Chapter 18 Electrochemistry

Chapter 18 Electrochemistry Definition The study of the interchange of chemical and electrical energy in oxidation-reduction (redox) reactions This interchange can occur in both directions: 1. Conversion of Chemical Energy into Electrical Energy: an electrical current is generated from a spontaneous chemical reaction (galvanic cell) 2. Conversion of Electrical Energy into Chemical Energy: a chemical reaction is induced by an electrical current (electrolysis) Oxidation-Reduction (Redox) Reactions Reactions in which electrons are transferred from one substance to another The substance that loses electrons undergoes oxidation and is said to be oxidized The substance that gains electrons undergoes reduction and is said to be reduced A simple way to remember: OIL RIG Oxidation Is Loss of electrons Reduction Is Gain of electrons Redox reactions most commonly occur between metals and non-metals in the formation of ionic compounds 1

2Mg(s) + O 2 (g) 2MgO(s) Oxidizing Agent A substance that contains an atom that accepts electrons (from an atom in another substance that is being oxidized) and is reduced in the process Electrons are transferred from Mg to O 2 Mg is oxidized to Mg 2+ O 2 is reduced to O 2- Reducing Agent A substance that contains an atom that donates electrons (to an atom in another substance that is being oxidized) and is oxidized in the process 2Mg(s) + O 2 (g) 2MgO(s) Summary Mg donates electrons so it is the reducing agent O 2 accepts electrons so it is the oxidizing agent Oxidation States (Numbers) Molecules can also undergo redox reactions: CH 4 (g) + 2O 2 (g) CO 2 (g)+ 2H 2 O(g) 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (aq) + 6O 2 (g) However, it is not clear how the electrons are transferred and which species are oxidized and which are reduced! Need a general way to follow the transfer of electrons in any reaction! A systematic way to keep track of electrons in redox reactions containing covalent compounds Numbers are assigned to each atom in a compound which represent the imaginary charge each would have if the electrons in the molecule were distributed according to their ability to attract electrons (electronegativity) For binary ionic compounds, the oxidation state of each ion is equal to its charge 2

For a neutral compound, the sum of oxidations states is zero For an ion, the sum of oxidation states is equal to the charge on the ion Application to Redox Reactions In all redox reactions: one atom loses electrons and has its oxidation increase (is oxidized) and another gains electrons and has its oxidation state decrease (is reduced) the number of electrons lost by the atom being oxidized must be equal to the number of electrons gained by the atom being reduced if the oxidation states do not change then the reaction is not a redox reaction! Identify the atoms or ions which are oxidized and reduced and identify the oxidizing and reducing agents: 16H + (aq) + 2Cr 2 O 2-7 (aq) + C 2 H 5 OH(l) 4Cr 3+ (aq) + 11H 2 O(l) + 2CO 2 (g) Half-Reactions A redox reaction can be split into two half-reactions, one oxidation, one reduction: Example: Mg(s) + Fe 2+ (aq) Mg 2+ (aq) + Fe(s) Oxidation half-reaction: Mg(s) Mg 2+ (aq) + 2e - Reduction half-reaction: Fe 2+ (aq) + 2e - Fe(s) 3

Luigi Galvani (1737-1798) Galvanic Cells If we physically separate one half-reaction from another other and then provide a wire path for the electrons to flow across, the redox reaction can still proceed Since the redox reaction is causing electrons to flow it is generating an electric current which can then be passed through a device (e.g. a light bulb) to produce useful work Such an arrangement is referred to a galvanic, voltaic or electrochemical cell Batteries are examples of galvanic cells Separate the two half-reactions and connect them with electrodes and a wire: Example: Reaction between MnO 4- and Fe 2+ in acidic solution: 5Fe 2+ (aq) + MnO 4- (aq) + 8H + (aq) 5Fe 3+ (aq) + Mn 2+ (aq) + 4H 2 O(l) Broken down into half-reactions: Fe 2+ (aq) Fe 3+ (aq) + e - MnO 4- (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O(l) Anode (-) Fe 2+ (aq) Fe 3+ (aq) + e - Oxidation takes place Electrons are produced Problem: electron flow would create a charge separation which would require an huge amount of energy to maintain Cathode (+) MnO 4- (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O(l) Reduction takes place Electrons are used up Electrons flow along the wire from anode (-) to cathode (+) To remember the correct polarity think: anode = anion = negative (-) cathode = cation = positive (+) 4

