1 Electrochemistry Oxidation-Reduction Review Topics Covered Oxidation-reduction reactions Balancing oxidationreduction equations Voltaic cells Cell EMF Spontaneity of redox reactions Batteries Electrolysis Reading Quiz Read Chapter 20 Know definitions for all words in bold Table 20.1 Sample exercises 20.1, 20.5, 20.6 Oxidation refers to the loss of electrons (LEO) Reduction refers to the gain of electrons (GER) Thus oxidation-reduction occurs when electrons are transferred from the atom that is oxidized to the atom that is reduced Oxidation-Reduction Reactions Oxidation-reduction reactions are often called redox reactions By writing the oxidation number of each element above or below the equation we can see the oxidation state changes that occur Oxidation-Reduction Reactions In any redox reaction both oxidation and reduction must occur The substance that is being reduced is called the oxidizing agent or oxidant and is gaining electrons in the reaction The substance being oxidized is called the reducing agent or reductant and is losing electrons to the oxidant The nickel-cadmium (nicad) battery uses the following redox reaction to generate electricity: Cd(s) + NiO 2 (s) + 2H 2 O(l) Cd(OH) 2 (s) + Ni(OH) 2 (s) Identify the oxidant and reductant in the following equation: 2H 2 O + Al + MnO 4 Al(OH) 4- + MnO 2 (s) Identify the substances that are oxidized and reduced, and indicate which are oxidizing agents and which are reducing agents.
2 Balancing Redox Reactions When balancing reactions we must obey the law of conservation of mass As we balance redox reactions we must also balance the loss and gain of electrons It is often easiest to do this by separating a reaction into oxidation and reduction halfreactions Sn 2+ + 2Fe 3+ Sn 4+ + 2Fe 2+ Oxidation: Sn 2+ Sn 4+ + 2e - Reduction: 2Fe 3+ + 2e - 2Fe 2+ Balancing by Half-Reaction We can use half-reactions to balance complicated redox reactions in acidic solution by using the following procedure: Divide the equation into two incomplete half reactions, one for oxidation and one for reduction Balance each half reaction First balance the elements other than H and O Next, balance the O atoms by adding H 2 O Then balance the H atoms by adding H + Finally, balance the charge by adding e - Balancing by Half-Reaction Multiply each half-reaction by an integer so that the number of electrons lost in one reaction equals the number of electrons gained in the other Add the two half-reactions and simplify by cancelling species that occur on both sides Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on each side s Balance the following reactions in acidic solution: MnO 4- + C 2 O 4 Mn 2+ + CO 2 Cr 2 O 7 + Cl - Cr 3+ + Cl 2 Cu + NO 3- Cu 2+ + NO 2 Balancing Redox Reactions in Basic Solution If a redox reaction occurs in basic solution, the equation must be balanced using OH - and H 2 O instead of H + and H 2 O The half-reactions can be balanced initially as if they occurred in acidic solution The H + ions can then be neutralized by adding an equal number OH - ions to each side and canceling the resulting water molecules s Complete and balance the following reactions in basic solution: CN - + MnO 4 - CNO - + MnO 2 NO + Al NH 3 + Al(OH) 4 - Cr(OH) 3 + ClO - CrO 4 + Cl 2
3 s The energy released in a spontaneous redox reaction can be used to perform electrical work This task is accomplished through a voltaic (or galvanic) cell This device transfers electrons in a redox reaction through an external pathway rather than directly between reactants Spontaneous Redox Reaction A strip of zinc metal is immersed in an aqueous copper sulfate solution The redox reaction takes place at the metal-solution interface and involves direct transfer of two electrons from Zn atoms to Cu 2+ ions As time passes, a darkcolored deposit of copper metal appears on the zinc, and the blue color due to Cu 2+ fades from the solution In this set-up, the Zn metal and the Cu 2+ ions are not in direct contact Instead, the Zn is in direct contact with Zn 2+ and the Cu is in contact with Cu 2+ in another compartment The reduction of the Cu 2+ can occur only through an external circuit The two solid metals that are connected by the external circuit are called electrodes The electrode at which oxidation occurs is called the anode The electrode at which reduction occurs is called the cathode Each of the two compartments is called a half-cell Anode (oxidation half-reaction): Zn(s) Zn 2+ (aq) + 2e - Cathode (reduction half-reaction): Cu 2+ (aq) + 2e - Cu(s)
