1 Slide 1 / 144 Electrochemistry Electrochemistry Slide 2 / 144 Electrochemistry deals with relationships between reactions and electricity In electrochemical reactions, electrons are transferred from one species to another. Provide insight into batteries, corrosion, electroplating, spontaneity of reactions Electrochemical Reactions Slide 3 / 144 In electrochemical reactions, electrons are transferred between various reactant and product species in reactions. As a result, oxidation state/number of one or more substances/species change Oxidation number is the formal charge on the atom when it is connected to other atoms. In order to keep track of what species loses electrons and what gains them, we assign oxidation numbers/oxidation states to individual atoms.
2 Oxidation Numbers Slide 4 / 144 Take a look at this reaction between Zn metal and acid with assigned oxidation numbers. How do we know what number goes with each atom? Where do these numbers came from? Zn(s) + 2H + (aq) Zn 2+ + H 2 (g) How do we assign oxidation numbers? Slide 5 / 144 Rules for Assigning Oxidation Numbers Elements Compounds Monoatomic ions Polyatomic ions Elements in their elemental form have an oxidation number of 0. The sum of the oxidation numbers in a neutral compound is 0. The oxidation number of a monatomic ion is the same as its charge. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. Rules for Assigning Oxidation Numbers Slide 6 / 144 Hydrogen Fluorine Other halogens -1 when bonded to a metal +1 when bonded to a nonmetal Fluorine always has an oxidation number of -1. Usually -1. May have positive oxidation numbers in oxyanions. For example, Cl has an oxidation number of +5 in ClO 3-.
3 Slide 7 / 144 Rules for Assigning Oxidation Numbers Nonmetals Nonmetals tend to have negative oxidation numbers although some are positive in certain compounds or ions. Oxygen Oxygen has a oxidation number of -2, except in the peroxide ion, when its oxidation number is What is the oxidation number of each oxygen atom in the compound MnO2? Slide 8 / 144 A -2 B -1 C 0 D +1 E +2 2 What is the oxidation number of the manganese atom in the compound MnO2? Slide 9 / 144 A +3 B +2 C +1 D +4 E +7
4 3 What is the oxidation number of oxygen atom in MnO 4 1-, the permanganate ion? Slide 10 / 144 A -2 B -1 C 0 D +2 E +4 4 What is the oxidation number of the manganese atom in MnO4 1-, the permanganate ion? Slide 11 / 144 A +1 B +2 C +5 D +4 E +7 5 What is the oxidation number of sulfur in HSO 4 1-, the hydrogen sulfate ion? Slide 12 / 144 A -2 B +1 C +2 D +4 E +6
5 Oxidation and Reduction Zn(s) + 2H + (aq) Zn 2+ + H 2 (g) Slide 13 / Oxidation-loss of electrons A species is oxidized when it loses electrons. Here, zinc loses two electrons to go from neutral Zn metal to the Zn 2+ ion. Zn is also a reducing agent- provides electrons (reductant) Reducing agent loses electrons. LEO The lion says OIL RIG GER Oxidation and Reduction Slide 14 / 144 Zn(s) + 2H + (aq) Zn 2+ + H 2 (g) Reduction- gaining of electrons A species is reduced when it gains electrons. Here, each of the H + gains an electron, and they combine to form H2. H is an oxidizing agent- accepts electrons (oxidant) An oxidizing agent gains electrons. Oxidation and Reduction Slide 15 / 144 Zn(s) + 2H + (aq) Zn 2+ + H 2 (g) What is reduced is the oxidizing agent. H + oxidizes Zn by taking electrons from it. What is oxidized is the reducing agent. Zn reduces H + by giving it electrons.
6 Oxidation and Reduction Redox Reactions Slide 16 / 144 Zn(s) + 2H + (aq) Zn 2+ + H 2 (g) An electrochemical reaction in which oxidation and reduction occurs is known as a REDOX reaction 6 Which of the following is/are an oxidationreduction (redox) reactions? (a) K 2 CrO 4 + BaCl 2 KCl + BaCrO 4 Slide 17 / 144 (b) Pb Br 1- PbBr 2 (c) Cu + S CuS A B C D E a only b only c only a and c b and c 7 Which substance is oxidized in the following reaction? (First, assign oxidation numbers.) Slide 18 / 144 Cu + S CuS A B C D E Cu S Cu and S CuS This is not a redox reaction.
