-252-ELECTROCHEMISTRY[MH5; Chapter 18] Electrochemistry is the study of therelationship between chemicalreactions and electrical energy. The transfer of electrons (what happens in an oxidation-reduction,or redox reaction) constitutes electrical energy. The reactions studied in electrochemistry are always redoxreactions. Electrochemical reactions occur experimentally in anELECTROCHEMICAL CELL, of which there are two basic types.Voltaic Cell Stored chemical energy is converted by a spontaneous reaction toelectrical energy; for example, a battery.Electrolytic Cell A chemical reaction is made to occur by passage of an electriccurrent. Electrical energy is converted to chemical energy, for example,electroplating. The reactions in an electrochemical cell take place on the surface ofthe electrodes. An electrode is an electronic conductor (usually a metal) in contactwith an ionically conducting medium (usually an electrolyte solution). Recall when we learned to balance redox reactions; we used halfreactions....... In an electrochemical cell, the Oxidation half reaction takes placeat one electrode, which we call the anode. The Reduction half reaction takes place at the other electrode,which we call the cathode.-253-VOLTAIC or GALVANIC CELLS [MH5; 18.1] Recall that a voltaic cell is one in which the redox reaction occursspontaneously and electrical energy is produced. Consider the following reaction: Zn (s) + Cu 2+ (aq) ! Zn 2+ (aq) + Cu (s) What happens if we combine just the reactants ? Electrical energy can only be produced if the two half reactions areactually separated in space..... What does this look like ? What actually happens when the cell is operating ?? The electrons produced in the oxidation half reaction at the anodetravel through a external circuit to the cathode where they areconsumed in the reduction half reaction.-254- A simple voltaic cell, the DANIELL cell[MH5, Figure 18.2]

Zn metal inCu metal inZn2+(aq) solution Cu2+ (aq) solution A voltaic cell consists of two half cells; in this cell there is a zinchalf cell and a copper half cell The Zn metal and the Cu metal are acting as the electrodes. Recall that oxidation occurs at the anode, which has a negativecharge; reduction occurs at the cathode, which has a positivecharge. The electrons must pass through an external circuit which connectsthe two half cells (electrons flow from anode to cathode). A salt bridge connects the solutions and prevents mixing; it allowsfor ions to flow between the two solutions. If the salt concentrations are 1 M, and a current is allowed to flowby completing the external circuit, a potential difference of 1.10 V(at 25EC) will be established; this is the Standard Cell Potential.-255-The Redox reactions occurring at the electrodes: at Zn electrode: at Cu electrode:Overall redox reaction: As this reaction occurs, there is a build-up in cation concentration atthe Zn anode, and a decrease in cation concentration at the Cucathode. So anions migrate through the salt bridge from Cu side to Zn sideand cations pass in the opposite direction to preserve the chargebalance. We abbreviate this cell as:-256- The voltmeter reads the potential difference between the twoelectrodes. This potential difference (or voltage) is the electromotive force(EMF) of the cell and is measured in Volts. The potential of the cell is independent of the number of electronspassing through the cell. We cannot divide the 1.10V cell potential between Zn and Cu,because we cannot study one electrode in isolation. Electrode Potentials for a half-cell are measured relative to theStandard Hydrogen Electrode reaction, on a Pt surface [catalyst]with [ H+(aq) ] = 1.0 M, at 25EC The reaction for this electrode is: 2 H+ (aq)+ 2 e ! H2 (g) The reduction potential for this cell has been defined as: 0.00 V Reduction Potentials for other half reactions can be measured bysetting up a cell using the half cell in question along with theStandard Hydrogen Electrode.............