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Ch. 20: Electrochemistry

Ch. 20: Electrochemistry. Electrochemistry is the branch of chemistry that deals with relationships between electricity and chemical reactions. We are going to study a type of reaction where electrons are transferred between reactants.

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Ch. 20: Electrochemistry

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  1. Ch. 20: Electrochemistry • Electrochemistry is the branch of chemistry that deals with relationships between electricity and chemical reactions. • We are going to study a type of reaction where electrons are transferred between reactants. • These reactions are called “oxidation-reduction reactions”, or redox reactions. • Before we begin studying redox reactions, we need to cover… • Oxidation Numbers • Oxidation numbers allow us to keep track of the electrons gained and lost during chemical reactions. • The oxidation #, (or oxidation state), of an element in a compound is a hypothetical charge based on a set of rules…

  2. Rules for Assigning Oxidation Numbers • Rule #1: An atom in its elemental form has an oxidation # of zero. • Examples: H2 , Ag, Br2, Pb…(Ox. #’s = zero for each element) • Rule #2: Monatomic ions have an Ox. # equal to its charge. • Examples: K+ (Ox. # for K = +1); S2−(Ox. # for S = −2) • Rule #3: Hydrogen is +1 when attached to nonmetals in a compound and −1 when attached to metals in a “hydride”. • Examples: HCl (Ox. # for H= +1); NaH (Ox. # for H= − 1) • (* Since the compounds are electrically neutral, we can then say that the Ox. # of Cl must be −1, and Na is +1.) • Rule #4: Fluorine is −1 in all compounds, and Oxygen has an Ox. # of −2 in all compounds except peroxides when it’s Ox. # is −1… O22− • Examples: Na2O… (Ox. # of oxygen = −2 while Na is +1) • H2O2 …(Ox. # of oxygen is −1 while H is +1) • Rule #5: The sum of the Ox. #’s in a compound is zero, and for polyatomic ions, the sum of the Ox. #’s equals the charge of the ion. • Example: HNO3…(H= +1, O= −2, so N = +5) SO42−…(O= −2, soS= +6)

  3. Redox Reactions • Now we can explore a redox reaction in detail… • Consider the reaction of zinc with an acid: • Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) • If we examine the oxidation state of zinc, we see that zinc started out at zero and ended up at +2…it lost 2 electrons. • − The process in which a substance increases its oxidation state (by losing electrons) is called “oxidation.” • If we examine the oxidation state of hydrogen, we see that hydrogen started out at +1 and ended up at zero…it gained an electron. • − The process in which a substance decreases its oxidation state (by gaining electrons) is called “reduction.” • Zinc reduced hydrogen, so zinc is called the reducing agent. (In other words, the substance that is oxidized is the “reductant”. • Hydrogen oxidized zinc, so hydrogen is called the oxidizing agent. (In other words, the substance that is reduced is called the “oxidant”.

  4. Redox Reactions • The easiest way to remember the difference between reduction and oxidation is… • “LEO goes GER” • Losing Electrons = Oxidation Gaining Electrons =Reduction • Balancing Redox Reactions • In order to balance redox reactions we need to remember the following: • 1) Conservation of Mass: the amount of each element present at the beginning of the reaction must be present at the end. • 2) Conservation of Charge: electrons are not lost in a chemical reaction. They are transferred from one reactant to another. • Half-reactions are a convenient way of separating oxidation and reduction reactions. • Let’s look at an easy example…

  5. Redox Reactions • Consider the reaction: • Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq) • • The oxidation half-reaction is: • Sn2+(aq) Sn4+(aq) + 2e– (LEO) • (*Note that electrons are shown as a product…they are lost.) • • The reduction half-reaction is: • 2Fe3+(aq) + 2e– 2Fe2+(aq) (GER) • (*Note that electrons are shown as a reactant…they are gained.) • Now we’ll look at one that is more complicated, but it is still simply a redox reaction.

