Slide 37 / 144 The Half-Reaction Method 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.
Slide 38 / 144 The Half-Reaction Method In acidic medium: 2− : − and C 2 O 4 Consider the reaction between MnO 4 − ( aq ) + C 2 O 4 2− ( aq ) ➝ Mn 2+ ( aq ) + CO 2 ( aq ) MnO 4 +7 +3 +2 +4 − ( aq ) + C 2 O 4 2− ( aq ) ➝ Mn 2+ ( aq ) + CO 2 ( aq ) MnO 4 · 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.
Slide 39 / 144 The Half-Reaction Method In acidic medium: +7 +3 +2 +4 − ( aq ) + C 2 O 4 2− ( aq ) ➝ Mn 2+ ( aq ) + CO 2 ( aq ) MnO 4 · Which substance gets reduced? · Which substance gets oxidized? · Which substance is the reducing agent? · Which substance is the oxidizing agent?
Slide 40 / 144 The Half-Reaction Method In acidic medium: +7 +3 +2 +4 − ( aq ) + C 2 O 4 2− ( aq ) ➝ Mn 2+ ( aq ) + CO 2 ( aq ) MnO 4 Since the manganese goes from +7 to +2, it is reduced. - ion is the oxidizing agent. The MnO 4 Since the carbon goes from +3 to +4, it is oxidized. The C 2 O 4 2- ion is the reducing agent.
Slide 41 / 144 Oxidation Half-Reaction In acidic medium: C 2 O 4 2− ➝ CO 2 To balance the carbon, we add a coefficient of 2: 2− ➝ 2 CO 2 C 2 O 4
Slide 42 / 144 Oxidation Half-Reaction In acidic medium: 2− ➝ 2 CO 2 C 2 O 4 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side. C 2 O 4 2− ➝ 2 CO 2 + 2 e −
Slide 43 / 144 Reduction Half-Reaction In acidic medium: − ➝ Mn 2+ MnO 4 The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side. − ➝ Mn 2+ + 4 H 2 O MnO 4
Slide 44 / 144 Reduction Half-Reaction In acidic medium: MnO 4 − ➝ Mn 2+ + 4 H 2 O To balance the hydrogen, we add 8 H + to the left side. − ➝ Mn 2+ + 4 H 2 O 8 H + + MnO 4
Slide 45 / 144 Reduction Half-Reaction In acidic medium: − ➝ Mn 2+ + 4 H 2 O 8 H + + MnO 4 To balance the charge, we add 5 e − to the left side. − ➝ Mn 2+ + 4 H 2 O 5 e − + 8 H + + MnO 4
Slide 46 / 144 Combining the Half-Reactions In acidic medium: Now we evaluate the two half-reactions together: 2− ➝ 2 CO 2 + 2 e − C 2 O 4 5 e − + 8 H + + MnO 4 − ➝ Mn 2+ + 4 H 2 O To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.
Slide 47 / 144 Combining the Half-Reactions In acidic medium: 5 C 2 O 4 2− ➝ 10 CO 2 + 10 e − 10 e − + 16 H + + 2 MnO 4 − ➝ 2 Mn 2+ + 8 H 2 O When we add these together, we get: 10 e − + 16 H + + 2 MnO 4 − + 5 C 2 O 4 2− --> 2 Mn 2+ + 8 H 2 O + 10 CO 2 +10 e −
Slide 48 / 144 Combining the Half-Reactions In acidic medium: 10 e − + 16 H + + 2 MnO 4 − + 5 C 2 O 4 2− ➝ 2 Mn 2+ + 8 H 2 O + 10 CO 2 +10 e − The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16 H + + 2 MnO 4 − + 5 C 2 O 4 2− ➝ 2 Mn 2+ + 8 H 2 O + 10 CO 2
Slide 49 / 144 The Half-Reaction Method In acidic medium: Practice 1 Cd(s) + NiO 2 (s) --> Cd(OH) 2 (s) + Ni(OH) 2 (s) 0 +4 -2 +2 -2 +1 +2 -2 +1 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.
