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Electrochemistry

Electrochemistry. Unit 14. Overview. Oxidation States Oxidation versus Reduction Redox reactions Identifying Half Reactions Balancing Voltaic (Galvanic) Cells Cell potential Standard reduction potentials Standard hydrogen electrode. Strength of Agents Nernst Equation Calculations

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Electrochemistry

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  1. Electrochemistry Unit 14

  2. Overview • Oxidation States • Oxidation versus Reduction • Redox reactions • Identifying • Half Reactions • Balancing • Voltaic (Galvanic) Cells • Cell potential • Standard reduction potentials • Standard hydrogen electrode • Strength of Agents • Nernst Equation • Calculations • Equilibrium • Gibbs Free Energy • Electrolytic cells • Electroplating • Stoichiometry

  3. Oxidation States (Numbers) • Actual charge of a monatomic ion • Rules for assigning oxidation numbers • For an atom in its elemental form (uncombined with other elements) the oxidation number is always 0. • Example: H2 = 0 • For any monatomic ion, the oxidation number equals the charge of the ion. • Alkali metals always = +1 • Alkaline earth metals always = +2 • Aluminum always = +3

  4. Oxidation States (Numbers) • Nonmetals usually have negative oxidation numbers (although they can be positive) • Oxygen is usually -2 (exception: peroxide) • Hydrogen is usually +1 (-1 when bonded to metals) • Fluorine is -1. • Other halogens have oxidation number of -1 in binary compounds. • In oxyanions, halogens usually have positive oxidation states. • The sum of oxidation numbers in neutral compounds is zero. The sum of oxidation numbers in polyatomic ions equals the charge of the ion.

  5. Examples (Oxidation States) • What is the oxidation state of S? • H2S = -2 • SCl2 = +2 • S8 = 0 • Na2SO3 = +4 • SO4-2 = +6

  6. Oxidation - Reduction • Oxidation – A process in which an element attains a more positive oxidation state (loses electrons) Na(s)  Na+ + e- • Reduction – A process in which an element attains a more negative oxidation state (gains electrons) Cl2 + 2e- 2Cl-

  7. Leo the Lion (Oxidation Reduction) LEOsaysGER LoseElectrons =Oxidation GainElectrons=Reduction

  8. Oxidation - Reduction OR…

  9. Redox Reactions Mg + S → Mg2+ + S2- (MgS) The magnesium atom (which has zero charge) changes to a magnesium ion by losing 2 electrons, and is oxidized to Mg2+ The sulfur atom (which has no charge) is changed to a sulfide ion by gaining 2 electrons, and is reduced to S2- • Redox Reaction: reaction in which oxidation and reduction occur • Oxidation and reduction occur spontaneously

  10. Redox Reactions Each sodium atom loses one electron (oxidation): Each chlorine atom gains one electron (reduction):

  11. Trends • Active metals: • Lose electrons easily • Are easily oxidized • Are strong reducing agents • Active nonmetals: • Gain electrons easily • Are easily reduced • Are strong oxidizing agents

  12. Half Reactions A common approach for listing species that undergo REDOX is as half-reactions. For 2Fe3+ + Zno(s) = 2Fe2+ + Zn2+ Fe3+ + e- Fe2+ (reduction) Zno(s) Zn2+ + 2e- (oxidation) They are individual equations showing just the oxidation or just the reduction.

  13. Balancing Redox • The amount of each element must be the same on both sides (like normal) • BUT the gain/loss of electrons must also be balanced • If an element loses a certain amount of electrons, another element must gain the same amount

  14. Balancing Redox #1 Divide the equation into two half reactions #2 Balance each half reaction • Balance all elements other than H and O • Balance the O atoms by adding H2O • Balance the H atoms by adding H+ • Balance the charge by adding e-

  15. Balancing Redox #3 Multiply the half reactions by integers so that the electrons lost in one half- reaction equals electrons gained in the other reaction #4 Add the two half-reactions together • Simplify by cancelling out species appearing on both sides of the equation

  16. Fe2+ + Cr2O72- Fe3+ + Cr3+ +2 +3 Fe2+ Fe3+ +6 +3 Cr2O72- Cr3+ Cr2O72- 2Cr3+ Balancing Redox Reactions Balance the following reaction in acidic solution. 1. Separate the equation into two half-reactions. Oxidation: Reduction: 2. Balance the atoms other than O and H in each half-reaction. 19.1

  17. Fe2+ Fe3+ + 1e- 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- 2Cr3+ + 7H2O Cr2O72- 2Cr3+ + 7H2O Balancing Redox Reactions 3. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. 4. Add electrons to one side of each half-reaction to balance the charges on the half-reaction. 5. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 19.1