4 Electrons become available as zinc metal is oxidized at the anode They flow through the external circuit to the cathode, where they are consumed as Cu 2+ is reduced The zinc electrode loses mass, the zinc solution becomes more concentrated The copper electrode gains mass and the copper solution becomes more dilute and loses color For the cell to work, the solutions must remain neutral As Zn 2+ ions enter solution, anions must flow from the cathode to keep the solution electrically neutral As Cu 2+ ions leave solution, cations must flow from the anode This migration of ions can be through a porous material or (as in this case) through a salt bridge A salt bridge consists of a U-shaped tube that contains an electrolyte solution whose ions will not react with other ions in the cell The solution is often a gel so that it stays in the tube Whatever means is used to allow ions to migrate between half-cells, anions always migrate towards the anode and cations toward the cathode In any voltaic cell, the electrons flow from the anode (labeled negative) to the cathode (labeled positive) The shorthand notation shows the anode half-cell on the left of the salt bridge and the cathode half-cell on the right A single vertical line indicates the phase boundary between the metal and the metal ion solution
5 Cell EMF The difference in potential energy per electric charge between two electrodes in measured in volts (V) 1 V = 1 J/C The potential difference (voltage) between two electrodes provides the driving force that pushes the electrons through the external circuit We call this potential difference the electromotive force or emf The emf of a cell, denoted E cell, is also called the cell potential or cell voltage Cell EMF The emf of a given cell changes with concentration and temperature Standard conditions for a cell are 25 C and 1 M solution concentrations (1 atm pressure for gases) Under these conditions, the emf is called the standard emf or the standard cell potential Standard emf is denoted Eº cell Standard Reduction Potentials The emf for a given volatic cell depend on the particular cathode and anode used Because there are so many different types of cells, it is easiest to calculate the emf from the cell by knowing the standard reduction potential, E red, for each half-cell The standard cell potential can then be calculated as follows: E cell = E red (cathode) - E red (anode) Standard Reduction Potentials Because it is impossible to measure E red directly, a reference point is needed for comparison The reference point used is the reduction of H + 2H + (aq, 1 M) + 2e - H 2 (g, 1 atm) E red = 0 V The electrode designed to produce this half-reaction is called a standard hydrogen electrode (SHE) E cell is the voltage of any cell produced when a SHE is the anode Calculate the standard cell potential of the galvanic cell shown above
6 Calculate the standard cell potential of the galvanic cell shown above Oxidizing and Reducing Agents The more positive the E red value for a half-reaction, the greater the tendency for reactant of the half-reaction to be reduced and, therefore, to oxidize another species The half reaction with the smallest reduction potential is most easily reversed as an oxidation s Rank the following ions in order of increasing strength as oxidizing agents: NO 3-, Ag +, Cr 2 O 7 Rank the following species from strongest to weakest reducing agent: I -, Fe, Al Spontaneity of Redox Reactions Any reaction that can occur in a voltaic cell to produce a positive emf must be spontaneous In general, we can write: E = E red (reduction process) - E red (oxidation process) A positive value of E indicates a spontaneous process and a negative value of E indicates a nonspontaneous one Determine whether or not the following reactions are spontaneous under standard conditions Cu(s) + 2H + (aq) Cu 2+ (aq) + H 2 (g) Cl 2 (g) + 2I - (aq) 2Cl - (aq) + I 2 (s) EMF and Free Energy Change Gibbs free energy and emf of a redox reaction are related in the following way: ΔG = -nfe or ΔG = -nfe (at standard conditions) n is a positive number that represents the number of electrons transferred in the reaction F is called Faraday s constant and is the quantity of charge on 1 mol of electrons 1 F = 96,500 C/mol = 96,500 J/V-mol A positive value of E and a negative value of ΔG both indicate that a reaction is spontaneous
7 s Calculate ΔG for the following reactions using standard reduction potentials. 4Ag(s) + O 2 (g) + 4H + (aq) 4Ag + (aq) + 2H 2 O(l) 3Ni 2+ + 2Cr(OH) 3 + 10OH - 3Ni + 2CrO 4 + 8H 2 O Effect of Concentration on Cell EMF As a voltaic cell discharges, reactants are consumed and products are generated, so the concentrations of these substances change The emf progressively drops until E = 0, at which point the cell is dead At that point the concentrations of reactants and products are in equilibrium The Nernst Equation To find the emf of a cell under nonstandard conditions we can use the Nernst equation: RT 0.0592 Ecell Ecell ln Q or Ecell Ecell log Q @ 298 K nf n Q is the reaction quotient Given the following reaction: Cr 2 O 7 (aq) + 14H + (aq) + 6I - (aq) 2Cr 3+ (aq) + 3I 2 (s) + 7H 2 O(l) Calculate the emf at 298 K when [Cr 2 O 7 ] = 2.0 M, [H + ] = 1.0 M, [I - ] = 1.0 M, and [Cr 3+ ] = 1.0 x 10-5 M. If the voltage of a Zn-H + cell is 0.45 V at 25 C when [Zn 2+ ] = 1.0 M and P H2 = 1.0 atm, what is the concentration of H +? Using standard reduction potentials, calculate the equilibrium constant for the oxidation of Fe 2+ by O 2 in acidic solution