7 8 Which substance is the reducing agent below? Slide 19 / 144 Cu + S CuS A B C D E Cu S Cu and S CuS This is not a redox reaction. 9 Which substance is oxidized in the following reaction? (First, assign oxidation numbers.) Slide 20 / 144 Ca + Fe 3+ Ca 2+ + Fe A Ca B Fe 3+ C Ca 2+ D Fe E This is not a redox reaction. 10 Which substance is the oxidizing agent below? Slide 21 / 144 Ca + Fe 3+ Ca 2+ + Fe A Ca B Fe 3+ C Ca 2+ D Fe E This is not a redox reaction.
8 11 Which substance is reduced in the following reaction? (First, assign oxidation numbers.) Slide 22 / K + Al(NO 3 ) 3 Al + 3 KNO 3 A B C D E K Al N O This is not a redox reaction. 12 Which substance is the reducing agent? Slide 23 / K + Al(NO 3 ) 3 Al + 3 KNO 3 A K B Al(NO 3 ) 3 C KNO 3 D This is not a redox reaction. Redox Practice 1 Slide 24 / 144 H 2S (g) + Cl 2 (g) --> 2HCl (g) + S (s) a) Assign oxidation numbers to each element above. b) Which element is oxidized? c) Which element is reduced? d) Name the reducing agent. e) Name the oxidizing agent.
9 Redox Practice 2 Slide 25 / 144 SnCl 2(aq) + 2HgCl 2 (aq) --> SnCl 4 (aq) + Hg 2Cl 2 (s) a) Assign oxidation numbers to each element above. b) Which element is oxidized? c) Which element is reduced? d) Name the reducing agent. e) Name the oxidizing agent. 13 Which element is oxidized in the reaction below? Slide 26 / 144 Fe 2+ + H + + Cr 2 O 7 2- Fe 3+ + Cr 3+ + H 2 O 14 H 2 S (g) + Cl 2 (g) --> 2HCl (g) + S (s) Slide 27 / 144 Which is oxidized? Which is reduced?
10 15 SnCl 2 (aq) + 2HgCl 2 (aq) --> SnCl 4 (aq) + Hg 2 Cl 2 (s) Slide 28 / 144 Which is oxidized? Which is reduced? Redox reactions in aqueous solutions Slide 29 / 144 A large number of redox reactions occur in aqueous solutions. Unlike acid base nutralization and precipitation reactions,most of the reaction proceed slowly. Each redox reaction is the sum of two half reactions: Consider the reaction of iodide ions and hydrogen peroxide. 2I- (aq) + H 2 O 2 (aq) + 2e- I2 + 2OH- (aq) + 2e- Redox reactions in aqueous solutions Slide 30 / 144 This reaction involves two parts as represented below. 2I- (aq) + H 2 O 2 (aq) + 2e- I 2 + 2OH- (aq) + 2e- 1. Oxidation half reaction 2. Reduction half reaction. 2I - (aq) I 2 + 2e - oxidation H 2 O 2 (aq) + 2e - 2OH - (aq) reduction Add the two half reactions to get the overall reaction. 2I- (aq) + H 2 O 2 (aq) + 2e- I 2 + 2OH- (aq) + 2e- How do we balance a redox reaction?
11 Balancing Redox reactions Slide 31 / 144 Half-reaction method (oxidation # method) Assign oxidation numbers to determine what is oxidized and what is reduced. Identify the oxidation and reduction process. Write down the individual oxidation and reduction equations. Balance these half reactions Combine them to attain the balanced equation for the overall reaction. This method can be used in general to balance any redox reaction unless any specific condition such as acidic or basic is mentioned Half-reaction method (oxidation # method) Slide 32 / 144 Let us consider the simple replacement reaction of Mg with AgCl Mg + AgCl Ag + MgCl Oxidation: Mg --> Mg e (1) Reduction: Ag e - --> Ag (2) Since all the atoms are balanced, we need to balance only electrons. Multiply equation (2) x 2 Oxidation: Mg --> Mg e (1) Reduction: 2Ag e- --> 2Ag (3) Half-reaction method (oxidation # method) Slide 33 / 144 Adding the half-reactions (1) and (3) yields the following: Oxidation: Mg --> MgCl 2 + 2e - Reduction: 2AgCl + 2e - --> 2Ag Overall: Mg + 2AgCl + 2e - --> MgCl 2 + 2e - + 2Ag and we cancel out electrons from both sides: Overall: Mg + 2AgCl + 2e - --> MgCl 2 + 2e - + 2Ag Net equation: Mg + 2AgCl --> MgCl 2 + 2Ag Since the original equation is given with chlorine you would keep it here in the final balanced equation too.
