Chemistry The Central Science 10 th edition Theodore

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Chemistry, The Central Science, 10 th edition Theodore L. Brown; H. Eugene Le. May,

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

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

LEO GER Losing Electrons is Oxidation. Gaining Electrons is Reduction Electrochemistry

OIL RIG • Oxidation • Is • Loss. • • Reduction Is Gain. Electrochemistry

OIL RIG • Oxidation • Is • Loss. • • Reduction Is Gain. Electrochemistry

REDOX REACTIONS • Are reduction – oxidation reactions. • Electrons are transferred. • When

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

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.

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

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,

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,

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

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

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

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)

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

• • 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.

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−

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

• • 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−

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

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

• 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

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

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 -

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

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.

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

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

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

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

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

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

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. •

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

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

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

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 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).

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

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

For the reaction to be SPONTANEOUS E has to be positive Electrochemistry

Cell Potentials • For the oxidation in this cell, Ered = − 0. 76

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 −

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

• 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

• 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:

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) +

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

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

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.

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

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

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 •

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

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

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

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 =

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

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

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

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

Applications of Oxidation-Reduction Reactions Electrochemistry

Batteries • A battery is a galvanic cell or, more commonly, a group of

Batteries • A battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series. Electrochemistry

Batteries Electrochemistry

Batteries Electrochemistry

A lead storage battery consists of a lead anode, lead Electrochemistry dioxide cathode, and

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

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

Common dry cell and its components. Electrochemistry

Alkaline Battery • Anode: Zn cap: Zn(s) Zn 2+(aq) + 2 e • Cathode:

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

• Anode: Zn powder mixed in a gel: Zn(s) Zn 2+(aq) + 2 e • Cathode: reduction of Mn. O 2. Electrochemistry

Electrochemistry

Electrochemistry

 • • Fuel Cells Direct production of electricity from fuels occurs in a

• • 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

Mercury battery used in calculators. Electrochemistry

Fuel Cells • . . . galvanic cells for which the reactants are continuously

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

Hydrogen Fuel Cells Electrochemistry

Ion selective electrodes are glass electrodes that measures Electrochemistry a change in potential when

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

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.

• 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

Electrochemistry

 • Zinc protects the iron since Zn is the anode and Fe the

• 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

Electrochemistry

Preventing Corrosion of Iron • To protect underground pipelines, a sacrificial anode is added.

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

Electrochemistry

Corrosion • Some metals, such as copper, gold, silver and platinum, are relatively difficult

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

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

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

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

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