[MH5; Figure 18.3] This example uses the Zn half cell. The measured cell potential is + 0.762 V; all of this voltage isassigned to theZn electrode. For:EE, the STANDARD REDUCTION POTENTIAL is - 0.762 V relative to the standard hydrogen electrode.-257- For the cell:EEcell is found to be+ 0.339V EE for the Cu2+/Cu reduction, must be+ 0.339 V To combine these two half-reactions to make a Daniell cell, one mustbe reversed, so its EE value changes sign.....Oxidation Half Reaction:Reduction Half Reaction:Overall:EEcell = When EE values for half-reactions are combined in this way, theunknown contribution of H+/H2 (which we assumed was 0.00 V)vanishes and the experimentally correct EEcell value is obtained.-258-STANDARD VOLTAGES [MH5; 18.2] In tables ofEE values, all half-reactions are given as reductions... So one reaction must be reversed (changing the sign of EE) when twohalfreactions are combined to give an overall redox reaction.Reaction will proceed spontaneously in the direction for whichEE cell is POSITIVE If we multiply equations for half-reactions to balance the overallredox equation, the EE values do not change. Half cell potentials are intensive properties, independent of theamount of reagent.EXAMPLE: Given the cell: AR (s) / AR 3+ (aq) // Fe 2+ (aq) , Fe 3+ (aq) / Pt (s)Will AR metal reduce Fe3+ to Fe2+ ? What is the anode half reaction ?What is the cathode half reaction ?Overall:EE cell =-259-What happens as reaction proceeds ? Ecell will change as reactant concentrations change. Eventually the battery or cell runs down asEcell has fallen to zero. The reaction stops; the system has reached chemical equilibrium. In practice, most redox reactions go virtually to completion; for theDaniell cell....essentially no Cu2+ remains (not responsible for calculation). Once EE values have been measured relative to the H+/H2 half-reaction, a table of standard reduction potentials may be drawn up(see tutorial problems, p. 17-18 or MH5; page 481) from which EEcellfor any overall redox reaction may be calculated........ A redox reaction and overall EEcell value can only be made up byreversing one half-reaction (and its EE value) and addingthem................Remember thatEEvalues are NOT measured per mole!!-260- How do we set up reactions which occur only in solution (no metals toact as electrodes) ?EXAMPLE: a)Will the reaction Ce3+ (aq) + Fe3+ (aq) ! Ce 4+ (aq) + Fe2+ (aq)proceed spontaneously ?b)Write the cell diagram for this reaction.Write the two half reactions... Look up Eo valuesThe Cell Diagram:What Can We Determine from Eo values ? The relative strengths of oxidants (those species which arereduced) and reductants (those species which are oxidized) may bedetermined from their positions in a table of Standard ReductionPotentials. [MH5; Table 18.1] or Electrochemistry Tutorial p. 17 &18.-261-Remember: Half-reactions with large+ve valuesoccur readily. A strong oxidizing agent has a high +veEovalue. A strong reducing agent has a high -ve Eo value.EXAMPLE: Which of the metals Ag, Cu, Sn and Fe will dissolve in aqueous acid ? (Recall that acid means H+)....The reduction half-reaction in each case is:Write each metal half-reaction as an oxidation: Those metals appearing above H2 in the table of reduction EE values(electro-chemical series) should dissolve in any aqueous acid.-262-Reversibility of Cell Reactions Can the electrode reactions in a voltaic cell be reversed by applyingan external potential ? Sometimes !EXAMPLE 1: Apply an external potential of > 1.10 V to the Daniellcell....... At the anode: At the cathode:# because H+ is reduced more easily than Zn2+ This is an example of an electrolytic cell - electrical energy is beingused to drive a non-spontaneous chemical reaction.