12 Redox reactions -balancing Slide 34 / 144 Practice Fe3O4 +C --> Fe + CO Fe3O4 + 4C --> 3Fe + 4CO Redox reactions -balancing Slide 35 / 144 Practice: SnO2 + C --> Sn + CO SnO2 + 2C --> Sn + 2CO The Half-Reaction Method In acidic medium: Slide 36 / 144 This diagram shows the steps involved in balancing half-reactions. Write down the individual half reaction. First balance atoms other than H and O. Balance oxygen atoms by adding H2O. Balance hydrogen atoms by adding H +. Balance charge by adding electrons. Multiply the half-reactions by integers so that the electrons gained and lost are the same. Other atoms O H e-
13 The Half-Reaction Method Slide 37 / 144 In acidic medium: Continued Add the half-reactions, subtracting things that appear on both sides. Make sure the equation is balanced according to mass. Make sure the equation is balanced according to charge. The Half-Reaction Method Slide 38 / 144 In acidic medium: Consider the reaction between MnO4 and C2O4 2 : MnO4 (aq) + C2O4 2 (aq) Mn 2+ (aq) + CO2 (aq) MnO 4 (aq) + C2 O 4 2 (aq) Mn 2+ (aq) + CO 2 (aq) First, we assign oxidation numbers. We only assign oxidation numbers to elements whose oxidation numbers CHANGES. Here, oxygen's oxidation number remains constant at -2. In acidic medium: The Half-Reaction Method Slide 39 / MnO 4 (aq) + C2 O 4 2 (aq) Mn 2+ (aq) + CO 2 (aq) Which substance gets reduced? Which substance gets oxidized? Which substance is the reducing agent? Which substance is the oxidizing agent?
14 In acidic medium: The Half-Reaction Method Slide 40 / MnO 4 (aq) + C2 O 4 2 (aq) Mn 2+ (aq) + CO 2 (aq) Since the manganese goes from +7 to +2, it is reduced. The MnO4 - ion is the oxidizing agent. Since the carbon goes from +3 to +4, it is oxidized. The C2O4 2- ion is the reducing agent. In acidic medium: Oxidation Half-Reaction Slide 41 / 144 C 2 O 4 2 CO 2 To balance the carbon, we add a coefficient of 2: C 2 O CO 2 In acidic medium: Oxidation Half-Reaction Slide 42 / 144 C2O4 2 2 CO2 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side. C2O4 2 2 CO2 + 2 e
15 In acidic medium: Reduction Half-Reaction Slide 43 / 144 MnO4 Mn 2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side. MnO4 Mn H2O Slide 44 / 144 In acidic medium: Reduction Half-Reaction MnO 4 Mn H 2 O To balance the hydrogen, we add 8 H + to the left side. 8 H + + MnO 4 Mn H 2 O Reduction Half-Reaction Slide 45 / 144 In acidic medium: 8 H + + MnO 4 Mn H 2 O To balance the charge, we add 5 e to the left side. 5 e + 8 H + + MnO 4 Mn H 2 O
16 Combining the Half-Reactions In acidic medium: Slide 46 / 144 Now we evaluate the two half-reactions together: C2O4 2 2 CO2 + 2 e 5 e + 8 H + + MnO4 Mn H2O To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2. In acidic medium: Combining the Half-Reactions Slide 47 / C2O CO e 10 e + 16 H MnO4 2 Mn H2O When we add these together, we get: 10 e + 16 H MnO4 + 5 C2O4 2 --> 2 Mn H2O + 10 CO2 +10 e Combining the Half-Reactions Slide 48 / 144 In acidic medium: 10 e + 16 H MnO C2O42 2 Mn H 2O + 10 CO e The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16 H MnO C2O42 2 Mn H 2O + 10 CO 2
17 Practice 1 The Half-Reaction Method In acidic medium: Cd(s) + NiO2 (s) --> Cd(OH) 2(s) + Ni(OH) 2(s) Slide 49 / 144 a) Write the oxidation half reaction. b) Write the reduction half reaction. c) Write the balanced net reaction. d) Identify the oxidizing agent. e) Identify the reducing agent. In acidic medium: - Practice 2 Cu + NO 3 --> NO 2 + Cu 2+ Slide 50 / 144 Practice 3 Cr 2 O Fe 2+ + H + --> Cr 3+ + Fe 3+ + H 2 O Slide 51 / 144
18 Slide 52 / 144 Practice 4 :MnO Br - --> Mn 2+ + Br 2 in acidic solution Slide 53 / 144 Practice 5 : Cr 2O C 2H 4O --> C 2H 4O 2 + Cr 3+ in acidic solution Cr 2O H+ + 3C 2H 4O --> 3C 2H 4O 2 + 2Cr H 2O Redox reaction in basic medium Some redox reactions requires basic medium to occur.in this case the following steps need to be performed to balance the reaction. Slide 54 / Assign the oxidation numbers 2- Balance the "other atoms" involved 3- Separate the half reactions 4- Add water molecules to balance oxygen atom whatever side deficient in O atoms 5- Add water molecules equal in number to the deficiency of H atoms. 6- Add same number of OH- to the other side. 7- Balance the charge by adding electrons on the appropriate side 8- Balance the electrons lost /gained by multiplying the reactions by integers 9- Add the two reactions removing any duplication if any of common species on either side. Can also be performed without splitting the two equations.
