Prof. T.L. Meeks, CHE Chapter Twenty Copyright Tyna L. Meeks All Rights Reserved Prof. T.L. Meeks, CHE Electrochemistry Redox reactions Oxidation is the loss of electrons Reduction is the gain of electrons The transfer of electrons that occurs is exothermic and can be used to produce electricity Prof. T.L. Meeks, CHE Electrochemistry At other times the reactions you want to occur may be nonspontaneous and require the addition of electricity to drive the reaction. Electrochemistry is the branch of chemistry that deals with the relationships between electricity and chemical reactions Prof. T.L. Meeks, CHE Oxidation - Reductions Reaction Determination of oxidation reduction reactions occurs by examining the following: 3assign oxidation numbers 3look for the change in oxidation states 3dont forget, something that oxidizes is also the reducing agent, something that reduces is also the oxidizing agent Prof. T.L. Meeks, CHE Oxidation - Reductions Reaction Sample Exercise: Identify the oxidizing and reducing agents in the following oxidation-reduction equation: 2H 2 O(l) + Al(s) + MnO 4 - (aq) Al(OH) 4 - (aq) + MnO 2 (s) Prof. T.L. Meeks, CHE Oxidation - Reductions Reaction Sample Exercise: Identify the oxidizing and reducing agents in the following oxidation-reduction equation: H 2 O(l) + Al(s) + MnO 4 - (aq) Al(OH) 4 - (aq) + MnO 2 (s) Al(OH) 4 - (aq) + MnO 2 (s) Prof. T.L. Meeks, CHE Oxidation - Reductions Reaction Sample Exercise: Identify the oxidizing and reducing agents in the following oxidation-reduction equation: H 2 O(l) + Al(s) + MnO 4 - (aq) Al(OH) 4 - (aq) + MnO 2 (s) Al(OH) 4 - (aq) + MnO 2 (s) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Whenever we balance a chemical equation, we must obey the conservation of mass. As we balance redox reactions we will see an additional rule: the balancing of charge as well. Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction 1. Assign oxidation numbers to all elements 2. Bridge the elements that are changing in oxidation state 3. Identify the number of electrons being lost and gained and balance them. 4. Balance all remaining elements besides H and O 5. Balance O by adding H 2 O to the deficient side Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction 6. Balance H by adding H + to the deficient side 7. Check to make sure your charge for each side is the same Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. A) Cu(s) + NO 3 - (aq) Cu 2+ (aq) + NO 2 (g) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. A) Cu(s) + NO 3 - (aq) Cu 2+ (aq) + NO 2 (g) Cu(s) + NO 3 - (aq) Cu 2+ (aq) + NO 2 (g) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. A) Leo = RA 2e Cu(s) + NO 3 - (aq) Cu 2+ (aq) + NO 2 (g) Cu(s) + NO 3 - (aq) Cu 2+ (aq) + NO 2 (g) Ger = OA 1e - Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. A) Leo = RA 2e Cu(s) + 2NO 3 - (aq) Cu 2+ (aq) + 2NO 2 (g) Cu(s) + 2NO 3 - (aq) Cu 2+ (aq) + 2NO 2 (g) Ger = OA 2(1e -) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. A) Leo = RA 2e Cu(s)+2NO 3 - (aq) Cu 2+ (aq)+2NO 2 (g)+2H 2 O Ger = OA 2(1e -) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. A) Leo = RA 2e Cu +2NO 3 - (aq)+4H + Cu 2+ (aq)+2NO 2 (g)+2H 2 O Ger = OA 2(1e -) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. B) Mn 2+ + NaBiO 3 Bi 3+ + MnO Na + Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. B) Mn 2+ + NaBiO 3 Bi 3+ + MnO Na + Mn 2+ + NaBiO 3 Bi 3+ + MnO Na + Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. B) Leo = RA 5e Mn 2+ + NaBiO 3 - Bi 3+ + MnO Na + Mn 2+ + NaBiO 3 - Bi 3+ + MnO Na + Ger = OA 2e - Ger = OA 2e - Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. B) Leo = RA 2(5e -) Mn 2+ + NaBiO 3 - Bi MnO Na + 2Mn 2+ + NaBiO 3 - Bi MnO Na + Ger = OA 2e - Ger = OA 2e - Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. B) Leo = RA 2(5e -) Mn NaBiO 3 - 5Bi MnO Na + 2Mn NaBiO 3 - 5Bi MnO Na + Ger = OA 5( 2e -) Ger = OA 5( 2e -) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. B) Leo = RA 2(5e -) Mn NaBiO 3 - 5Bi MnO Na + 2Mn NaBiO 3 - 5Bi MnO Na + Ger = OA 5( 2e -) Ger = OA 5( 2e -) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample Exercise: Complete and balance the following oxidation-reduction equations in acidic solutions. B) 2Mn BiO H + 5Bi MnO H 2 O Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction At times the solution in which a redox reaction occurs is a basic one. The balancing of an equation in a basic solution is slightly different. Instead of balancing with H 2 O and H +, you must balance with H 2 O and OH -. Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. A) NO Al NH 3 + Al(OH) 4 - Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. A) NO Al NH 3 + Al(OH) 4 - NO Al NH 3 + Al(OH) 4 - Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. A) Ger = OA 6e NO Al NH 3 + Al(OH) 4 - NO Al NH 3 + Al(OH) 4 - Leo = RA 3e - Leo = RA 3e - Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. A) Ger = OA 6e NO Al NH 3 + 2Al(OH) 4 - NO Al NH 3 + 2Al(OH) 4 - Leo = RA 2(3e -) Leo = RA 2(3e -) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. A) NO Al + 5H 2 O + OH - NH 3 + 2Al(OH) 4 - NO Al + 5H 2 O + OH - NH 3 + 2Al(OH) 4 - Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. B) Cr(OH) 3 + ClO - CrO Cl 2 Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. B) Cr(OH) 3 + ClO - CrO Cl 2 Cr(OH) 3 + ClO - CrO Cl 2 Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. B) Leo = RA 3e Cr(OH) 3 + ClO - CrO Cl 2 Cr(OH) 3 + ClO - CrO Cl 2 Ger = OA 1e - Ger = OA 1e - Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. B) Leo = RA 3e Cr(OH) 3 + 6ClO - CrO Cl 2 Cr(OH) 3 + 6ClO - CrO Cl 2 Ger = OA 6(1e -) Ger = OA 6(1e -) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. B) Leo = RA 2(3e -) Cr(OH) 3 + 6ClO - 2CrO Cl 2 2Cr(OH) 3 + 6ClO - 2CrO Cl 2 Ger = OA 6(1e -) Ger = OA 6(1e -) Prof. T.L. Meeks, CHE Balancing Oxidation - Reductions Reaction Sample exercise: Complete and balance the following equations for oxidation and reduction reactions that occur in basic solutions. B) 2Cr(OH) 3 + 6ClO - 2CrO Cl 2 + 2H 2 O + 2OH - Prof. T.L. Meeks, CHE Voltaic Cells The energy released in a spontaneous redox reaction can be used to perform electrical work. This task is accomplished in a voltaic cell, a device in which the transfer of electrons takes place through an external pathway rather than directly between reactants. Prof. T.L. Meeks, CHE Voltaic Cells Voltaic Cells The spontaneous reaction occurs in a single cell, but splitting the cell into two halves allows for the external pathway of electrons to be created. Prof. T.L. Meeks, CHE Voltaic Cells Prof. T.L. Meeks, CHE Voltaic Cells Prof. T.L. Meeks, CHE Voltaic Cells Parts of Voltaic Cell: 3the two solid metals connected by the external circuit are called electrodes. 3the electrode at which oxidation occurs is called the anode 3the electrode at which reduction occurs is called the cathode. Prof. T.L. Meeks, CHE Voltaic Cells Parts of Voltaic Cell: 3the anode metal is the substance that undergoes oxidation to form more positive ions. 3the positive metallic ion in the beaker that contains the cathode is the substance that undergoes reduction to form more of the metal electrode. Prof. T.L. Meeks, CHE Voltaic Cells Parts of Voltaic Cell: 3As the oxidation half cell gets a larger concentration of positive ions, negative ions flow through the semipermeable membrane that forms the salt bridge to balance the charges in the beaker. 3As the reduction half cell gets a smaller concentration of positive ions, positive ions flow through the membrane to replace the loss. Prof. T.L. Meeks, CHE Voltaic Cells Parts of Voltaic Cell: 3in any voltaic cell the electrons flow from the anode (-) to the cathode (+) through the external circuit 3The collection of lost electrons at the anode causes it to be negative, and the cathode is positive so it can attract the electrons and pull them through the wire. Prof. T.L. Meeks, CHE Voltaic Cells Prof. T.L. Meeks, CHE Voltaic Cells Sample exercise: The two half-reactions in a voltaic cell are Zn(s) Zn 2+ (aq) + 2e - ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) a) Indicate which reaction occurs at the anode and the cathode Prof. T.L. Meeks, CHE Voltaic Cells Sample exercise: The two half-reactions in a voltaic cell are Zn(s) Zn 2+ (aq) + 2e - ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) Prof. T.L. Meeks, CHE Voltaic Cells Sample exercise: The two half-reactions in a voltaic cell are Zn(s) Zn 2+ (aq) + 2e - ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) b) Which electrode is consumed first in the cell reaction? Prof. T.L. Meeks, CHE Voltaic Cells Sample exercise: The two half-reactions in a voltaic cell are Zn(s) Zn 2+ (aq) + 2e - ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) b) Which electrode is consumed first in the cell reaction? The anode is consumed Prof. T.L. Meeks, CHE Voltaic Cells Sample exercise: The two half-reactions in a voltaic cell are Zn(s) Zn 2+ (aq) + 2e - ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) c) Which electrode is positive? Prof. T.L. Meeks, CHE Voltaic Cells Sample exercise: The two half-reactions in a voltaic cell are Zn(s) Zn 2+ (aq) + 2e - ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) ClO 3 - (aq) + 6H + + 6e - Cl - (aq) + 3H 2 O(l) c) Which electrode is positive? The cathode is positive Prof. T.L. Meeks, CHE Voltaic Cells An atomic view of what occurs in a single cell in which both oxidation and reduction are occurring may give better insight. Prof. T.L. Meeks, CHE Voltaic Cells A molecular view allows a better understanding of what is occurring in each half cell. Prof. T.L. Meeks, CHE Cell EMF The reason that electrons flow spontaneously is do to the driving force behind certain redox reactions. 3The driving force is a difference in potential energy. 3The anode has a higher potential energy, and electrons flow to the cathode to achieve a lower potential energy. 3As in many natural wonders, the flow from higher concentration to lower continues Prof. T.L. Meeks, CHE Cell EMF The difference in potential energy per charge is measured in volts. 31 V (volt) is the potential difference required to impart 1 J (joule) of energy to a charge of 1 C (coulomb) 3the potential difference is called an electromotive force or emf 3emf, or E cell, is also called the cell potential Prof. T.L. Meeks, CHE Cell EMF The difference in potential energy per charge is measured in volts. 3The emf of a particular cell is dependent on the reaction that is occurring there. 3We will always assume standard conditions, and the cell potential should be denoted as such, E cell Prof. T.L. Meeks, CHE Cell EMF Standard reduction potential: 3the cell potential will depend on the particular cathode and anode used 3the cell potential is the difference between two electrode potentials E cell = E cathode + E anode E cell = E cathode + E anode 3the amount of moles of each DOES NOT change the potentials of cells Prof. T.L. Meeks, CHE Cell EMF Standard reduction potential: 3when finding the potentials, a reduction table will be used. 3Finding the potential for cathodes is straightforward, simply read the table 3Finding the potential for anodes means reading the table BACKWARDS and the sign of the potential should be switched 3Table 20.1 is a standard reduction table Prof. T.L. Meeks, CHE Cell EMF Sample exercise: Calculate the standard emf for a cell that employs the following overall reaction: Ni(s) + 2Fe 3+ (aq) 2Fe 2+ (aq) + Ni +2 (aq) Ni(s) + 2Fe 3+ (aq) 2Fe 2+ (aq) + Ni +2 (aq) Prof. T.L. Meeks, CHE Cell EMF Sample exercise: Calculate the standard emf for a cell that employs the following overall reaction: Ni(s) + 2Fe 3+ (aq) 2Fe 2+ (aq) + Ni +2 (aq) Ni(s) + 2Fe 3+ (aq) 2Fe 2+ (aq) + Ni +2 (aq) Prof. T.L. Meeks, CHE Cell EMF Sample exercise: Calculate the standard emf for a cell that employs the following overall reaction: Ger = cathode Ni(s) + 2Fe 3+ (aq) 2Fe 2+ (aq) + Ni +2 (aq) Ni(s) + 2Fe 3+ (aq) 2Fe 2+ (aq) + Ni +2 (aq) Leo = anode Prof. T.L. Meeks, CHE Cell EMF Sample exercise: Calculate the standard emf for a cell that employs the following overall reaction: Ger = cathode Ni(s) + 2Fe 3+ (aq) 2Fe 2+ (aq) + Ni +2 (aq) Ni(s) + 2Fe 3+ (aq) 2Fe 2+ (aq) + Ni +2 (aq) Leo = anode E cell = E cathode + E anode Prof. T.L. Meeks, CHE Cell EMF Sample exercise: Calculate the standard emf for a cell that employs the following overall reaction: Ger = cathode Ni(s) + 2Fe 3+ (aq) 2Fe 2+ (aq) + Ni +2 (aq) Ni(s) + 2Fe 3+ (aq) 2Fe 2+ (aq) + Ni +2 (aq) Leo = anode E cell = E cathode + E anode = = = V = V Prof. T.L. Meeks, CHE Cell EMF Sample exercise: Calculate the standard emf for a cell that employs the following overall reaction: 2Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) Prof. T.L. Meeks, CHE Cell EMF Sample exercise: Calculate the standard emf for a cell that employs the following overall reaction: Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) Prof. T.L. Meeks, CHE Cell EMF Sample exercise: Calculate the standard emf for a cell that employs the following overall reaction: leo = anode leo = anode Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) ger = cathode ger = cathode Prof. T.L. Meeks, CHE Cell EMF Sample exercise: Calculate the standard emf for a cell that employs the following overall reaction: leo = anode leo = anode Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) ger = cathode ger = cathode Ecell = = 2.20 V Prof. T.L. Meeks, CHE Cell EMF The more positive the potential difference, the greater the driving force. Prof. T.L. Meeks, CHE Batteries A battery is a portable, self-contained electrochemical power source that consists of one or more voltaic cells. The common 1.5 V battery is a single voltaic cell, where a 12 V battery uses multiple voltaic cells in one case. When cells are connected in series, with the cathode of one attached to the anode to another, the battery produces a voltage that is the sum of the emfs of the single cells. Prof. T.L. Meeks, CHE Batteries Batteries connected in series... Prof. T.L. Meeks, CHE Batteries Commerical batteries that has specific performance characteristics can require considerable ingenuity. The emf of a battery is determined by the substances used as the cathode and anode, and the usable life of the battery depends on the quantities of the substances packaged. Keep in mind the anode and cathode need to be separated by a porous barrier similar to a salt bridge. Prof. T.L. Meeks, CHE Batteries The materials used to construct the battery must be stable under the conditions in which it is to be used and must be chosen to minimize health and environmental concerns upon use and disposal. Different applications require batteries with different properties. Prof. T.L. Meeks, CHE Batteries Lead-Acid Battery Battery used in cars is one of the most common batteries, with over 100 million produced annually. A 12-V battery consists of six voltaic cells in series, each producing 2-V. Cathode: lead dioxide packed on a metal grid Anode: lead Prof. T.L. Meeks, CHE Batteries Lead-Acid Battery Both electrodes are immersed in sulfuric acid. Both reactants are solids so there is no need to separate the cell into anode and cathode compartments. To keep the electrodes from touching, wood or fiber glass spacers are placed between them Prof. T.L. Meeks, CHE Batteries Lead-Acid Battery Using solids offers another advantage, no concentration changes occur and the emf stays mostly constant, fluctuations occuring with variations in [H 2 SO 4 ]. One major advantage with this battery is the ability to be recharged Prof. T.L. Meeks, CHE Batteries Prof. T.L. Meeks, CHE Batteries Alkaline Battery: The most common non-rechargeable battery is the alkaline battery. More than are produced annually. The anode of this battery consists of powdered zinc metal immobilized in a gel in contact with a concentrated solution of KOH. The cathode is a mixture of MnO 2 and graphite Prof. T.L. Meeks, CHE Batteries Prof. T.L. Meeks, CHE Batteries Nickel-Cadmium, Nickel-Metal-Hydride, and Lithium-Ion Batteries All of these batteries are lightweight, easily rechargeable batteries for cell phones, notebook computers and video recorders. Prof. T.L. Meeks, CHE Batteries Nickel-Cadmium Cadmium is oxidized while nickel oxyhydroxide is reduced The solid reaction products adhere to the electrodes, which permits the reactions to be reversed during charging. Drawbacks: cadmium is a toxic heavy metal Prof. T.L. Meeks, CHE Batteries Nickel-Metal-Hydride Nickel oxyhydroxide is still reduced, but now the anode is a metal alloy that has the ability to absorb hydrogen atoms. The hydrogen atoms are released as water Prof. T.L. Meeks, CHE Batteries Lithium-Ion Batteries Lithium is a very light-weight element, and the technology is unique. Li+ ions insert themselves into certain layered solids; during discharge lithium ions migrate between two different layered materials that serve as the anode and cathode