-263-ELECTROLYSIS [MH5; 18.5] Electrolysis is the process by which a non spontaneous chemicalreaction is forced to occur by pumping electrical energy into thesystem. This process takes place in an electrolytic cell. A battery provides a source of direct electrical current. Two wires from the battery lead to the electrodes; the anode andthe cathode. The battery is the electron pump; it pushes electrons into thecathode and pulls them away from the anode. If electrical neutrality is to be maintained, electrons must somehowbe consumed at the cathode and released at the anode. A redox reaction will accomplish this; the oxidation reaction (whichoccurs at the anode) releases electrons, while the reduction reaction(which occurs at the cathode) consumes electrons. This whole process is what is known as Electrolysis.Quantitative Electrolysis Faradays Law defines the relationship between the amount ofelectricity passed in an electrochemical cell and the amount ofreactants used or products formed in the corresponding redoxreaction.................-264-EXAMPLES:Half Reactions Electrons React / ProdAg+(aq) + e- ! Ag(s)Fe(s)! Fe2+(aq) + 2 e-AR3+(aq) + 3 e- ! AR(s)Cu2+(aq) + 2 e- ! Cu(s) Only chemists use moles for measuring electrons......... The physicists and engineers use Coulombs for measuring charge andAmperes for measuring current. One Coulomb is the amount of charge produced when a current ofone Ampere flows for one second......... So..........the amount of charge (denoted by Q) is equal to the amountof current (in amperes, denoted by i) that flows multiplied by thetime it flows for in seconds (denoted by t). This relationship is usually expressed by:Q = i x t-265- The link between the charge and the number of moles of electrons isknown as the Faraday Constant. This constant tells us that each mole of electrons that are passedcarries a charge of 96490 Coulombs.EXAMPLE 1:What mass, in g, of silver, Ag,can be plated out from a solution ofAgNO3 if acurrent of 0.750 amperes flows for 1 hour ?-266-EXAMPLE 2:We wish to electroplate 6.50 g of copper, Cu,from a solution of CuCR2. How long will the electrolysis take if we use a current of 3.50 Amperes?-267-EXAMPLE 3: Electroplating is the deposition of a thin layer of metal on aconducting surface. The object to be electroplated is used as the cathode; the anode isthe electroplating metal. Chromium metal has been used for chrome plating; suppose we wishto chrome plate an object using a solution of CrCR3. What happens at each electrode when a current is passed ?At the anodeAt the cathode If a current of 1.65 amperes is passed for 3 hours, what mass, ingrams, of chromium metal would be plated ?-268-EXAMPLE 4: Two electrolysis cells were connected in series as follows: Cu*Cu2+*CuCCCAg*Ag+*AgCell 1Cell 2 After a period of electrolysis, 0.295 g of Cu was deposited on thecathode of the Cu cell (Cell 1) . What mass of Ag, in grams, wasdeposited in the other cell (Cell 2)?-269-Electrolysis of Water While we know that pure water does not conduct electricity verywell; it can participate in a redox reaction if there are some ionspresent.............. At the anode; the oxidation half reaction:2 H2O(l) ! O2 (g) + 4 H+(aq) + 4 eEo = - 1.23 V At the cathode; the reduction half reaction:2 H2O(l) + 2 e! H2(g) + 2 OH(aq) Eo = - 0.83 V The overall reaction is:6 H2O (l)! 2 H2 (g) + 4 OH (aq) + 4 H+(aq) + O2 (g)OR: 2 H2O (l) ! 2 H2 (g) + O2 (g)Eocell = What does this mean in terms of performing electrolysis usingaqueous solutions ? It depends on the Standard Reduction Potentials of the solutespresent in the solution........... If quite a bit of energy is required to react the solute species, thewater will react first, using up the energy that was intended for thesolute ions.-270- There are two possibilities for reduction at the cathode: 1) Cation to metal, characteristic of easily-reduced transition metalcations and electroplating...... 2) Water to hydrogen gas, occurring when the cation in solution isdifficult to reduce (such as Na + and K +)..... There are also two possibilities for the reaction occurring at theanode:1) The anion may be oxidized to the corresponding non metal....2) If the anion cannot be oxidized (for example, NO3or SO42, then water will be oxidized to oxygen gas......