19 Balancing in Basic Solution Slide 55 / 144 Zn + NO > Zn 2+ + NH 4 + in basic medium: Oxidation half reaction: Zn ----> Zn e- Reduction half reaction: NO > NH 4 + NO > NH H 2O 10H 2O + NO > NH H 2O 10H 2O + NO > NH H 2O + 10OH - 8e- 10H 2O + NO > NH H 2O + 10OH - 4Zn ----> 4 Zn2+ + 8e- 4Zn + 1NO H 2 O--> 4Zn NH OH - Balancing in Basic Solution ** Slide 56 / 144 Can also be performed without splitting the two equations. Zn + NO > Zn 2+ + NH 4 + in basic medium: 1. Assigh oxidation #s: Increases by 2 Zn + NO > Zn 2+ + NH decreases by 8 2. Balance the change in Oxidation # change on either side. increases by 8 4Zn + 1NO > 4Zn NH 4 + decreases by 8 Balancing in Basic Solution 4Zn + 1NO > 4Zn NH Balance O atoms by adding H 2O molecules to the side deficient in O atoms. 3 O atoms on the LHS so add 3 water on the RHS Slide 57 / 144 4Zn + 1NO > 4Zn NH H 2O 4. The H atoms are then balanced by adding H 2O to the side lacks H. 10 H on the RHS, so add 10 water on the LHS. 4Zn + 1NO H 2O--> 4Zn NH H 2O 5. Add 10 OH- on the other side of the reaction to balance the extra H and O. 4Zn + 1NO H 2O--> 4Zn NH H 2O + 10 OH - 6. If this produces water on both sides, you might have to subtract water from each side. 4Zn + 1NO H 2O--> 4Zn NH OH -
20 Balancing in Basic Solution Slide 58 / 144 Practice: 1 Fe(OH)2 + H2O2 --> Fe(OH)3 + H2O in basic solution Oxidation: decrease by 1 for each O atom, total 2 e- taken Fe(OH) 2 + H 2O 2 --> Fe(OH) 3 + H 2O in basic solution increase by 1, 1 e - given 2 Fe(OH) 2 + H 2O 2 --> 2Fe(OH) 3 + H 2O in basic solution Balance O atoms by adding 2 H2O to LHS 2 Fe(OH) 2 + 2H 2O --> 2Fe(OH) 3 Balance H atoms by adding 2 H2O to RHS 2 Fe(OH) 2 + 2H 2O --> 2Fe(OH) 3 + 2H 2O Add 2 OH- on the LHS 2 Fe(OH) 2 + 2H 2O + 2OH - --> 2Fe(OH) 3 + 2H 2O Balancing in Basic Solution Practice 2: Fe(OH)2 + H2O2 --> Fe(OH)3 + H2O in basic solution Slide 59 / 144 Reduction H 2O 2 --> H 2O Add 1 H2O on RHS H 2O 2 --> H 2O + H 2O Add A 2H 2O on LHS to balance H atoms 2H 2O + H 2O 2 --> H 2O + H 2O Add 2 OH- to RHS 2H2O + H2O2 --> H2O + H2O +2OH - Add the two equations: 2Fe(OH)2 + H2O2 --> 2Fe(OH)3 Balancing in Basic Solution Slide 60 / 144 Practice 3 : Bi(OH) 3 + SnO 2 --> Bi + SnO 3 2Bi(OH)3 + 3SnO2 --> 2Bi + 3SnO3 + 3H2O
21 Balancing in Basic Solution Slide 61 / 144 Practice 4: Cr(OH) H 2O 2 --> (CrO 4) 2- + H 2O 2Cr(OH) OH - + 3H2O2 --> 2CrO H2O Voltaic Cells Slide 62 / 144 The energy released in a spontaneous reaction can be used to perform electrical work. Such a set up through which we can transfer electrons is called a voltaic cell or galvanic cell or electrochemical cell Voltaic Cells Zn + Cu 2+ Zn 2+ + Cu Slide 63 / 144 single replacement Zn 2+ 2e- Cu atom Zn metal Cu 2+ Zn metal Cu 2+ Zn metal strip placed in CuSO 4 In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released. In the above stup, lectron transfer takes place inside the beaker Note that the blue color fades as more Cu is reduced to metallic copper
22 Voltaic Cells Slide 64 / 144 This shows what is occurring on an atomic level at the anode and the cathode. ZnCutransfer.html Voltaic Cells Slide 65 / 144 Here the Cu and Zn strips are in two different beakers Cu/CuNO 3 Cu e- Cu RED- Half reaction Zn/ZnNO 3 Zn Zn e- OXD- Half reaction We can use the energy to do work if we make the electrons flow through an external device. Voltaic Cells Slide 66 / 144 Here the Cu and Zn strips are in two different beakers e- salt bridge Cu/CuNO3 Cu e- Cu RED- Half reaction Zn/ZnNO 3 Zn Zn e- OXD- Half reaction The salt bridge allows the migration of the ions to keep electrical neutrality Electrons are generated at the anode and flows through the external line to the cathode.