Slide 50 / 144 In acidic medium: - --> NO 2 + Cu 2+ Practice 2 Cu + NO 3
Slide 51 / 144 Practice 3 2- + Fe 2+ + H + --> Cr 3+ + Fe 3+ + H 2 O Cr 2 O 7
Slide 52 / 144 - + Br - --> Mn 2+ + Br 2 in acidic solution Practice 4 :MnO 4
Slide 53 / 144 2- + C 2 H 4 O --> C 2 H 4 O 2 + Cr 3+ in acidIc Practice 5 : Cr 2 O 7 solution 2- + 8H+ + 3C 2 H 4 O --> 3C 2 H 4 O 2 + 2Cr 3+ + 4H 2 O Cr 2 O 7
Slide 54 / 144 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. 1- 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.
Slide 55 / 144 Balancing in Basic Solution - --> Zn 2+ + NH 4 + in basic medium: Zn + NO 3 Oxidation half reaction: Zn ----> Zn 2+ + 2e- - ---> NH 4 + Reduction half reaction: NO 3 - ---> NH 4+ + 3H 2 O NO 3 - ---> NH 4 + + 3H 2 O 10H 2 O + NO 3 - ---> NH 4 + + 3H 2 O + 10OH - 10H 2 O + NO 3 - ---> NH 4 + + 3H 2 O + 10OH - 8e- 10H 2 O + NO 3 4Zn ----> 4 Zn2+ + 8e- - + 7H 2 O--> 4Zn 2+ + 1NH 4 + + 10 OH - 4Zn + 1NO 3
Slide 56 / 144 ** Balancing in Basic Solution Can also be performed without splitting the two equations. - --> Zn 2+ + NH 4 + in basic medium: Zn + NO 3 Increases by 2 1. Assigh oxidation #s: - --> Zn 2+ + NH 4 + Zn + NO 3 0 +5 2- +2 -3 +1 decreases by 8 2. Balance the change in Oxidation # change on either side. increases by 8 - --> 4Zn 2+ + 1NH 4 + 4Zn + 1NO 3 decreases by 8
Slide 57 / 144 Balancing in Basic Solution - --> 4Zn 2+ + 1NH 4 + 4Zn + 1NO 3 3. Balance O atoms by adding H 2 O molecules to the side deficient in O atoms. 3 O atoms on the LHS so add 3 water on the RHS - --> 4Zn 2+ + 1NH 4 + + 3H 2 O 4Zn + 1NO 3 4. The H atoms are then balanced by adding H 2 O to the side lacks H. 10 H on the RHS, so add 10 water on the LHS. - + 10H 2 O--> 4Zn 2+ + 1NH 4 + + 3H 2 O 4Zn + 1NO 3 5. Add 10 OH- on the other side of the reaction to balance the extra H and O. - + 10H 2 O--> 4Zn 2+ + 1NH 4 + + 3H 2 O + 10 OH - 4Zn + 1NO 3 6 . If this produces water on both sides, you might have to subtract water from each side. - + 7H 2 O--> 4Zn 2+ + 1NH4 + + 10 OH - 4Zn + 1NO 3
Slide 58 / 144 Balancing in Basic Solution Fe(OH) 2 + H 2 O 2 --> Fe(OH) 3 + H 2 O in basic solution Practice: 1 Oxidation: decrease by 1 for each O atom, total 2 e- taken +2 -1 +3 -2 Fe(OH) 2 + H 2 O 2 --> Fe(OH) 3 + H 2 O in basic solution increase by 1, 1 e - given 2 Fe(OH) 2 + H 2 O 2 --> 2Fe(OH) 3 + H 2 O in basic solution Balance O atoms by adding 2 H 2 O to LHS 2 Fe(OH) 2 + 2H 2 O --> 2Fe(OH) 3 Balance H atoms by adding 2 H2O to RHS 2 Fe(OH) 2 + 2H 2 O --> 2Fe(OH) 3 + 2H 2 O Add 2 OH- on the LHS 2 Fe(OH) 2 + 2H 2 O + 2OH - --> 2Fe(OH) 3 + 2H 2 O
Slide 59 / 144 Balancing in Basic Solution Practice 2: Fe(OH) 2 + H 2 O 2 --> Fe(OH) 3 + H 2 O in basic solution Reduction H 2 O 2 --> H 2 O Add 1 H 2 O on RHS H 2 O 2 --> H 2 O + H 2 O Add 2H 2 O on LHS to balance H atoms A 2H 2 O + H 2 O 2 --> H 2 O + H 2 O Add 2 OH- to RHS 2H 2 O + H 2 O 2 --> H 2 O + H 2 O +2OH - Add the two equations: 2Fe(OH) 2 + H 2 O 2 --> 2Fe(OH) 3
Slide 60 / 144 Balancing in Basic Solution --> Bi + SnO 3 Practice 3 : Bi(OH) 3 + SnO 2 --> 2Bi + 3SnO 3 + 3H 2 O 2Bi(OH) 3 + 3SnO 2