Prof. T.L. Meeks, CHE Batteries Fuel Cells The direct production of electricity from fuels by a voltaic cell could, in principle, yield a higher rate of conversion of the chemical energy of the reaction. Voltaic cells that perform this conversion using conventional fuels are called fuel cells. Fuel cells are not true batteries because they are not self-contained. Prof. T.L. Meeks, CHE Batteries Fuel Cells The most promising fuel cell system involves the reaction of H 2 and O 2 to form H 2 O(l) as the only product. Cathode: 4e- + O 2 + 2H 2 O 4OH- Cathode: 4e- + O 2 + 2H 2 O 4OH- Anode: 2H 2 + 4OH- 4H 2 O + 4e- Anode: 2H 2 + 4OH- 4H 2 O + 4e- Prof. T.L. Meeks, CHE Corrosion Undesirable redox reactions that lead to the corrosion of metals are the other side to spontaneous reactions. For nearly all metals, oxidation is a thermodynamically favorable process in air at room temperature. With certain metals, when oxidation occurs a protective oxide layer is formed - example: aluminum Prof. T.L. Meeks, CHE Corrosion Corrosion of iron From an economic standpoint, this is a major problem: 20% of the iron produced is used to replace iron objects that have been discarded because of rust damage. The rusting of iron requires both oxygen and water. Other factors can accelerate rusting - pH, salts, contact with metals and stress Prof. T.L. Meeks, CHE Corrosion Corrosion of iron Corrosion can occur anywhere, but the best place is the spot where the iron has the most access to oxygen. Prof. T.L. Meeks, CHE Corrosion Preventing the Corrosion of Iron Covering with paint or another metal can protect the iron from corrosion. Paint offers a layer against the iron coming into contact A second metal that is easier to oxidize then iron will corrode first offering protection. Prof. T.L. Meeks, CHE Corrosion Prof. T.L. Meeks, CHE Corrosion Prof. T.L. Meeks, CHE Electrolysis It is possible to use electrical energy to cause nonspontaneous redox reactions to occur. Processes which are driven by an outside source of electrical energy are called electrolysis reactions and take place in electrolytic cells. Prof. T.L. Meeks, CHE Electrolysis Electrolytic cells 3consists of two electrodes in a molten salt or a solution. 3driven by a battery or some other source of direct electrical current. 3the battery acts as an electron pump, pushing electrons into one electrode and pulling them from the other 3the electrodes are inert, acting only as the site for oxidation and reduction Prof. T.L. Meeks, CHE Electrolysis Electrolytic cells reduction still occurs at the cathode oxidation still occurs at the anode Prof. T.L. Meeks, CHE Electrolysis Electrolytic cells Na+ reduces at cathode to form Na Cl- oxidizes at anode to form Cl 2 Prof. T.L. Meeks, CHE Electrolysis Electrolytic cells 3the electrode connected to the negative terminal is now the cathode due to the battery supplying the electrons for reduction 3the electrode connected to the positive terminal is now the anode to draw electrons off the metal and cause the metal to oxidize and form ions Prof. T.L. Meeks, CHE Electrolysis Electrolysis of Aqueous Solutions 3because of the high melting points of ionic substances, the electrolysis of molten salts requires very high temperatures 3we can produce ions from soluble salts at room temperatures by dissolving the salt in water 3the electrolysis of an aqueous solution is complicated by the presence of water Prof. T.L. Meeks, CHE Electrolysis Electrolysis of Aqueous Solutions 3we must consider if the water is oxidized or reduced 3if the water is oxidized, O 2 and H + ions are formed 3if the water is reduced, H 2 and OH - ions are formed 3the more positive the value of E red, the more favorable the reduction is Prof. T.L. Meeks, CHE Electrolysis Electrolysis with Active Electrodes 3several practical applications of electrochemistry are based on active electrodes 3electroplating involves using electrolysis to deposit a thin layer of one metal on top of another in order to improve beauty or resistance to corrosion Prof. T.L. Meeks, CHE Electrolysis Electrolysis with Active Electrodes Prof. T.L. Meeks, CHE Electrolysis Quantitative Aspects of Electrolysis 3for any half reaction the amount of a substance that is reduced or