiched between the two electrodes and the electrolyte between them is sodium hydroxide solution. Hydroxide ions produced at the cathode travel across the membrane to the anode.

Anode reactionThe oxidation of hydrogen is catalysed by the layer of platinum:

H2(g) + 2OH- → 2H2O(l) + 2e-

The electrons released by this reaction flow through the external circuit. The hydrogen ions pass through the proton exchange membrane to the cathode.

Cathode reactionOxygen is reduced to water vapour. It reacts with the hydrogen ions that pass through the proton exchange membrane and electrons from the external circuit:

2H2O(aq) + 4e- + O2(g) → 4OH-(aq)

Overall reaction

The overall reaction is: 2H2(g) + O2(g) → 2H2O(g)

Benefits and drawbacks of electrochemical cells


Electrochemical cells are a very convenient, portable sources of electricity. They reduce the need for expensive cabling and bring electricity supplies to remote places. Spacecraft use hydrogen fuel cells to provide electricity and drinking water.

Non-rechargeable cells are cheap, but are usually thrown away. This wastes the energy and resources needed to manufacture them and they usually go to landfill.

The use of rechargeable cells reduces the total number of batteries thrown away each year. Rechargeable cells also improve the performance of solar cells. These convert sunlight directly into electricity but they do not work at night. Rechargeable cells store electricity generated during the day and release it at night. The cadmium in nickel–cadmium cells is very toxic. These cells should be recycled rather than disposed of in a landfill site.

Water vapour is the only waste product of hydrogen–oxygen fuel cells. These cells may replace petrol and diesel engines in the future, reducing the amount of carbon dioxide produced by vehicles. This will help to reduce the release of greenhouse gases, if the hydrogen is produced without fossil fuels. But most industrial hydrogen is produced using these fuels at the moment. It could be produced by electrolysis of water but this uses electricity which has been generated by burning of fossil fuels. Hydrogen itself is highly flammable, and is difficult to store and handle safely.


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