23 Slide 67 / 144 Voltaic Cells animations/cuzncell.html Voltaic Cells Slide 68 / 144 A typical cell looks like this. The oxidation occurs at the anode. The reduction occurs at the cathode. NO3 - Zn 2+ Slide 69 / 144 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. more Zn 2+ are produced more NO3- are created in solution
24 Voltaic Cells Therefore, we use a salt bridge, usually a U- shaped tube that contains a gel of a salt solution, to keep the charges balanced. Cations move toward the cathode. Anions move toward the anode. Slide 70 / 144 more Zn 2+ are produced more NO3 - are in solution The increase in Zn 2+ and NO3 - ions in the two compartment create electrical imbalance. The salt bridge ions will neutralize these ions and create neutrality. Voltaic Cells Slide 71 / 144 e- 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 Zn 2+ Cu atom Zn metal Cu 2+ Zn metal Cu 2+ Voltaic Cells Slide 72 / 144 As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. e- Zn 2+ 2e- 2e- Cu atom Zn metal Cu 2+ Zn metal Cu 2+
25 Voltaic Cells Slide 73 / 144 e- The electrons are taken by the cation, and the neutral metal atoms are deposited onto the cathode. Zn 2+ 2e- Cu atom Zn metal Cu 2+ Zn metal Cu 2+ Voltaic Cells Slide 74 / 144 This shows how a typical voltaic cell works flashfiles/electrochem/volticcell.html 16 The electrode at which oxidation occurs is called the. Slide 75 / 144
26 17 In a voltaic cell, electrons flow from the to the. Slide 76 / Which element is oxidized in the reaction below? Slide 77 / 144 Fe 2+ + H + + Cr2O7 2- Fe 3+ + Cr 3+ + H2O 19 Fe 2+ + H + + Cr 2 O 7 2- Fe 3+ + Cr 3+ + H 2 O If a voltaic cell is made with Fe and Cr electrode in contact with their own solution, the electrons will flow from to electrode. Slide 78 / 144
27 20 The purpose of the salt bridge in an electrochemical cell is to. Slide 79 / 144 A) maintain electrical neutrality in the half-cells via migration of ions. B) provide a source of ions to react at the anode and cathode. C) provide oxygen to facilitate oxidation at the anode. D) provide a means for electrons to travel from the anode to the cathode. E) provide a means for electrons to travel from the cathode to the anode. 21 A cell was made with Mg and Cu as two electrodes. The electrons will flow from to electrode. Slide 80 / The electrode where reduction is taking place is the Slide 81 / 144
28 23 The cation concentration increases in the solution where oxidation occurs. Slide 82 / 144 Yes No 24 the cations move towards the anode and anions move towards the cathode in a voltaic cell. Slide 83 / 144 True False 25 The salt bridge ions may react with the Ions in the cell compartments to form a precipitate. Slide 84 / 144 Yes No
29 26 Which of the following substances would NOT provide a suitable salt bridge? Slide 85 / 144 A KNO 3 B Na 2 SO 4 C LiC 2 H 3 O 2 D PbCl 2 27 Which of the following substances would provide a suitable salt bridge? Slide 86 / 144 A B AgBr KCl C BaF 2 D CuS 28 In a Cu-Zn voltaic cell, which of the following is true? Slide 87 / 144 A B C D E Both strips of metal will increase in mass. Both strips of metal will decrease in mass. Cu will increase in mass; Zn will decrease. Cu will decrease in mass; Zn will increase. Neither metal will change its mass, since electrons have negligible mass. Cu/CuNO 3 Cu e- Cu RED- Half reaction Zn/ZnNO3 Zn Zn e- OXD- Half reaction
30 29 In any voltaic cell, which of the following is true? Slide 88 / 144 A B C D The cathode will always increase in mass. The anode strip will always decrease in mass. The anode strip will always increase in mass. Both A and B Electro motive force Slide 89 / 144 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. The accumulation of large number of electrons at the anode create higher potential at the anode. Natural flow will occur to cathode where there is less potential Higher - to - lower Slide 90 / 144 Electro motive force 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.
