Chemistry The Central Science 10 th edition Theodore
- Slides: 85
Chemistry, The Central Science, 10 th edition Theodore L. Brown; H. Eugene Le. May, Jr. ; and Bruce E. Bursten Chapter 20 Electrochemistry John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc. Electrochemistry
Electrochemical Reactions In electrochemical reactions, electrons are transferred from one species to another. Metals tend to lose electrons and are oxidized, non metals tend to gain electrons and are reduced. Electrochemistry
LEO GER Losing Electrons is Oxidation. Gaining Electrons is Reduction Electrochemistry
OIL RIG • Oxidation • Is • Loss. • • Reduction Is Gain. Electrochemistry
REDOX REACTIONS • Are reduction – oxidation reactions. • Electrons are transferred. • When an atom is losing electrons its O. N. increases. It is being oxidized. • When an atom gains electrons its O. N. decreases. It is being reduced Electrochemistry
Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers. Electrochemistry
Assigning Oxidation Numbers 1. Elements in their elemental form have an ON= 0. 2. The oxidation number of a monatomic ion is its charge. 3. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. Ø Oxygen has an oxidation number of − 2, except in the peroxide ion in which it has an oxidation number of − 1. Ø Hydrogen is +1 except in metal hydrides when is − 1. Ø Fluorine always has an oxidation number of − 1. Electrochemistry
Assigning Oxidation Numbers Ø The other halogens have an oxidation number of − 1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions 4. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. 5. The sum of the oxidation numbers in a neutral compound is 0. Electrochemistry
Oxidation and Reduction • A species is oxidized when it loses electrons. Ø Here, zinc loses two electrons to go from neutral zinc metal to the Zn 2+ ion. Electrochemistry
Oxidation and Reduction • A species is reduced when it gains electrons. Ø Here, each of the H+ gains an electron and they combine to form H 2. Electrochemistry
Oxidation and Reduction • What is reduced is the oxidizing agent. Ø H+ oxidizes Zn by taking electrons from it. • What is oxidized is the reducing agent. Ø Zn reduces H+ by giving it electrons. Electrochemistry
Oxidation-Reduction Reactions • Zn added to HCl yields the spontaneous reaction Zn(s) + 2 H+(aq) Zn 2+(aq) + H 2(g). • The oxidation number of Zn has increased from 0 to 2+. • The oxidation number of H has reduced from 1+ to 0. • Zn is oxidized to Zn 2+ while H+ is reduced to H 2. • H+ causes Zn to be oxidized and is the oxidizing agent. • Zn causes H+ to be reduced and is the reducing agent. • Note that the reducing agent is oxidized and the oxidizing Electrochemistry agent is reduced.
Balancing Oxidation. Reduction Reactions • Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end. • Conservation of charge: electrons are not lost in a chemical reaction. Half Reactions • Half-reactions are a convenient way of separating oxidation and reduction reactions. Electrochemistry
Half Reactions • The half-reactions for Sn 2+(aq) + 2 Fe 3+(aq) Sn 4+(aq) + 2 Fe 2+(aq) are Sn 2+(aq) Sn 4+(aq) +2 e 2 Fe 3+(aq) + 2 e- 2 Fe 2+(aq) • Oxidation: electrons are products. • Reduction: electrons are reactants. Loss of Gain of Electrons is Oxidation Reduction Electrochemistry
• • Balancing Equations by the Method of Half Reactions Consider the titration of an acidic solution of Na 2 C 2 O 4 (sodium oxalate, colorless) with KMn. O 4 (deep purple). Mn. O 4 - is reduced to Mn 2+ (pale pink) while the C 2 O 42 - is oxidized to CO 2. The equivalence point is given by the presence of a pale pink color. If more KMn. O 4 is added, the solution turns purple due to the excess KMn. O 4. Electrochemistry