Solution: solutions must be connected (without mixing) so that ions can also flow to keep the charge neutral This is accomplished by connecting the two solutions with a salt bridge (a U-shaped tube filled with a strong electrolyte) or a porous disk in a connecting tube which allows ion flow but prevents mixing This allows the electrons to flow through the wire from the anode (-) to the cathode (+), while the ions flow from one compartment to the other to keep the net charge zero Direction of Electron Flow The direction of electron flow can be predicted by the activity series Electrons always flow from the most active species (most powerful reducing agent) to the least active species (least powerful reducing agent) Therefore the electrode containing the most active metal becomes the anode (-) and the electrode containing the least active metal becomes the cathode (+) Zn is more active than Cu Metal Activity Series anode(-) cathode(+) 5

Mg is more active than Zn anode(-) cathode(+) Cell Potential In a galvanic cell, we can think of the reducing agent at the anode literally pushing the electrons through the wire connecting the electrodes to the oxidizing agent at the cathode The strength of this push on the electrons is referred to as the cell potential (E cell ) or electromotive force (emf) of the cell and is determined by the difference in activity (potential difference) between the substances at the anode and the cathode The unit of electrical potential is the volt (V), which measures the number of joules of work produced per coulomb (C) of charge transferred Alessandro Volta (1745-1827) Cell potentials are commonly measured with a digital voltmeter: Half-reaction Potentials It is possible to determine the potential of a half-reaction individually by constructing a galvanic cell in which one of the electrodes is a standard hydrogen electrode consisting of a chemically inert platinum electrode in contact with 1 M H + ions (e.g. 1 M HCl) bathed in H 2 gas at 1 atm pressure The potential of a half-reaction is measured by arbitrarily assigning a voltage of exactly 0 volts to the standard hydrogen electrode If the constituents of the half-reaction are also in their standard states (molar concentrations, 1 atm pressures) then: E cell0 = E H20 + E half-reaction0 = 0 + E half-reaction 0 E half-reaction0 = E cell 0 In this way the potential (relative to the standard hydrogen electrode) of any half-reaction can be easily measured 6

2H + (aq) + Zn(s) Zn 2+ (aq) + H 2 (g) Standard Reduction Potentials (E 0 ) By convention, the potentials of half-reactions are always written as reduction processes The E 0 values corresponding to these reduction reactions under standard conditions (solutes at 1 M and gases at 1 atm) are referred to as standard reduction potentials Elements, compounds or ions which are reduced more easily than hydrogen have positive E 0 values while elements, compounds or ions which are reduced less easily than hydrogen have negative E 0 values The more positive the reduction potential, the easier the species is to reduce and the more powerful an oxidizing agent it is E cell0 = 0.76 V = E H20 + E Zn Zn2+0 = 0.00 + E Zn Zn2+ 0 E Zn Zn2+0 = 0.76 V The more negative the reduction potential, the more difficult the species is to reduce and the more powerful a reducing agent it is The order of standard reduction potentials is opposite to the activity series which are always written as oxidation processes Value of Cell Potential The value of the cell potential in any galvanic cell is equal to the sum of standard reduction potentials of the two half reactions E cell0 = E Zn Zn2+0 + E Cu2+ Cu0 = 0.76 V + 0.34 V = 1.10 V Calculating Cell Potentials 1. Write the two half-reactions and their standard potentials 2. The half-reaction with the largest positive potential is left unchanged while the other half-reaction is written in reverse (since redox reactions must involve both oxidation and reduction) and the sign of its potential changed. This is done because a cell will always run spontaneously to produce a positive cell potential 3. Balance the number of electrons lost and the number of electrons gained by multiplying by integers (the potentials are not changed since they are intensive) 4. Add the two balanced half-reactions to give the balanced cell reaction 5. Add the two potentials together to obtain the overall cell potential. Since reduction occurs at the cathode and oxidation at the anode we can write: E cell0 = E 0 (cathode) + [-E 0 (anode)] 7