oxidized in an electrolytic cell is directly proportional to the number of electrons passed into the cell 3Fig 20.31 Prof. T.L. Meeks, CHE Electrolysis Sample exercise: The half reaction for formation of magnesium metal upon electrolysis of molten MgCl 2 is Mg e - Mg. Calculate the mass of magnesium formed upon passage of current of 60.0 A for a period of 4.00 x 10 3 s. Prof. T.L. Meeks, CHE Electrolysis Sample exercise: Mg e - Mg. Calculate the mass of magnesium formed upon passage of current of 60.0 A for a period of 4.00 x 10 3 s. Coulombs= A x s = 60.0 A x 4.00 x 10 3 s = 2.4 x 10 5 C Prof. T.L. Meeks, CHE Electrolysis Sample exercise: Mg e - Mg. Calculate the mass of magnesium formed upon passage of current of 60.0 A for a period of 4.00 x 10 3 s. 2.4 x 10 5 C 1 mole e - 1 mole Mg g Mg 96,500 C 2 mole e - 1 mole Mg Prof. T.L. Meeks, CHE Electrolysis Sample exercise: Mg e - Mg. Calculate the mass of magnesium formed upon passage of current of 60.0 A for a period of 4.00 x 10 3 s. 2.4 x 10 5 C 1 mole e - 1 mole Mg g Mg 96,500 C 2 mole e - 1 mole Mg 30.2 g Mg 30.2 g Mg Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Any reaction that can occur in a voltaic cell to produce a positive emf must be spontaneous. 3E cell = E red + E ox 3a + emf is spontaneous 3a - emf is not spontaneous Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Using the standard reduction potentials, determine which of the following reactions are spontaneous. A) I 2 (s) + 5Cu 2+ (aq) + 6H 2 O(l) 2IO 3 - (aq) + 5Cu(s) + 12H + (aq) 2IO 3 - (aq) + 5Cu(s) + 12H + (aq) Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Using the standard reduction potentials, determine which of the following reactions are spontaneous. A) I 2 (s) + 5Cu 2+ (aq) + 6H 2 O(l) 2IO 3 - (aq) + 5Cu(s) + 12H + (aq) 2IO 3 - (aq) + 5Cu(s) + 12H + (aq) Red: Cu 2+ (aq) Cu(s) Ox: I 2 (s) + 6H 2 O 2IO 3 - (aq) + 12H + (aq) Ox: I 2 (s) + 6H 2 O 2IO 3 - (aq) + 12H + (aq) Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Using the standard reduction potentials, determine which of the following reactions are spontaneous. A) I 2 (s) + 5Cu 2+ (aq) + 6H 2 O(l) 2IO 3 - (aq) + 5Cu(s) + 12H + (aq) 2IO 3 - (aq) + 5Cu(s) + 12H + (aq) Red: Cu 2+ (aq) Cu(s) Ox: I 2 (s) + 6H 2 O 2IO 3 - (aq) + 12H + (aq) Ox: I 2 (s) + 6H 2 O 2IO 3 - (aq) + 12H + (aq) Ecell = Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Using the standard reduction potentials, determine which of the following reactions are spontaneous. A) I 2 (s) + 5Cu 2+ (aq) + 6H 2 O(l) 2IO 3 - (aq) + 5Cu(s) + 12H + (aq) 2IO 3 - (aq) + 5Cu(s) + 12H + (aq) Red: Cu 2+ (aq) Cu(s) Ox: I 2 (s) + 6H 2 O 2IO 3 - (aq) + 12H + (aq) Ox: I 2 (s) + 6H 2 O 2IO 3 - (aq) + 12H + (aq) Ecell = = NOT Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Using the standard reduction potentials, determine which of the following reactions are spontaneous. B) Hg 2+ (aq) + 2I - (aq) Hg(l) + I 2 (s) Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Using the standard reduction potentials, determine which of the following reactions are spontaneous. B) Hg 2+ (aq) + 2I - (aq) Hg(l) + I 2 (s) Red: Hg 2+ (aq) Hg(l) Ox: 2I - (aq) I 2 (s) Ox: 2I - (aq) I 2 (s) Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Using the standard reduction potentials, determine which of the following reactions are spontaneous. B) Hg 2+ (aq) + 2I - (aq) Hg(l) + I 2 (s) Red: Hg 2+ (aq) Hg(l) Ox: 2I - (aq) I 2 (s) Ox: 2I - (aq) I 2 (s) Ecell = Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Using the standard reduction potentials, determine which of the following reactions are spontaneous. B) Hg 2+ (aq) + 2I - (aq) Hg(l) + I 2 (s) Red: Hg 2+ (aq) Hg(l) Ox: 2I - (aq) I 2 (s) Ox: 2I - (aq) I 2 (s) Ecell = = SPON Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Looking at the standard reduction table, a general rule can be established: 3any metal in the series is able to spontaneously react with any ion beneath it there is also a relationship between emf and Gibbs Free Energy (remember a negative G is spontaneous) G = -nFE Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions G = -nFE n = number of electrons transferred in reaction F = Faradays constant 1 F = 96,500 J/V-mol E = emf of cell units of G are J/mol Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Consider the following reaction: 3Ni 2+ (aq) + 2Cr(OH) 3 (s) + 10OH - 3Ni(s) + 2CrO 4 2- (aq) + 8H 2 O(l) a) what is the value of n for this reaction? Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Consider the following reaction: 3Ni 2+ (aq) + 2Cr(OH) 3 (s) + 10OH - 3Ni(s) + 2CrO 4 2- (aq) + 8H 2 O(l) a) what is the value of n for this reaction? Each Ni 2+ gains 2e - and there are 3 ions, so 6e- are involved n = 6 Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Consider the following reaction: 3Ni 2+ (aq) + 2Cr(OH) 3 (s) + 10OH - 3Ni(s) + 2CrO 4 2- (aq) + 8H 2 O(l) b) Calculate G G = -nFE Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Consider the following reaction: 3Ni 2+ (aq) + 2Cr(OH) 3 (s) + 10OH - 3Ni(s) + 2CrO 4 2+ (aq) + 8H 2 O(l) b) Calculate G G = -nFE n = 6 F = 96,500 J/V-mol F = 96,500 J/V-mol E = = V E = = V Prof. T.L. Meeks, CHE Spontaneity of Redox Reactions Sample exercise: Consider the following reaction: 3Ni 2+ (aq) + 2Cr(OH) 3 (s) + 10OH - 3Ni(s) + 2CrO 4 2+ (aq) + 8H 2 O(l) b) Calculate G G = -nFE -(6)(96,500 J/V-mol)(-0.15 V) J/mol NOT Spon Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF As a voltaic cell is discharged, the reactants of the reaction are consumed and the products are generated, so the concentrations of these substances change. The emf progressively drops until E = 0, at which point we say the cell is dead. At that point the concentrations in the cell are at equilibrium. Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF The dependence of cell emf on concentration can be obtained from the dependence of the free energy change on concentration Nernst Equation: E = E - RT (lnQ) nF = E V (log Q) n Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF The dependence of cell emf on concentration can be obtained from the dependence of the free energy change on concentration Nernst Equation: E = E - RT (lnQ) nF * Q is the reaction= E V (log Q) quotient n Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Calculate the emf generated by the cell 2Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) when [Al 3+ ] = 4.0 x M and [I - ] = M. Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Calculate the emf generated by the cell 2Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) when [Al 3+ ] = 4.0 x M and [I - ] = M. E = E V (log Q) n Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Calculate the emf generated by the cell 2Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) when [Al 3+ ] = 4.0 x M and [I - ] = M. E = E V (log Q) n E = = 2.22 V Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Calculate the emf generated by the cell 2Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) when [Al 3+ ] = 4.0 x M and [I - ] = M. E = E V (log Q) n n = 6 e - Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Calculate the emf generated by the cell 2Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) when [Al 3+ ] = 4.0 x M and [I - ] = M. E = E V (log Q) n Q = [Al 3 +] 2 [I - ] 6 = [4.0 x ] 2 [0.010] 6 = 1.6 x Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Calculate the emf generated by the cell 2Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) when [Al 3+ ] = 4.0 x M and [I - ] = M. E = E V (log Q) n = (log 1.6 x ) 6 = (log 1.6 x ) 6 Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Calculate the emf generated by the cell 2Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) when [Al 3+ ] = 4.0 x M and [I - ] = M. E = E V (log Q) n = (log 1.6 x ) 6 = 2.36 V = (log 1.6 x ) 6 = 2.36 V Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Concentration cells: a voltaic cell with an emf of ZERO can be constructed if the same species is used as both the anode and the cathode if concentrations are the same; using two different concentrations of the same species will create a non-zero cell. A cell based solely on the emf generated because of a difference in a concentration is called a concentration cell Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF The standard emf for this cell would have to zero, but the cell operates due to the differences in concentration. The driving force for the cell is provided by the difference in concentration. Operation of the cell proceeds to equalize the concentrations: the more dilute concentration acts as the anode, the solid metal electrode will oxidize to turn into more ions. Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF The more concentrated solution acts as the cathode, where the larger number of ions reduce to plate the electrode and thereby reduce there number. The nernst equation can also be used here to calculate the E cell under non-standard conditions. Anode: X(s) X + (dilute) + e - Cathode: X + (conc.) + e - X(s) Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Anode: X(s) X + (dilute) + e - Cathode: X + (conc.) + e - X(s) X + (conc.) + X(s) X + (dilute) + X(s) X + (conc.) + X(s) X + (dilute) + X(s) Q = X + (dilute) / X + (conc.) Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: A concentration cell is constructed with two Zn(s) Zn 2+ (aq) half-cells. The first cell has [Zn 2+ ] = 1.35 M and the second cell has [Zn 2+ ] = 3.75 x M. a) Which half cell is the anode of the cell? Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: A concentration cell is constructed with two Zn(s) Zn 2+ (aq) half-cells. The first cell has [Zn 2+ ] = 1.35 M and the second cell has [Zn 2+ ] = 3.75 x M. a) Which half cell is the anode of the cell? The cell with [Zn 2+ ] = 3.75 x M is the anode. Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: A concentration cell is constructed with two Zn(s) Zn 2+ (aq) half-cells. The first cell has [Zn 2+ ] = 1.35 M and the second cell has [Zn 2+ ] = 3.75 x M. b) What is the emf of the cell? Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: A concentration cell is constructed with two Zn(s) Zn 2+ (aq) half-cells. The first cell has [Zn 2+ ] = 1.35 M and the second cell has [Zn 2+ ] = 3.75 x M. b) What is the emf of the cell? E = E V (log Q) n Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: A concentration cell is constructed with two Zn(s) Zn 2+ (aq) half-cells. The first cell has [Zn 2+ ] = 1.35 M and the second cell has [Zn 2+ ] = 3.75 x M. b) What is the emf of the cell? E = E V (log Q) n = (log 3.75x10 -4 /1.35) 2 = (log 3.75x10 -4 /1.35) 2 Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: A concentration cell is constructed with two Zn(s) Zn 2+ (aq) half-cells. The first cell has [Zn 2+ ] = 1.35 M and the second cell has [Zn 2+ ] = 3.75 x M. b) What is the emf of the cell? E = E V (log Q) n = (log 3.75x10 -4 /1.35) 2 = (log 3.75x10 -4 /1.35) 2 = V = V Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Cell emf and Chemical Equilibrium: the nernst equation helps us understand why the emf drops as the cell is discharged: as the reactants are converted into products, the value of Q increases, so the value of E decreases. The cell emf eventually reaches 0. When G = 0, E = 0, and an E cell = 0 shows a cell at equilibrium. Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF If 0 = E V (log Q) n Then log K = nE Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Using standard reduction potentials, calculate the equilibrium constant at 25C for the reaction Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Using standard reduction potentials, calculate the equilibrium constant at 25C for the reaction Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) log K = nE Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Using standard reduction potentials, calculate the equilibrium constant at 25C for the reaction Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) log K = nE = 2( ) = 2( ) Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Using standard reduction potentials, calculate the equilibrium constant at 25C for the reaction Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) log K = nE = 2( ) = 2( ) = = -9.93 Prof. T.L. Meeks, CHE Effect of Concentration on Cell EMF Sample exercise: Using standard reduction potentials, calculate the equilibrium constant at 25C for the reaction Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) Br 2 (l) + 2Cl - (aq) Cl 2 (g) + 2Br - (aq) K = 10 (-9.93) K = 10 (-9.93) = 1.2 x = 1.2 x Prof. T.L. Meeks, CHE Electrolysis Work done by electrical current can be calculated now based on the same formula as Gibbs Free Energy. w max = -nFE * voltaic cells w = nFE ext *electrolytic cells