What is the balanced chemical equation? 1. Write down the two half reactions. 2. Balance each half reaction: a. First with elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H+. d. If it is in basic solution, remove H+ by adding OHe. Finish by balancing charge by adding electrons. 3. Multiply each half reaction to make the number of electrons equal. 4. Add the reactions and simplify. 5. Check! Electrochemistry
Half-Reaction Method Consider the reaction between Mn. O 4− and C 2 O 42− : Mn. O 4−(aq) + C 2 O 42−(aq) Mn 2+(aq) + CO 2(aq) Electrochemistry
• • Balancing Equations by the Method of Half Reactions Consider the titration of an acidic solution of Na 2 C 2 O 4 (sodium oxalate, colorless) with KMn. O 4 (deep purple). Mn. O 4 - is reduced to Mn 2+ (pale pink) while the C 2 O 42 - is oxidized to CO 2. The equivalence point is given by the presence of a pale pink color. If more KMn. O 4 is added, the solution turns purple due to the excess KMn. O 4. Electrochemistry
Half-Reaction Method First, we assign oxidation numbers. +7 +3 +2 +4 Mn. O 4− + C 2 O 42 - Mn 2+ + CO 2 Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized. Electrochemistry
For KMn. O 4 + Na 2 C 2 O 4: • The two incomplete half reactions are Mn. O 4 -(aq) Mn 2+(aq) C 2 O 42 -(aq) 2 CO 2(g) 2. Adding water and H+ yields 8 H+ + Mn. O 4 -(aq) Mn 2+(aq) + 4 H 2 O • There is a charge of 7+ on the left and 2+ on the right. Therefore, 5 electrons need to be added to the left: 5 e- + 8 H+ + Mn. O 4 -(aq) Mn 2+(aq) + 4 H 2 O Electrochemistry
• In the oxalate reaction, there is a 2 - charge on the left and a 0 charge on the right, so we need to add two electrons: C 2 O 42 -(aq) 2 CO 2(g) + 2 e 3. To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplying gives: 10 e- + 16 H+ + 2 Mn. O 4 -(aq) 2 Mn 2+(aq) + 8 H 2 O 5 C 2 O 42 -(aq) 10 CO 2(g) + 10 e. Electrochemistry
4. Adding gives: 16 H+(aq) + 2 Mn. O 4 -(aq) + 5 C 2 O 42 -(aq) 2 Mn 2+(aq) + 8 H 2 O(l) + 10 CO 2(g) 5. Which is balanced! Electrochemistry
Balancing in Basic Solution • If a reaction occurs in basic solution, one can balance it as if it occurred in acid. • Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place. • If this produces water on both sides, you might have to subtract water from each side. Electrochemistry
Examples – Balance the following oxidationreduction reactions: 1. Cr (s) + NO 3 - (aq) Cr 3+ (aq) + NO (g) (acidic) 2. Al (s) + Mn. O 4 - (aq) Al 3+ (aq) + Mn 2+ (aq) (acidic) 3. PO 33 - (aq) + Mn 4 - (aq) PO 43 - (aq) + Mn. O 2 (s) (basic) 4. H 2 CO (aq) + Ag(NH 3)2+ (aq) HCO 3 - (aq) + Ag (s) + NH 3 (aq) (basic) Electrochemistry
Section 20 -3 20 -4 • • • Voltaic Cells – Spontaneous reactions ELECTRODES – POLARITIES SALT BRIDGE DRIVING FORCE- EMF STANDARD REDUCTION POTENTIALS HW Q 23, 25, 31, 33 (a, b ) , 35 Electrochemistry
Voltaic Cells In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released. Electrochemistry
Voltaic Cells • We can use that energy to do work if we make the electrons flow through an external device. • We call such a setup a voltaic cell. Electrochemistry
Voltaic Cells • A typical cell looks like this. • The oxidation occurs at the anode. • The reduction occurs at the cathode. Electrochemistry
Voltaic Cells Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop. Electrochemistry
Voltaic Cells • Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. Ø Cations move toward the cathode. Ø Anions move toward the anode. Electrochemistry
Voltaic Cells • In the cell, then, electrons leave the anode and flow through the wire to the cathode. • As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment. Electrochemistry
Voltaic Cells • As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. • The electrons are taken by the cation, and the neutral metal is deposited on the cathode. Electrochemistry
Electromotive Force (emf) • Water only spontaneously flows one way in a waterfall. • Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy. Electrochemistry