General Rule: The anode compartment (-) will always contain the species with the smallest standard reduction potential since it is most easily oxidized and wants to give up electrons The cathode (+) compartment will always contain the species with the largest standard reduction potential since it is most easily reduced and wants to accept electrons This is why electrons flow from the anode to cathode! 8

Line Diagrams Substitute Electrodes In some half-cells there is no component that can be used as an electrode, e.g. if there are two ionic components. In such cases chemically inert electrodes made from platinum or graphite are used Example: 2MnO 4- (aq) + 6H + (aq) + 5ClO 3- (aq) 2Mn 2+ (aq) + 3H 2 O(l) + 5ClO 4- (aq) Half-reactions: MnO 4- (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O(l) ClO 3- (aq) + H 2 O(l) ClO 4- (aq) + 2H + (aq) + 5e - Line diagram: Pt(s) ClO 3- (aq),clo 4- (aq),h + (aq) MnO 4- (aq),h + (aq),mn 2+ (aq) Pt(s) or Pt ClO 3-, ClO 4-, H + MnO 4-, H +, Mn 2+ Pt Double vertical lines indicate the salt bridge or the porous disk while a single vertical line represents a phase boundary Note: coefficients, water and spectator ions are never written in line diagrams! State symbols are sometimes not included! A galvanic cell consists of a graphite electrode in an acidified plumbic oxide and plumbous sulfate solution and a graphite electrode in a solution of ferrous chloride and ferric chloride. Assume all components are in standard states. Write the half reactions, balanced reaction, line diagram and calculate the potential for the cell Write the half-reactions and the balanced reaction from the line diagram of the galvanic cell below: Ag(s) Cl - (aq), AgCl(s) NO 3- (aq), H + (aq), NO(g) C(s) Complete Description of a Galvanic Cell Given its half-reactions, a complete description of a galvanic cell includes a diagram showing the following information: 1. The cell potential 2. The direction of electron flow 3. Designation of the anode and cathode 4. The nature of each electrode (an inert conductor must be used if none of the substances in the half-reaction is a conducting solid) and the ions present in each compartment 5. A balanced equation for the redox reaction 6. A line diagram for the cell 9

Electrical Work from Galvanic Cells The amount of work that can be extracted from a galvanic cell depends on the on the potential difference between the anode and cathode and the amount of charge transferred: w = -qe where w is the work (in Joules), q the charge transferred (in Coulombs) and E is the cell potential (in Volts: 1 V = JC -1 ) For a cell to produce a current, the cell potential, E must be positive and the work, w must be negative (since useful work must flow out of the system into the surrondings) Michael Faraday (1791-1867) The maximum possible amount of work, w max will always be obtained at the maximum cell potential, E max : w max = -qe max However, the actual work obtained is always less than the maximum due to resistive losses as the current flows through the wire Cell Potential and Free Energy At constant temperature and pressure (Chapter 17): The Faraday (F) A constant equal to the charge on 1 mole of electrons 1 F = 96,485 Cmol -1 From this definition, the charge transferred in a galvanic cell can be expressed as follows: charge transferred = moles of electrons transferred x 1 Faraday q = nf w max = G For a galvanic cell: w max = -qe max so: G = -qe max since q = nf and omitting the subscript on E max : G = -nfe Under standard conditions: G 0 = -nfe 0 G = -nfe Conclusions: 1. The maximum cell potential is directly related to the free energy difference between the reactants and products in the cell 2. G for a cell can be determined experimentally from the measured cell potential, E 3. Consistent with the fact that cells run in the direction of positive E, since this corresponds to a negative G (the condition for spontaneity) 4. This formula also applies to half-reactions 10