-271-EXAMPLE 1:Can Sodium metal be produced by the electrolysis ofaqueous brine (a solution of NaCR) ? The desired reaction is: Na+ (aq)+e !Na (s)E o=- 2.714 V But instead, at the cathode, the reaction is:2 H2O + 2 e ! H2 (g) + 2 OH Eo= - 0.83 V At the anode:2 CR !CR2 (g) + 2 e E o = - 1.36 V Overall:2 CR +2 H2O! H2 (g)+CR2 (g)+2 OH EXAMPLE 2: What reactions would occur at each electrode duringthe electrolysis of aqueous CuSO4 ? -272-COMMERCIAL CELLS [MH5; 18.6] Consider the electrolysis of aqueous NaCR ; (EXAMPLE 1, p. 107) At the cathode:2 H2O + 2 e ! H2 (g) + 2 OH Eo= - 0.83 V At the anode:2 CR !CR2 (g) + 2 e E o = - 1.36 V There are a variety of uses for the products formed in thiselectrolysis reaction........Batteries Batteries are voltaic, or galvanic cells; they undergo spontaneousreactions. These reactions are exothermic (heat producing); but when reactionhappens in a cell, the energy is released in the form of electricity. There are two main types of batteries: dry cells (or primarybatteries) and rechargeable (secondary batteries). A dry cell consists of a zinc container which acts as the anode, agraphite (carbon) cathode and a paste of MnO2, NH4CR and ZnCR2which acts as the electrolyte.....-273- The half reactions are:At the anode:Zn (s) ! Zn2+ (aq) + 2 e At the cathode:2 MnO2 (aq)+ 2 NH4+ (aq)+ 2 e !Mn2O3 (s)+ H2O (R) + 2NH3(aq)Overall: E ocell = 1.5 V (not at standard conditions) and delivers ~ 0.5 Acurrent for about 6 hours. The voltage decreases as Zn 2+ accumulates; eventually goes dead;cannot be recharged. In alkaline cells the NH4CR is replaced by KOH and the anode is a gelmade of Zn powder and KOH. The voltage is also 1.5 V; but is delivered consistently and for alonger period of time; these batteries are not rechargeable either. The half reactions are: At the anode:Zn (s)+ 2 OH (aq) !ZnO (s)+ H2O (R) + 2 e At the cathode:2MnO2 (s)+H2O (R) + 2 e !Mn2O3 (s)+2 OH (aq) Overall:-274- The mercury cell is a very small primary cell used in watches, hearingaids, cameras and some calculators. The anode is a zinc - mercury amalgam, with the reacting speciesbeing zinc. The cathode is mercury (II) oxide, HgO.At the anode: Zn (s)+ 2 OH (aq) !Zn(OH)2 (s)+2 e At the cathode:HgO (s) + H2O + 2 e ! Hg (R) + 2 OH (aq) Overall:Storage (Rechargeable) Voltaic Cells Rechargeable batteries, or storage cells can be repeatedlyrecharged.......... The products of the reaction are deposited directly on theelectrodes when the battery is being used; the battery is acting as agalvanic cell (or is said to be discharging). When recharging, a current is applied (battery is acting as anelectrolytic cell) and the discharge reaction is reversed. The lead storage battery is commonly used as a car battery; this isa 12 V battery. It consists of six 2 V cells, connected in series. The anode consists of lead plates filled with spongy lead; thecathode is lead plates filled with PbO2; these grills alternatethrough the cell. The large surface area supplied by these plates allows large-275-currents to flow for short periods of time (like when starting thecar). The electrolyte is sulfuric acid, H2SO4. During the initial charge, the battery is acting as an electrolytic cell:2 PbSO4+ 2 H2O !Pb (s) + PbO2 (s) + 4 H+ + 2 SO42(anode) (cathode) The reactions when battery is operating (starting the car, poweringthe headlights etc) are:Anode:

Pb (s)+SO4 2 (aq)!PbSO4 (s)