  18. 14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O Balancing Redox Reactions • Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. Oxidation: Reduction: 19.1

  19. More Examples • Cu + NO3-  Cu+2 + NO2 Answer: Cu + 4H+ + 2NO3-  Cu+2 + 2NO2 + 2H2O • Cr2O7-2 + Cl- Cr+3 + Cl2 Answer: 14H+ + Cr2O7-2 + 6Cl-  2Cr+3 + 7H2O + 3Cl2

  20. Balancing Redox • If redox reaction occurs in “basic solution”… • Follow all steps as usual • Before half-reactions are combined, add 1 OH- to BOTH sides of the equation for every 1 H+ in the equation • Combine H+ and OH+ to form H2O when they appear on the same side of the equation

  21. Examples • The following reactions occur in basic solutions • CN- + MnO4-  CNO- + MnO2 Answer: 3CN- + H2O + 2MnO4-  3CNO- + 2MnO2 + 2OH- • NO2- + Al  NH3 + Al(OH)4- Answer: NO2- + 2Al + 5H2O + OH- NH3 + 2Al(OH)4-

  22. Electrochemical Cells • There are two general ways to conduct an oxidation-reduction reaction • #1 Mixing oxidant and reductant together • Cu2+ + Zn(s) Cu(s) + Zn2+ • This approach does • not allow for • control of thereaction.

  23. Electrochemical cells • Electrochemical cells • Each half reaction is put in a separate ‘half cell.’ They can then be connected electrically. • Called a “voltaic” or “galvanic” cell • This permits better control over the system.

  24. Voltaic cells Cu2+ + Zn(s) Cu(s) + Zn2+ Electrons are transferred from one half-cell to the other using an external metal conductor. e- e- Cu Zn Zn2+ Cu2+

  25. Voltaic Cells • Electrodes • Two solid metals that are connected by the external circuit • Anode • The electrode at which oxidation occurs • Cathode • The electrode at which reduction occurs

  26. Voltaic Cells Just remember the… Red Cat “Reduction occurs at the cathode”

  27. Voltaic Cells • This means that electrons flow from the anode to the cathode through the external circuit Oxidation occurs at the anode Reduction occurs at the cathode

  28. Voltaic cells e- e- To complete the circuit, a salt bridge is used. salt bridge

  29. KCl Cl- K+ K+ is released as Cu+2is converted to Cu Cl- is released to Zn side as Zn is converted to Zn+2 Voltaic Cells • Salt bridge • Allows ions to migrate without mixing electrolytes (keeps solutions neutral). • It can be a simple porous disk or a gel saturated with a non-interfering, strong electrolyte like KCl.

  30. Voltaic Cells • Salt bridge • Consists of a U-shaped tube that contains the electrolyte solution (ex. NaNO3) • Usually solution is a paste or gel • Ions of electrolyte solution will not react with other ions in the cell or with the electrode materials • Ions migrate to neutralize the cells • Anions migrate toward the anode • Cations migrate toward the cathode

  31. Voltaic Cell

  32. Cell Diagrams • Rather than drawing an entire cell, a type of shorthand can be used. • For our copper - zinc cell, it would be: Zn | Zn2+ (1M) || Cu2+ (1M) | Cu • The anode is always on the left. | = boundaries between phases || = salt bridge • Other conditions like concentration are listed just after each species.

  33. Cell Potential • Electromotive Force (EMF): Driving force that pushes electrons through the external circuit • Measures how willing a species is to gain or lose electrons • Also referred to as “cell potential” or “cell voltage” • Denoted Ecell • Spontaneous cell reaction = positive (+) Ecell • Measured in volts (1 V = 1Joule/Coulomb)

  34. Standard Cell Potential • EMF under standard state conditions (Eocell) • All soluble species are at 1 M • Slightly soluble species must be at saturation. • Any gas is constantly introduced at 1 atm • Any metal must be in electrical contact • Other solids must also be present and in contact.