31 Electro motive force Slide 91 / 144 The difference in potential energy /electon charge is measured in volts. 1 volt is the potential required to impart 1joule energy to a charge of 1coulomb 1v = 1J / 1C The potential difference between the electrodes is the driving force that pushes the electrons - so called EMF In a voltaic cell, EMF = Ecell Electromotive Force (emf) Slide 92 / 144 Cell potential is measured in volts (V). 1V = 1J/C In a spontaneous reaction, Ecell is positive EMF depends on the cell reaction involved Standard condition: 1M, 1atm and 25 C E cell = standard cell potential Standard Reduction Potentials Slide 93 / 144 Electrode potential: The tendency of an electrode to lose or gain electrons is called electrode potential ( oxidation or reduction potential) Reduction potentials for many electrodes have been measured and tabulated. By convention, the process is viewed as a reduction and the values are reported as reduction potential Li + (aq) + e- Li(s) Na + (aq) + e- Na(s) Al 3+ (aq) + 3e- Al(s) H + (aq) + 2e- H2(g 0 Cu 2+ (aq) + 2e- Cu(s) F2(g) + 2e- 2F - (aq) The more negative value indicate that, reduction is unlikely at that electrode The more positive the value is, reduction is highly likely at that electrode. This parallels their activity in single replacement reaction.
32 Standard Reduction Potentials Slide 94 / 144 Standard Hydrogen Electrode ( SHE) By definition, the reduction potential for hydrogen is 0 V: 2 H + (aq, 1M) + 2 e H2 (g, 1 atm) H2, 1 atm Pt HCl, 1M Standard Reduction Potentials Slide 95 / 144 How did we measure the reduction potential of all lements? Their values are referenced to a Standard Hydrogen Electrode (SHE). The metal electrode will be connected to the SHE By definition, the reduction potential for hydrogen is 0 V: The reduction potential measured will be that of the metal H2, 1 atm Zn HCl, 1M Pt Zn(NO3)2 30 If a volatic cell is made with iron and zinc, which metal will be reduced? Slide 96 / 144 Use the reduction potential table and compare the values. The more positive the value is, that is where reduction takes place, is the cathode. Oxidation - at Anode (vowels) Reduction - at Cathode (consonants) Zn Fe 0.1M Zn(NO3)2 0.1M Fe(NO3)2
33 31 If a volatic cell is made with Cu and Na, which metal will be the cathode? Slide 97 / 144 A B C Cu Na Cu and Na cannot make a voltaic cell. F2(g) + 2e- 2F - (aq) Cu 2+ (aq) + 2e- Cu(s) H + (aq) + 2e- H2(g 0 Al 3+ (aq) + 3e- Al(s) Na + (aq) + e- Na(s) Li + (aq) + e- Li(s) If a volatic cell is made with Li and Al, which metal will be the anode? Slide 98 / 144 A B C Li Al Li and Al cannot make a voltaic cell. F2(g) + 2e- 2F - (aq) Cu 2+ (aq) + 2e- Cu(s) H + (aq) + 2e- H2(g 0 Al 3+ (aq) + 3e- Al(s) Na + (aq) + e- Na(s) Li + (aq) + e- Li(s) Cell Potentials Slide 99 / 144 The cell potential at standard conditions can be found through this equation: E cell = E o red.pot (cathode) Eo red.pot (anode) Because cell potential is based on the potential energy per unit of charge, it is an intensive property. This means that it does not depend on the amount of substance (e.g. mass or moles).