8 Calculate the equilibrium constant for the following reaction: Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) Batteries A battery is a portable, self-contained electrochemical power source that consists of one or more voltaic cells Smaller batteries, like AAs or AAAs, contain a single voltaic cell and produce around 1.5 V Larger batteries consist of a series of voltaic cells and can produce greater voltages, such as a 12 V car battery Lead-Acid Battery Lead-Acid Battery A 12 V lead-acid battery consists of six voltaic cells in series Cathode: PbO 2 (s) + HSO 4- (aq) + 3H + (aq) + 2e - PbSO 4 (s) + 2H 2 O(l) Anode: Pb(s) + HSO 4- (aq) PbSO 4 (s) + H + (aq) + 2e - Complete: PbO 2 (s) + Pb(s) + 2HSO 4- (aq) + 2H + (aq) 2PbSO 4 (s) + 2H 2 O(l) Eº cell = 2.041 V Because the reactants are solids, there is no need to separate the anode and cathode into separate compartments The reverse reaction plus supplied electrical energy recharges the battery Alkaline Battery Dry Cell Batteries The most common primary (non-rechargeable) battery is the alkaline battery The name comes from the fact that anode consists of powdered metal suspended in a concentrated KOH gel The emf of an alkaline battery is about 1.55 V at room temperature Alkaline batteries provide far superior performance over older dry cell batteries
9 Fuel Cells Fuel Cells In a hydrogen-oxygen fuel cell, gaseous H 2 is oxidized to water at the anode, and gaseous O 2 is reduced to hydroxide ion at the cathode The net reaction is the conversion of H 2 and O 2 to water Fuel cells are not technically batteries because they are not self-contained Corrosion Corrosion reactions are spontaneous redox reactions in which metals are converted into an unwanted compound by some substance in the environment For nearly all metals, oxidation is a thermodynamically favorable process and can be very destructive The rusting of iron is one of the most common and expensive forms of corrosion 20% of all iron used each year is to replace rust-damaged items Corrosion The metal and a surface water droplet constitute a tiny galvanic cell in which iron is oxidized to Fe 2+ in a region of the surface (anode region) remote from atmospheric O 2 O 2 is reduced near the edge of the droplet at another region of the surface (cathode region) Electrons flow from anode to cathode through the metal, while ions flow through the water droplet Dissolved O 2 oxidizes Fe 2+ further to Fe 3+ before it is deposited as rust (Fe 2 O 3 ) Corrosion Corrosion A layer of zinc protects iron from oxidation, even when the zinc layer becomes scratched The zinc (anode), iron (cathode), and water droplet (electrolyte) constitute a tiny galvanic cell Oxygen is reduced at the cathode, and zinc is oxidized at the anode, thus protecting the iron from oxidation Protecting a metal from corrosion by making it the cathode in an electrochemical cell is called cathodic protection The metal that is oxidized while protecting the cathode is called the sacrificial anode Underground pipes, ship hulls, and even coffins are often protected by sacrificial anodes
10 Electrolysis It is possible to use electrical energy to cause nonspontaneous redox reactions to occur Such processes that are driven by an outside energy source are called electrolysis reactions and take place in electrolytic cells An electrolytic cell consists of two electrodes in a molten salt or aqueous solution Just as in voltaic cells, oxidation occurs at the anode and reduction at the cathode Electrolysis Electrolysis Electrolysis of molten sodium chloride Chloride ions are oxidized to Cl 2 gas at the anode, and Na + ions are reduced to sodium metal at the cathode Electrolysis of AgF(aq) in an acidic solution leads to the formation of silver metal and oxygen gas. Write the half-reaction that occurs at each electrode and calculate the minimum emf needed for this process to occur under standard conditions. Electroplating uses electrolysis to deposit a thin layer of metal on another metal This is used for beautification or protection from corrosion Electroplating Quantitative Aspects of Electrolysis For any half reaction, the amount of a substance that is reduced or oxidized is directly proportional to the number of electrons passed into the cell The charge of one mol of electrons is a faraday (F) The quantity of charge passing through an electrical circuit is measured in coulombs (C) Remember 1 F = 96,500 C/mol e - The number of coulombs passing through a cell can be calculated as follows: Coulombs = amperes x seconds
11 Calculate the number of grams of aluminum produced in 1.00 hour by the electrolysis of molten AlCl 3 if the current in 10.0 A. s The half-reaction for the formation of magnesium metal upon electrolysis of molten MgCl 2 is Mg 2+ + 2e - Mg. Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 x 10 3 s. How many seconds would be required to produce 50.0 g of Mg if the current is 100.0 A?