Electromotive Force (emf) • The potential difference between the anode and cathode in a cell is called the electromotive force (emf). • It is also called the cell potential, and is designated Ecell. Electrochemistry
Cell Potential Cell potential is measured in volts (V). One volt is the potential difference required to impart 1 J of energy to a charge of 1 coulomb. (1 electron has a charge of 1. 6 x 10 -19 Coulombs). The potential difference between the 2 electrode provides the driving force that pushes the electron through the external circuit. Electrochemistry
Cell Potential or Electromotive Force (emf) • The “pull” or driving force on the electrons. • ELECTROMOTIVE FORCE EMF • CAUSES THE ELECTRON MOTION! J 1 V=1 C Electrochemistry
Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated. Electrochemistry
Standard Hydrogen Electrode • Their values are referenced to a standard hydrogen electrode (SHE). • By definition, the reduction potential for hydrogen is 0 V: 2 H+ (aq, 1 M) + 2 e− H 2 (g, 1 atm) Electrochemistry
Standard Cell Potentials The cell potential at standard conditions can be found through this equation: Ecell (cathode) − Ered (anode) = Ered Because cell potential is based on the potential energy per unit of charge, it is an intensive property. Electrochemistry
For the reaction to be SPONTANEOUS E has to be positive Electrochemistry
Cell Potentials • For the oxidation in this cell, Ered = − 0. 76 V • For the reduction, Ered = +0. 34 V Electrochemistry
Cell Potentials Ecell = Ered (cathode) − Ered (anode) = +0. 34 V − (− 0. 76 V) = +1. 10 V Electrochemistry
• Since E red = -0. 76 V we conclude that the reduction of Zn 2+ in the presence of the SHE is not spontaneous. • The oxidation of Zn with the SHE is spontaneous. • Changing the stoichiometric coefficient does not affect E red. • Therefore, 2 Zn 2+(aq) + 4 e- 2 Zn(s), E red = -0. 76 V. • Reactions with E red > 0 are spontaneous reductions relative to the SHE. Electrochemistry
• Reactions with E red < 0 are spontaneous oxidations relative to the SHE. • The larger the difference between E red values, the larger E cell. • In a voltaic (galvanic) cell (spontaneous) E red(cathode) is more positive than E red(anode). • Recall Electrochemistry
Cell Potential Calculations • • • To Calculate cell potential using Standard Reduction Potentials: 1. One reaction and its cell potential’s sign must be reversed--it must be chosen such that the overall cell potential is positive. 2. The half-reactions must often be multiplied by an integer to balance electrons--this is not done for the cell Electrochemistry potentials.
Cell Potential Calculations Continued • • Fe 3+(aq) + Cu(s) ----> Cu 2+(aq) + Fe 2+(aq) Fe 3+(aq) + e- ----> Fe 2+(aq) Eo = 0. 77 V Cu 2+(aq) + 2 e- ----> Cu(s) Eo = 0. 34 V Reaction # 2 must be reversed. Electrochemistry
Cell Potential Calculations Continued • • • 2 (Fe 3+(aq) + e- ----> Fe 2+(aq)) Eo = 0. 77 V Cu(s) ----> Cu 2+(aq) + 2 e. Eo = - 0. 34 V 2 Fe 3+(aq) + Cu(s) ----> Cu 2+(aq) + 2 Fe 2+(aq) • Eo = 0. 43 V Electrochemistry
Oxidizing and Reducing Agents The greater the difference between the two, the greater the voltage of the cell. REMEMBER TO REVERSE THE SIGN OF THE SPECIE THAT GETS OXIDIZED (THE ONE BELOW!!!!!) Electrochemistry
Oxidizing and Reducing Agents • The strongest oxidizers have the most positive reduction potentials. • O. A. PULL electrons • The strongest reducers have the most negative reduction potentials. • R. A. PUSH their electrons Electrochemistry
Examples – Sketch the cells containing the following reactions. Include E°cell, the direction of electron flow, direction of ion migration through salt bridge, and identify the anode and cathode: 1. Cu 2+ + Mg (s) Mg 2+ + Cu (s) 2. Zn (s) + Ag+ Zn 2+ + Ag Electrochemistry