  35. Standard Reduction Potentials • Eocellis determined from the reduction potentials of the two half reactions Eocell = Eored(cathode) – Eored(anode) • Reduction potentials can be looked up for all half reactions (table given on AP test)

  36. Standard Reduction Potentials • Half reaction Eo, V • F2 (g) + 2H+ + e- 2HF (aq) 3.053 • Ce4+ + e- Ce3+(in 1M HCl) 1.28 • O2 (g) + 4H+ + 4e- 2H2O (l) 1.229 • Ag+ + e- Ag (s) 0.7991 • 2H+ + 2e- H2 (g) 0.000 • Fe2+ + 2e- Fe (s) -0.44 • Zn2+ + 2e- Zn (s) -0.763 • Al3+ + 3e- Al (s) -1.676 • Li+ + e- Li (s) -3.040

  37. Note • The table of standard reduction potentials relates to the activity series of metals • Flip the table upside down and only look at the metals (solids), and it is the activity series list • The AP test does not give you an activity series list so you have to use the standard reduction potentials to predict products of reactions

  38. Standard Reduction Potentials • E0 is for the reaction as written • The more positive E0 the greater the tendency for the substance to be reduced • The half-cell reactions are reversible • The sign of E0changes when the reaction is reversed (do this if you need the “oxidation potential” of an element) • Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

  39. Standard Reduction Potentials • Reversing the sign • For the reduction of Zn+2 + 2e- Zn Ered = -0.76 • The reduction potential of Zn is -0.76 • If we are looking at the reverse reaction and want the oxidizing potential, the sign is reversed • Zn  of Zn+2 + 2e-Eox = +0.76 • We will do redox calculations using only “reduction potentials”

  40. Standard Hydrogen Electrode • Standard Hydrogen Electrode (SHE) • We can’t measure the standard reduction potential of a half reaction directly • SHE used as a reference point • Reduction potential = 0 V • Consists of… • Platinum wire connected to piece of platinum foil • Electrode encased in a glass tube so that hydrogen at standard conditions can bubble over the platinum

  41. Standard Cell Potential To solve for the Ecell, subtract the reduction potential for the half-reaction at the cathode from the reduction potential for the half-reaction at the anode. Eocell = Eocathode – Eoanode

  42. Cd2+(aq) + 2e-Cd(s)E0 = -0.40 V Cr3+(aq) + 3e- Cr (s)E0 = -0.74 V E0 = E0 cathode- E0 anode E0 = -0.40 – (-0.74) E0 = 0.34 V cel l cell cell 2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M) Standard Cell Potential (Example 1) What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 MCd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?

  43. E0 = E0 cathode- E0 anode E0 = 0.34 V cel l cathode Zn(s) + Cu2+(aq,1 M) Cu(s) + Zn+2(aq,1 M) E0cell = 1.10 V Standard Cell Potential (Example 2) 1.10 = E0 cathode - (-0.76) Given that the standard reduction potential of Zn+2 to Zn(s) is -0.76 V, calculate the E0red for the reduction of Cu+2 to Cu(s).

  44. Standard Cell Potential • The more positive the value of E0red, the greater the driving force for reduction under standard conditions • Meaning…the reaction at the cathode has a more positive value of E0red than the reaction at the anode

  45. Cd2+(aq) + 2e-Cd(s) Sn2+(aq) + 2e-Sn(s) Standard Cell Potential (Example 3) A voltaic cell is based on the following two standard half reactions. Determine which half-reaction occurs at the cathode and which occurs at the anode as well as the standard cell potential. E0red of (Cd+2/Cd) is -0.403 and E0red of (Sn+2/Sn) is -0.136 Reaction of Sn+2/Sn is more positive so it occurs at the cathode

  46. Cd2+(aq) + 2e-Cd(s) Sn2+(aq) + 2e-Sn(s) E0 = E0 cathode- E0 anode E0 = -0.136 – (-0.403) E0 = 0.267 V cel l cell cell Standard Cell Potential (Example 3) • Cathode reaction is reduction/Anode reaction is oxidation Cathode: Sn2+(aq) + 2e-Sn(s) E0red = -0.136 Anode: Cd(s) Cd2+(aq) + 2e-E0red = -0.403

  47. Strengths of Oxidizing Agents • The more readily an ion is reduced (the more positive its E0red value), the stronger it is as an oxidizing agent. • The half-reaction with the smallest reduction potential (most negative E0red) is most easily reversed as an oxidation • Smallest reduction potential is strongest reducing agent • The stronger the oxidizing agent, the weaker it acts as a reducing agent (and vice versa)

  48. Strengths of Oxidizing Agents

  49. “E” versus… • E versus E0 • E means not at standard conditions • E versus Ecell • E doesn’t have to occur in a voltaic cell so there is no cathode and anode • E = Ereduction - Eoxidation

  50. Nernst Equation R= 8.31 J/(molK) T= Temperature in K n=moles of electrons transferred in balanced redox equation F = Faraday constant = 96,500 coulombs/mol e- (Charge of one mole of electrons) Q= reaction quotient Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M.

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