Cell Potential and Concentration If the concentration of an ion in a galvanic cell is changed from its standard concentration (1 M), the cell potential will change from its standard cell potential Example: Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s)z E cell 0 = 0.78 V Increase [Cu 2+ ]: Forward reaction is favored so E cell > E cell 0 Increase [Fe 2+ ]: Reverse reaction is favored so E cell < E cell 0 Walther Hermann Nernst (1864-1941) 11

Cell Potential and Concentration The potential of a galvanic cell can be related to the concentrations of their components using the relationship between free energy and concentration: G = G 0 + RTln(Q) Since G = -nfe and G 0 = -nfe 0 : -nfe = -nfe 0 + RTln(Q) Dividing each side by nf: E =E 0 -(RT/nF)ln(Q) At 25 0 C, changing from natural to base 10 logarithms: The Nernst Equation For the general cell reaction: aa + bb cc + dd The Nernst equation at 25 o Cis: E = E 0 - (0.0591/n)log(Q) or: E = E 0 - (0.0591/n)log([C] c [D] d / [A] a [B] b ) E =E 0 - (0.0591/n)log(Q) Concentration Cells Galvanic cells in which both compartments have the same components but at different concentrations The current flows in the direction which will equalize the concentrations of the ions in the two cells - low concentration (anode) to high concentration (cathode): Anode (-): Ag Ag + + e - [Ag + ] increases Cathode (+): Ag + + e - Ag [Ag + ] decreases Ag Ag + (0.1 M) Ag + (1 M) Ag 12

The Nernst Equation for Concentration Cells For the general concentration cell: M(s) + M n+ (aq)(m 1 ) M n+ (aq)(m 2 ) + M(s) M 1 >M 2 The Nernst equation for a concentration cell at 25 o Cis: E = E 0 - (0.0591/n)log(Q) E 0 = 0.00 V so: E = - (0.0591/n)log([M n+ (aq)](m 2 ) / [M n+ (aq)](m 1 )) Note: the smallest concentration is always on top! Ion Selective Electrodes Since the cell potential depends on the concentration of reactants and products, measured potentials can be used to measure the concentration of ions The concentrations of other ions can be measured in a similar manner: Example a ph meter: Changes in Potential with Time The potential calculated from the Nernst equation is the maximum potential before any current flow has occurred As the cell discharges and current flows from anode to cathode, the concentrations will change (products increase, reactants decrease and Q will decrease) and therefore E cell will gradually decrease This will continue until the cell reaches equilibrium: Q = K and E cell = 0 A dead battery is one in which the cell reaction has reached equilibrium so there is no longer any chemical driving force pushing the electrons through the wire and the cell no longer has the ability to do work At equilibrium, the components in the two cell compartments have the same free energy ( G = 0) Equilibrium Constants from Cell Potentials For a cell at equilibrium, Q = K and E cell = 0 Applying this to the Nernst equation at 25 0 C: E =E 0 - (0.0591/n)log(Q) gives: 0 =E 0 - (0.0591/n)log(K) so: log(k) = ne 0 /0.0591 13

Lead Storage (Car) Battery Batteries A Galvanic cell or a group of galvanic cells connected in series used as a source of direct current Anode: Pb(s) + HSO 4 - (aq) PbSO 4 (s) + H + (aq) + 2e - Cathode: PbO 2 (s) + HSO 4 - (aq) + 3H + (aq) + 2e - PbSO 4 (aq) + 2H 2 O(l) 14