Free Energy G for a redox reaction can be found by using the equation G = −n. FE where n is the number of moles of electrons transferred, and F is a constant, the Faraday. 1 F = 96, 485 C/mol = 96, 485 J/V-mol Electrochemistry
Free Energy and Cell Potential Under standard conditions • G = n. FE • • • n = number of moles of electrons F = Faraday = 96, 485 coulombs per mole of electrons = 96, 485 J/Vmol E=V Electrochemistry
Find the free energy for the reaction of Zn + Cu 2+ ---> Zn 2+ + Cu Electrochemistry
Effect of Concentration on Cell EMF The Nernst Equation • A voltaic cell is functional until E = 0 at which point equilibrium has been reached. • The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction. • The Nernst equation relates emf to concentration using and noting that Electrochemistry
The Nernst Equation • This rearranges to give the Nernst equation: • The Nernst equation can be simplified by collecting all the constants together using a temperature of 298 K: • (Note that change from natural logarithm to base-10 log. ) • Remember that n is number of moles of electrons. Electrochemistry
Cell EMF and Chemical Equilibrium • A system is at equilibrium when G = 0. • From the Nernst equation, at equilibrium and 298 K (E = 0 V and Q = Keq): Electrochemistry
Examples 1. Calculate E°cell for the following reaction: Zn (s) + Cu 2+ Zn 2+ + Cu (s) 2. Calculate Ecell if [Zn 2+] = 1. 88 M and [Cu 2+] = 0. 020 M 3. Calculate E°cell for the following reaction: IO 3 - (aq) + Fe 2+ (aq) Fe 3+ (aq) + I 2 (aq) 4. Calculate Ecell if [IO 3 -] = 0. 20 M, [Fe 2+] = 0. 65 M, [Fe 3+] = 1. 0 M, and [I 2] = 0. 75 M Electrochemistry
Concentration Cells • Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. would be 0, but Q would not. • For such a cell, Ecell • Therefore, as long as the concentrations are different, E will not be 0. Electrochemistry
PROBLEM • Use the Nernst equation to determine the (emf) at 25 o. C of the cell. Zn(s)│Zn 2+(0. 00100 M ║Cu 2+(10. 0 M)│Cu(s). • The standard emf of this cell is 1. 10 V. Electrochemistry
Applications of Oxidation-Reduction Reactions Electrochemistry
Batteries • A battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series. Electrochemistry
Batteries Electrochemistry
A lead storage battery consists of a lead anode, lead Electrochemistry dioxide cathode, and an electrolyte of 38% sulfuric acid.
Lead Storage Battery • Anode reaction: Pb(s) +H 2 SO 4(aq) ---> Pb. SO 4(aq) + 2 H+(aq) + 2 e • Cathode reaction: Pb. O 2(s)+H 2 SO 4(aq)+ 2 e-+2 H+(aq)->Pb. SO 4(aq)+2 H 2 O(l) • Overall reaction: Pb(s)+ Pb. O 2(s) + 2 H 2 SO 4(aq)-->Pb. SO 4(aq)+ 2 H 2 O (l) Electrochemistry
Common dry cell and its components. Electrochemistry
Alkaline Battery • Anode: Zn cap: Zn(s) Zn 2+(aq) + 2 e • Cathode: Mn. O 2, NH 4 Cl and C paste: 2 NH 4+(aq) + 2 Mn. O 2(s) + 2 e- Mn 2 O 3(s) + 2 NH 3(aq) + 2 H 2 O(l) • The graphite rod in the center is an inert cathode. • For an alkaline battery, NH 4 Cl is replaced with KOH. Electrochemistry
• Anode: Zn powder mixed in a gel: Zn(s) Zn 2+(aq) + 2 e • Cathode: reduction of Mn. O 2. Electrochemistry
Electrochemistry
• • Fuel Cells Direct production of electricity from fuels occurs in a fuel cell. On Apollo moon flights, the H 2 -O 2 fuel cell was the primary source of electricity. Cathode: reduction of oxygen: 2 H 2 O(l) + O 2(g) + 4 e- 4 OH-(aq) Anode: 2 H 2(g) + 4 OH-(aq) 4 H 2 O(l) + 4 e. Electrochemistry
Mercury battery used in calculators. Electrochemistry
Fuel Cells • . . . galvanic cells for which the reactants are continuously supplied. • 2 H 2(g) + O 2(g) 2 H 2 O(l) • • anode: 2 H 2 + 4 OH 4 H 2 O + 4 e cathode: 4 e + O 2 + 2 H 2 O 4 OH Electrochemistry
Hydrogen Fuel Cells Electrochemistry
Ion selective electrodes are glass electrodes that measures Electrochemistry a change in potential when [H+] varies. Used to measure p. H.