34 Cell Potentials A cell with Cu and Zn electrodes Slide 100 / 144 For the oxidation in this cell, E red = v For the reduction, E red = v Zn Cu 1M Zn(NO3)2 1M Cu(NO3)2 Zn(s) Zn e- Cu 2+ (aq) + 2e- Cu(s) Cell Potentials Slide 101 / 144 E cell = E red(cathode) - E red(anode) = +0.34V - (-0.76V) E red = +1.10V Cell Potentials Slide 102 / 144 The greater the difference between the two electrode potential, the greater the voltage of the cell. More positive Cu e- --> Cu E 0 cell = +0.34V --(-0.76) (-0.76V) cell = V 1.10v Zn --> Zn e-
35 33 Which of the following volatic cells would yield the greatest voltage (E o cell)? Slide 103 / 144 A B C D Cu-Al Cu-Na Al-Li F 2 - Cu F2(g) + 2e- 2F - (aq) Cu 2+ (aq) + 2e- Cu(s) H + (aq) + 2e- H2(g 0 Al 3+ (aq) + 3e- Al(s) Na + (aq) + e- Na(s) Li + (aq) + e- Li(s) Which of the following volatic cells would yield the lowest voltage (E o cell)? Slide 104 / 144 A B C Cu-Al Al-Na Na-Li F2(g) + 2e- 2F - (aq) Cu 2+ (aq) + 2e- Cu(s) H + (aq) + 2e- H2(g 0 Al 3+ (aq) + 3e- Al(s) Na + (aq) + e- Na(s) Li + (aq) + e- Li(s) Oxidizing and Reducing Agents Slide 105 / 144 Increasing strength of oxidizing agent Most positive values F 2 (g) + 2e- --> 2F- 2.87v Cl 2 (g) + 2e- --> 2Cl- 1.36v I 2 (s) + 2e- --> 2I v Rb + + e- --> Rb(s) -2.92v Most negative values Increasing strength of reducing agent The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials. F is a strong oxidizing agent than Cl
36 35 The more the value of E red, the greater the driving force for reduction. Slide 106 / 144 Class Practice: Slide 107 / 144 Identify: Cathode Anode Oxidation half reaction Reduction half reaction Combined cell reaction (balanced) E cell = Ni 1M Ni (NO3)2 Sn 1M Sn(NO3)2 Class Practice 2 Identify: Cathode Anode Oxidation half reaction Reduction half reaction Combined cell reaction (balanced) E cell Fe 1M Fe(NO3)3 Sn 1M Sn(NO3)2 Slide 108 / 144
37 36 Calculate E for the following reaction: Sn 4+ (aq) + 2K(s) --> Sn 2+ (aq) + 2K + (aq) A) V B) V C) V D) V E) V Slide 109 / 144 Free Energy Slide 110 / 144 ΔG for a redox reaction can be found by using the equation ΔG = nfe where n is the number of moles of electrons transferred, and F is a constant, the Faraday. 1 F = 96,500 C/mol = 96,485 J/V-mol Free Energy Slide 111 / 144 Under standard conditions, ΔG = -nfe Standard condition: 25 0 C, 1 atm and 1M
38 Slide 112 / 144 Nernst Equation Remember that ΔG = ΔG + RT ln Q This means nfe = nfe + RT ln Q Slide 113 / 144 Nernst Equation Dividing both sides by nf, we get the Nernst equation: E = E - RT nf lnq or, using base-10 logarithms, E = E - [8.31 x 298] x log Q [n x ] Nernst Equation Slide 114 / 144 E = E - [8.31 x 298] x log Q [n x ] At room temperature (298 K), RT F = V Thus, the equation becomes E = E n logq
39 Equilibrium Constant Slide 115 / 144 When E=0, -nfe = 0 ΔG = ΔG + RT ln Q E = E n 0= E logk (0.0592/n) E = logk (0.0592/n) logk = ne / Equilibrium constant for a redox reaction can be calculated using the above. logq 37 The relationship between the change in Gibbs free energy and the emf of an electrochemical cell is given by. Slide 116 / 144 A) ΔG = -nf/e B) ΔG = -E/nF C) ΔG = -nfe D) ΔG = -nrtf E) ΔG = -nf/ert Concentration Cells ** Slide 117 / 144 Ni Ni Ni Ni 0.001M [Ni 2+ ] 1M [Ni 2+ ] 0.5M [Ni 2+ ] 0.5M [Ni 2+ ] Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. For such a cell, E E = E - would be 0, but Q would not n logq [dilute] log Q = log [concent] Therefore, as long as the concentrations are different, E will not be 0. voltaiccellemf.html
40 38 A cadmium rod is placed in a 0.010M solution of cadmium sulfate at 298K. Calculate the potential of the electrode at the is temperature. Slide 118 / E = E - logq n = /2 log0.01 = Elcectrolytic cell/ Electrolysis Slide 119 / 144 Voltaic cells work as a result of a spontaneous reaction We can use electricity from outside source to make a nonspontaneous reaction to become spontaneous. A chemical reaction by using outside electricity is known as electrolysis. Such a cell is known as an electrolytic cell Elcectrochemical/voltaic cell Slide 120 / 144 Electrolytic Cell Energy is absorbed to drive a nonspontaneous redox reaction Surroundings (battery or power supply) do work on the system (cell) Voltaic (Electrochemical) Cell Energy is released from a spontaneous redox reaction System (cell) does work on the surroundings (e.g. light bulb)