Dry Cell Battery Alkaline Battery Anode: Zn(s) Zn 2+ (aq)+ 2e - Cathode:2NH 4 + (aq) + 2MnO 2 (s) + 2e - Mn 2 O 3 (s)+ 2NH 3 (aq) + H 2 O(l) Anode: Zn(s) + 2OH - (aq) ZnO(s) + H 2 O(l) + 2e - Cathode: 2MnO 2 (s) + H 2 O(l) + 2e - Mn 2 O 3 (s) + 2NH 3 (aq) + 2OH - (aq) Fuel Cells A galvanic cell for which the reactants are continuously supplied Anode: H 2 (g) + 2OH - (aq) 2H 2 O(l) + 2e - Cathode: O 2 (g) + 2H 2 O(l) + 4e - 4OH - (aq) Corrosion Involves oxidation of a metal in air (to form oxides and sulfides) Most metals corrode easily because they have standard reduction potentials which are less positive than oxygen gas so when these half-reactions are reverse and combined with the reduction halfreaction for oxygen, a positive cell potential results This process is often slowed by the formation of a thin oxide coating which prevents further corrosion Metals with large standard positive reduction potentials (for example gold) do not corrode and are referred to as noble metals 15

Corrosion of Iron Since steel (an alloy of iron and carbon) has an inhomogeneous surface, regions form where the iron is more easily oxidized (anodic regions) than at others (cathodic regions) In the cathodic regions, the ferrous ions react with oxygen to form rust (a hydrated ferric oxide of variable composition): Prevention of Corrosion This is achieved by applying a coating (usually paint or a metal plating) to protect the metal from oxygen and moisture Chromium and tin are often used to plate steel (via electrolysis) which then form a protective oxide coating Zinc can also be used to coat steel in a process called galvanizing which forms a protective mixed oxide-carbonate coating Since zinc is a more active metal than iron, any oxidation that does occur involves the zinc rather than the iron. Zinc is said to act as a sacrificial metal Corrosion can also be prevented by alloying For example, stainless steel contains chromium and nickel which form oxide coatings that increases the steel s reduction potential so it is less likely to corrode Galvanic Cell Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Electrolysis The forcing of a current through a cell to produce a chemical change for which the cell potential is negative (non-spontaneous) An electrolytic cell therefore uses electrical energy to produce a chemical change Anode: Zn(s) Zn 2+ (aq) + 2e - Cathode: Cu 2+ (aq) + 2e - Cu(s) 16

Electrolytic Cell Cu(s) + Zn 2+ (aq) Cu 2+ (aq) + Zn(s) Calculating Chemical Change from Current Flow Given the amount of current flowing (in amperes, A = Cs -1 ) for a specified time period it is possible to calculate the amount of neutral metal deposited at the cathode of an electrolytic cell as follows: 1. Multiply the current (in amperes) by time in seconds to obtain the total charge (in coulombs) passed into the solution at the cathode 2. Calculate the number of moles of electrons required to carry this charge by dividing by the Faraday (96,485 Cmol -1 ) 3. Use the half-reaction for the reduction to calculate the number of moles of neutral metal deposited from the moles of electrons 4. Convert the moles of neutral metal into grams by multiplying by the average atomic mass Anode: Cu(s) Cu 2+ (aq) + 2e - Cathode: Zn 2+ (aq) + 2e - Zn(s) 17

Caution! Standard reduction potentials must be used carefully in predicting the order of oxidation or reduction in electrolytic cells since some species require a much higher potential then expected (called an overvoltage) due to difficulties in transferring electrons from species in the solution to atoms on the electrode across the electrodesolution interface Example: Electrolysis of NaCl(aq) Two processes are possible at the anode: As voltage is increases we would expect to see O 2 produced at the anode first since it is easier to oxidize H 2 O However, in practice the Cl 2 is seen at the anode first! The Electrolysis of Water 2H 2 O(l) 2H 2 (g) + O 2 (g) 18

Electroplating Anode: Ag(s) Ag + (aq) + e - Cathode: Ag + (aq) + e - Ag(s) Electrolysis of Sodium Chloride Anode: 2Cl - (l) Cl 2 (g) + 2e - Cathode: Na + (l) + e - Na(l) 19