Slide 61 / 144 Balancing in Basic Solution -1 + H 2 O 2 --> (CrO 4 ) 2- + H 2 O Practice 4: Cr(OH) 4 -1 + 2OH - + 3H 2 O 2 --> 2CrO 4 2- + 8H 2 O 2Cr(OH)
Slide 62 / 144 Voltaic Cells 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
Slide 63 / 144 Voltaic Cells Zn + Cu 2+ ➝ Zn 2+ + Cu single replacement Zn 2+ 2e- Cu atom 2+ Cu Cu 2+ Zn metal Zn metal 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
Slide 64 / 144 Voltaic Cells This shows what is occurring on an atomic level at the anode and the cathode. http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/ ZnCutransfer.html
Slide 65 / 144 Voltaic Cells Here the Cu and Zn strips are in two different beakers Cu/CuNO 3 Zn/ZnNO 3 Cu 2+ + 2e- ➝ Cu Zn ➝ Zn 2+ + 2e- RED- Half reaction OXD- Half reaction We can use the energy to do work if we make the electrons flow through an external device.
Slide 66 / 144 Voltaic Cells Here the Cu and Zn strips are in two different beakers e- salt bridge Cu/CuNO 3 Zn/ZnNO 3 Cu 2+ + 2e- ➝ Cu Zn ➝ Zn 2+ + 2e- RED- Half reaction 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.
Slide 67 / 144 Voltaic Cells http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/ animations/CuZncell.html
Slide 68 / 144 Voltaic Cells · A typical cell looks like this. · The oxidation occurs at the anode. · The reduction occurs at the cathode. 2+ Zn - NO 3
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. - are more NO 3 2+ are more Zn created in solution produced
Slide 70 / 144 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. more Zn 2+ are - are in more NO 3 produced solution The increase in Zn 2+ and NO 3 - ions in the two compartment create electrical imbalance. The salt bridge ions will neutralize these ions and create neutrality.
Slide 71 / 144 Voltaic Cells 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 2+ Zn 2e- Cu atom Cu 2+ 2+ Cu Zn metal Zn metal
Slide 72 / 144 Voltaic Cells e- · As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. Zn 2+ 2e- Cu atom Cu 2+ Cu 2+ Zn metal Zn metal
Slide 73 / 144 Voltaic Cells e- · The electrons are taken by the cation, and the neutral metal atoms are deposited onto the cathode. 2+ Zn 2e- Cu atom Cu 2+ 2+ Cu Zn metal Zn metal
Slide 74 / 144 Voltaic Cells This shows how a typical voltaic cell works http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/ flashfiles/electroChem/volticCell.html
Slide 75 / 144 16 The electrode at which oxidation occurs is called the _______________.
Slide 76 / 144 17 In a voltaic cell, electrons flow from the ______ to the ________.
Slide 77 / 144 18 Which element is oxidized in the reaction below? 2- ➝ Fe 3+ + Cr 3+ + H 2 O Fe 2+ + H + + Cr 2 O 7
Slide 78 / 144 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 79 / 144 The purpose of the salt bridge in 20 an electrochemical cell is to ________________. 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.
Slide 80 / 144 21 A cell was made with Mg and Cu as two electrodes. The electrons will flow from ------- to --------- electrode.