Corrosion • • • Corrosion of Iron Since E red(Fe 2+) < E red(O 2) iron can be oxidized by oxygen. Cathode: O 2(g) + 4 H+(aq) + 4 e- 2 H 2 O(l). Anode: Fe(s) Fe 2+(aq) + 2 e-. Dissolved oxygen in water usually causes the oxidation of iron. Fe 2+ initially formed can be further oxidized to Fe 3+ which forms rust, Fe 2 O 3. x. H 2 O(s). Electrochemistry
• Oxidation occurs at the site with the greatest concentration of O 2. Preventing Corrosion of Iron • Corrosion can be prevented by coating the iron with paint or another metal. • Galvanized iron is coated with a thin layer of zinc. Electrochemistry
Electrochemistry
• Zinc protects the iron since Zn is the anode and Fe the cathode: Zn 2+(aq) +2 e- Zn(s), E red = -0. 76 V Fe 2+(aq) + 2 e- Fe(s), E red = -0. 44 V • With the above standard reduction potentials, Zn is easier to oxidize than Fe. Electrochemistry
Electrochemistry
Preventing Corrosion of Iron • To protect underground pipelines, a sacrificial anode is added. • The water pipe is turned into the cathode and an active metal is used as the anode. • Often, Mg is used as the sacrificial anode: Mg 2+(aq) +2 e- Mg(s), E red = -2. 37 V Fe 2+(aq) + 2 e- Fe(s), E red = -0. 44 V Electrochemistry
Electrochemistry
Corrosion • Some metals, such as copper, gold, silver and platinum, are relatively difficult to oxidize. These are often called noble metals. • About 1/5 of all iron and steel produced each year is used to replace rusted metal. Electrochemistry
Self-protecting Metals • Some metals such as aluminum, copper, and silver form a protective coating that keeps them from corroding further. • The protective coating for iron and steel flakes away opening new layers of metal to corrosion. Electrochemistry
Prevention of Corrosion • • Coating--painting or applying oil to keep out oxygen and moisture. Galvanizing--dipping a metal in a more active metal -- galvanized steel bucket. Alloying -- mixing metals with iron to prevent corrosion -- stainless steel. Cathodic protection -- attaching a more active metal. Serves as sacrificial metal--used to protect ships, gas lines, and gas tanks. Electrochemistry
Redox Titrations • Same as any other titration. • the permanganate ion is used often because it is its own indicator. Mn. O 4 - is purple, Mn+2 is colorless. When reaction solution remains clear, Mn. O 4 - is gone. • Chromate ion is also useful, but color change, orangish yellow to green, is harder to detect. Electrochemistry
Example • The iron content of iron ore can be determined by titration with standard KMn. O 4 solution. The iron ore is dissolved in excess HCl, and the iron reduced to Fe+2 ions. This solution is then titrated with KMn. O 4 solution, producing Fe+3 and Mn+2 ions in acidic solution. If it requires 41. 95 m. L of 0. 205 M KMn. O 4 to titrate a solution made with 10. 613 g of iron ore, what percent of the ore was iron? Electrochemistry
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