41 39 One of the differences between a voltaic cell and an electrolytic cell is that in an electrolytic cell,. Slide 121 / 144 A) an electric current is produced by a chemical reaction B) electrons flow toward the anode C) a nonspontaneous reaction is forced to occur D) gas is produced at the cathode E) oxidation occurs at the cathode Electroplating Slide 122 / 144 Uses an active electrode to deposit a thin layer of one metal to another metal object Item to be coated is cathode (metal ions get reduced at the (-) electrode) e- Ag Ag + Ag + Electrolysis Slide 123 / 144 This flowchart shows the steps relating the quantity of electrical charge used in electrolysis to the amounts of substances oxidized or reduced. Quantity of charge (Coulombs) Moles of substance oxidized or reduced Moles of electrons (Faradays) Current (A) and time Grams of substance A typical problem will give the current (amperes) that is applied for a specific amount of time (seconds). You would be asked to solve for the mass of metal that can be produced through electroplating. Alternatively, you might be asked for either the time or amount of current that is needed to produce a specific amount (given mass) of metal.
42 Electrolysis Slide 124 / 144 The quantity of charge passing through is measured in coulombs 1 mole of electrons passage = 96500C = 1Faraday 1coulomb = 1 ampere passing in 1 second Coulombs (C ) = ampere x seconds 40 The quantity of charge passing a point in a circuit in one second when the current is one ampere is called a. Slide 125 / 144 A) joule B) coulomb C) calorie D) Newton E) Mole 41 How many coulombs result from a current of 50 amps (A) applied for 20 seconds? Slide 126 / 144 A 2.5 B 10 C 70 D 700 E 1000 A = C s
43 42 How many seconds must a current of 25 A be applied in order to produce a charge of 100 C? Slide 127 / 144 A 0.25 B 0.4 C 4 D 75 E 125 A = C s 43 What amount of charge is required to release one mole of electrons? Slide 128 / 144 A B C D E 1 atm 25 o C L-atm/mol-K 96,500 C 760 mm Hg 44 How many moles of electrons would be released by a charge of 158,000 C? Slide 129 / 144 A 96,500 / 158,000 B 158,000 / 96,500 C 158,000*96,500 D 158,000-96,500
44 45 How many moles of electrons would be released by a charge of 48,250 C? Slide 130 / 144 A 0.25 B 0.5 C 1 D 48,250 E 96,500 Quantitative aspect Electrolysis If 10.0 A passes through molten AlCl3 for 60 minutes, how much of Al will be deposited? Slide 131 / 144 total charge C = 10.0 A x 60min x 60sec. = 3.6 x10 4 C Remember!! 1mole e- = 96500C How many moles of e - are we talking about in here????? Moles of e- = 3.6 x 10 4 /96500 = moles of e - Al e - Al 1 mol Al = 3 mols e - Moles of Al = x 1 mole Al/3 mole e - = mol Al How many grams of Al? = 27g x mol = 3.36g Al Electrolysis Slide 132 / 144 practice:1 Calculate the number of grams of aluminum produced in 30.0 minutes by electrolysis of at a current of 12.0 A.
45 Electrolysis Slide 133 / 144 practice:2 How many minutes will it take to plate out 6.36 g of Cu metal from a solution of Cu 2+ using a current of 12 amps in an electrolytic cell? 46 Plating out 1 mol of chromium requires of electrons. Slide 134 / 144 A B C D 0.33 mol 0.5 mol 1.0 mol 3.0 mol Cr 3+ (aq) + 3 e - --> Cr (s) 47 One mole of electrons would allow electroplating of mol of zinc. Slide 135 / 144 A B C D Zn cannot be electroplated. 0.5 mol 1.0 mol Zn 2+ (aq) + 2 e - --> Zn (s) 2.0 mol
46 48 How many minutes will it take to plate out g of Al metal from a solution of Al 3+ using a current of 12.9 amps in an electrolytic cell? Slide 136 / 144 A B C D E 60.1 min 74.9 min 173 min 225 min 13,480 min Electrochemistry -Applied Slide 137 / 144 Batteries Hydrogen fuel cells Corrosion Corrosion prevention Biology Batteries Slide 138 / 144
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