Slide 81 / 144 22 The electrode where reduction is taking place is the ----------
Slide 82 / 144 23 The cation concentration increases in the solution where oxidation occurs. Yes No
Slide 83 / 144 24 the cations move towards the anode and anions move towards the cathode in a voltaic cell. True False
Slide 84 / 144 25 The salt bridge ions may react with the Ions in the cell compartments to form a precipitate. Yes No
Slide 85 / 144 26 Which of the following substances would NOT provide a suitable salt bridge? A KNO 3 Na 2 SO 4 B LiC 2 H 3 O 2 C PbCl 2 D
Slide 86 / 144 27 Which of the following substances would provide a suitable salt bridge? A AgBr KCl B BaF 2 C CuS D
Slide 87 / 144 28 In a Cu-Zn voltaic cell, which of the following is true? A Both strips of metal will increase in mass. Both strips of metal will decrease in mass. B Cu will increase in mass; Zn will decrease. C Cu will decrease in mass; Zn will increase. D Neither metal will change its mass, since E electrons have negligible mass. Cu/CuNO 3 Zn/ZnNO 3 Cu 2+ + 2e- ➝ Cu Zn ➝ Zn 2+ + 2e- RED- Half reaction OXD- Half reaction
Slide 88 / 144 29 In any voltaic cell, which of the following is true? A The cathode will always increase in mass. The anode strip will always decrease in mass. B The anode strip will always increase in mass. C Both A and B D
Slide 89 / 144 Electro motive force · 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 E cell .
Slide 91 / 144 Electro motive force 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 = E cell
Slide 92 / 144 Electromotive Force (emf) Cell potential is measured in volts (V). 1V = 1J/C In a spontaneous reaction, E cell is positive EMF depends on the cell reaction involved Standard condition: 1M, 1atm and 25°C E° cell = standard cell potential
Slide 93 / 144 Standard Reduction Potentials 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) -3.05 Na + (aq) + e- ➝ Na(s) -2.71 Al 3+ (aq) + 3e- ➝ Al(s) -1.66 2H + (aq) + 2e- ➝ H 2 (g 0 Cu 2+ (aq) + 2e- ➝ Cu(s) + 0.34 F 2 (g) + 2e- ➝ 2F - (aq) + 2.87 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.
Slide 94 / 144 Standard Reduction Potentials 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) H 2 , 1 atm Pt HCl, 1M
Slide 95 / 144 Standard Reduction Potentials 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 H 2 , 1 atm Zn Pt Zn(NO 3 ) 2 HCl, 1M
Slide 96 / 144 If a volatic cell is made with iron and 30 zinc, which metal will be reduced? 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 Fe(NO 3 ) 2 0.1M Zn(NO 3 ) 2
Slide 97 / 144 31 If a volatic cell is made with Cu and Na, which metal will be the cathode? A Cu Na B Cu and Na cannot make a voltaic cell. C F 2 (g) + 2e- ➝ 2F - (aq) + 2.87 Cu 2+ (aq) + 2e- ➝ Cu(s) + 0.34 2H + (aq) + 2e- ➝ H 2 (g 0 Al 3+ (aq) + 3e- ➝ Al(s) -1.66 Na + (aq) + e- ➝ Na(s) -2.71 Li + (aq) + e- ➝ Li(s) -3.05
Slide 98 / 144 32 If a volatic cell is made with Li and Al, which metal will be the anode? A Li Al B Li and Al cannot make a voltaic cell. C F 2 (g) + 2e- ➝ 2F - (aq) + 2.87 Cu 2+ (aq) + 2e- ➝ Cu(s) + 0.34 2H + (aq) + 2e- ➝ H 2 (g 0 Al 3+ (aq) + 3e- ➝ Al(s) -1.66 Na + (aq) + e- ➝ Na(s) -2.71 Li + (aq) + e- ➝ Li(s) -3.05
Slide 99 / 144 Cell Potentials The cell potential at standard conditions can be found through this equation: E cell = E o red.pot (cathode) − E o 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).
Slide 100 / 144 Cell Potentials A cell with Cu and Zn electrodes · For the oxidation in this cell, · E ° red = - 0.76v · For the reduction, · E ° red = + 0.34v Cu Zn 1M Cu(NO 3 ) 2 1M Zn(NO 3 ) 2 Cu 2+ (aq) + 2e- ➝ Cu(s) Zn(s) ➝ Zn 2+ + 2e-
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