  6. Balancing Redox Reactions • Consider the titration of an acidic solution of Na2C2O4 with KMnO4... • MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(g) • MnO4− is reduced to Mn2+…Manganese starts out at an oxidation state of +7 and is reduced to +2. • C2O42− is oxidized to CO2…Carbon starts out at an oxidation state of +3 and ends up at +4. • First, we have to write the 2 half-reactions: • MnO4−(aq) Mn2+(aq) (Reduction) • C2O42−(aq) CO2(g)(Oxidation) • Then we balance each half-reaction by following these steps… +7 -2 +3 -2 +2 +4 -2

  7. Balancing Redox Reactions • MnO4−(aq) Mn2+(aq) • C2O42−(aq) CO2(g) • a. First, balance the elements other than H and O… • (Nothing needs to be done yet for this half reaction….) MnO4−(aq) Mn2+(aq) • (Balance the carbon…) C2O42−(aq) 2CO2(g) • b. Then balance O by adding water… • MnO4−(aq) Mn2+(aq) + 4H2O(l) • (The oxygen is already balanced in this half-reaction…) C2O42−(aq) 2CO2(g)

  8. (Continued…) MnO4−(aq) Mn2+(aq) + 4H2O(l) • C2O42−(aq) 2CO2(g) • c. Then balance H by adding H+ (if in an acidic environment.) • 8H++ MnO4−(aq) Mn2+(aq) + 4H2O(l) • (Nothing needs to be done yet to this half reaction…) C2O42−(aq) 2CO2(g) • d. Finish by balancing charge by adding electrons…. • 8H+ + MnO4−(aq) Mn2+(aq) + 4H2O(l)(GER) • The left side Mn is +7 while the right side Mn is +2. We add the electrons to the left side of the equation…(How many are gained?) • 5e− + 8H+ + MnO4−(aq) Mn2+(aq) + 4H2O(l) • C2O42−(aq) 2CO2(g) (LEO) • The left side is −2 and the right side is 0. We put the electrons on the product side which shows the electrons being “lost.” • C2O42−(aq)  2CO2(g) + 2e− (Each carbon lost one electron.)

  9. (We are not done yet!!!) • The half-reactions need to be multiplied so that the # of electrons gained and lost is equal. • 2 x [5e− + 8H+ + MnO4−(aq) Mn2+(aq) + 4H2O(l)] • 5 x [C2O42−(aq)  2CO2(g) + 2e−] • Then you end up with… • 10e− + 16H+ + 2MnO4−(aq) 2Mn2+(aq) + 8H2O(l) • 5C2O42−(aq) 10CO2(g) + 10e− • Now you add the reactions together and cancel out items wherever possible… • 16H+ + 2MnO4−(aq) + 5C2O42−(aq) 2Mn2+(aq) + 8H2O(l) + 10CO2(g) • It is wise to double-check and see if everything is balanced, including the charge!

  10. What if the reaction were in a basic environment? • Add [OH−] to both sides of the reaction to cancel out the [H+] ions… • 16OH− + 16H+ + 2MnO4−(aq)+ 5C2O42−(aq)2Mn2+(aq) + 8H2O(l) +10CO2(g) + 16OH− • This results in water being formed… • 16H2O(l) + 2MnO4−(aq) + 5C2O42−(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g) + 16OH− • Now we can simplify the equation by canceling out 8H2O(l) that appears on both sides of the equation… • 8H2O(l) + 2MnO4−(aq) + 5C2O42−(aq)2Mn2+(aq)+10CO2(g) + 16OH−(aq) • Balancing redox reactions is one of the most important aspects in Ch. 20! • We will practice more examples later!

  11. Voltaic Cells • The energy released in a spontaneous redox reaction is used to perform electrical work. • Voltaic or galvanic cells are devices in which electron transfer occurs via an external circuit. • Voltaic cells are spontaneous. • Example: If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves by forming Zn2+. • Zn is spontaneously oxidized into Zn2+ by the Cu2+. • The Cu2+ is spontaneously reduced into Cu0 by the Zn. • The entire process is spontaneous!