+2 eE o = + 0.36 VCathode: PbO2 (s)+4 H +(aq)+ SO4 2 (aq)+ 2 e ! PbSO4 (s) + 2 H2O (R) E o = + 1.68 VOverall: When the cell is recharged, a current is applied to the cell whichprovides the energy required for the non spontaneous reversereaction. In the car, this happens when the car is running (this is what thealternator does). The extent of the charge of the battery is a function of the densityof the electrolyte; if the density falls too low, the battery cannotusually be recharged.-276- A discharged battery may also freeze, because the density of theelectrolyte is too low; if this happens the plates can warp, come incontact with each other and short out the battery......... These batteries cannot be recharged forever; too much repeatedquick charging (when battery goes dead) can result in Pb, PbO2 andPbSO4 flaking off the electrodes. These speciescollect as sludge at the bottom of the battery andcan cause short circuiting. These batteries are heavy, the lead compounds can be toxic and thesulfuric acid is corrosive. NICAD batteries are also rechargeable and are used widely inportable electronic equipment. As the name suggests, these batteries use Nickel and Cadmium.....The anode reaction is: Cd (s)+2 OH (aq) !CdO (s)+H2O (R) + 2 e The cathode reaction is:

NiO2 (s)+H2O (R)+2 e ! NiO (s)+2 OH (aq)Overall: The reaction is reversed when battery is being charged. These batteries are lighter than the lead storage battery and havea longer rechargeable life.-277-EXAMPLE 1:Consider the mercury cell described on p. 274. Supposethis battery contains 0.030 g of Zn and 0.150 g of HgO. If thebattery can deliver a current of 0.00100 milliamps, how long, in hours,will it take for the battery to completely discharge (go dead) ?-278-EXAMPLE 2: Now consider the NICAD battery described on p. 276. Initially, this battery contains 1.00 g of Cd and 0.900 g of NiO2. When this battery dies it can be recharged. Assuming that a fullrecharge is necessary, how long, in hours, would it take if a current of1.50 amperes was used.-279-Fuel Cells In a fuel cell, a fuel (usually hydrogen) is oxidized at the anode. In the reduction reaction at the cathode, oxygen is reduced. In fuel cells currently in use, the reaction takes place in alkalinesolution.Anode:Cathode:Overall: This reaction is very exothermic; the energy produced can be usedfor power. A catalyst is necessary for this reaction to occur at lowtemperatures (40 oC); platinum is usually used. Only very pure hydrogen can be used....if any carbon monoxide ispresent it can poison the catalyst. Research is looking at replacing the internal combustion engine inautomobiles with fuel cells. The burning of hydrogen produces only water (not CO, CO2 and NOsuch as the car engine does at present).........so this would be muchbetter for the environment. Drawbacks to using fuel cells for car engines.....-280-Redox Chemistry of some Transition Metals Transition metals often exhibit several different oxidation states. This makes their redox chemistry more complex and often colourful,as some metal cations have different colours associated with theirdifferent oxidation numbers. Transition metals can form cations by themselves at lower oxidationnumbers ( +1, +2, +3). When these metals have higher oxidation numbers, (+4, +5, +6, +7),they are only found in covalent compounds or complex ions. A charge of +4 or greater on an ion would make it highly reactive, soit would not be stable for long in an aqueous solution. These different oxidation states are possible because the highestenergy electrons are easily lost to form cations.............EXAMPLE: The relative stabilities of these oxidation states are shown byEo values for interconversion.-281-EXAMPLES:[MH 5; table20.3, p. 542]- 0.41 V - 0.91 VCr3+! Cr2+! Cr(s)1.56 V- 1.18 VMn3+!Mn2+! Mn(s)0.77 V- 0.41 VFe3+ ! Fe2+! Fe(s)1.95 V- 0.28 VCo3+ ! Co2+! Co(s)0.16 V 0.52 VCu2+! Cu+! Cu(s)1.40 V1.70 VAu3+! Au+! Au(s)-282- Now lets look at some questions that can be answered using theinformation on the previous page.............1) Which is strongest oxidizing agent ?2) Which is the strongest reducing agent ?3) Which should disproportionate ? Look for an intermediateoxidation state where.....-283-4) Given:O2 + 4 H+ + 4 e! H2OEE =+ 1.23 V Which ions should oxidize water ?5) Which cations should reduce O2 to H2O?6) Which cations should react with H+ ?Recall:2 H++2 e! H2Eo = 0.00V