  12. Voltaic Cells • In order for the redox reaction to do work, the half-reactions need to be separated, and the electrons need to be able to flow from one half-reaction to the other. • Voltaic cells consist of • Anode (where oxidation occurs): Zn(s) Zn2+(aq) + 2e− • Cathode (where reduction occurs): Cu2+(aq) + 2e− Cu(s) • Salt bridge (used to complete the electrical circuit): cations move from anode to cathode, anions move from cathode to anode. • The two solid metals are the electrodes… [cathode, (+), and anode, (–).] • Next, we will look at a picture of this voltaic cell and then we will analyze what is going on…

  13. Voltaic Cells • As oxidation occurs, Zn is converted to Zn2+ and 2e−. These electrons flow towards the cathode where they are used in the reduction reaction which turns Cu2+ into Cu0. • We expect the Zn electrode, (the anode), to lose mass and the Cu electrode, (the cathode), to gain mass. • Electrons flow from the anode (−) to the cathode (+). • Electrons cannot flow through the solution; they have to be transported through an external wire. • Anions and cations move through a porous barrier or salt bridge.

  14. Voltaic Cells

  15. Voltaic Cells • Anions in the salt bridge move into the anode compartment to neutralize the excess Zn2+ ions formed by oxidation. • In the same way, cations from the salt bridge move into the cathode compartment to neutralize the excess negative charge formed during reduction. • Summary:Anode, (-), LEO, anions move towards anode • Cathode, (+), GER, cations move towards cathode • Electrons move from the anode to the cathode. • A way to remember them… ANO CPR • anode negative oxidation cathode positive reduction

  16. Voltaic Cells The zinc anode gets smaller and the copper cathode gets larger. ANO LEO CPR GER

  17. Voltage • The flow of electrons from anode to cathode is spontaneous. What is the “driving force”? • Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode. • Potential difference: difference in electrical potential. • The potential difference is measured in volts… • One volt (V) is the potential difference required to impart one joule (J) of energy to a charge of one coulomb (C). • Volts=Joules/Coulomb • A Coulomb of charge is ≈ 6.25 x 1018 electrons. • One mole of electrons is defined as having 96,500 C of charge. This is called a faraday, F… • 1 F = 96,500 C/mole of e–

  18. EMF • Electromotive force (emf) is the force required to push electrons through the external circuit. • Cell potential: Ecell is the emf of a cell. • This is known as the cell voltage. • Ecell is > 0 for a spontaneous reaction. • For 1Molar solutions & 1 atm pressure for gases, at 25º C (standard conditions), the standard emf (standard cell potential) is called Eºcell. • Example: For the zinc/copper voltaic cell: Eºcell = +1.10 V • The emf of a cell depends on the particular cathode and anode half-cells that are used.

  19. Standard Reduction (Half-Cell) Potentials • In order to determine the emf of a particular cell, you need to know the Eºred of each half cell involved. • • Standard reduction potentials, Eºred , are measured relative to a standard. • • We use the following half-reaction as our standard: • 2H+(aq), (1 M) + 2e– H2(g), (1 atm) Eºcell = 0V. • This electrode is called a standard hydrogen electrode, (SHE). • The SHE is assigneda standard reduction potential of zero. • The Eºred for other half-reactions can be measured relative to the SHE. • The emf of any cell can then be calculated from all of the standard reduction potentials that have been tabulated… • Eºcell = Eºred (cathode) – Eºred (anode)

  20. SHE and Zinc Anode Zinc is oxidized. Since the cell’s emf is + 0.76 V, that means the Eºred for [Zn2+(aq) + 2e−Zn(s)] would be –0.76 V.

  21. Eºcell • Practice Problem: The standard emf for the following cell is 1.46 V. • In+(aq) + Br2(l) In3+ (aq) + Br−(aq) • Calculate the Eºred for the reduction of In3+ to In+. • First, let’s break this up into the two half-reactions… • (LEO ANO) In+(aq) In3+(aq) + 2e− Eºred = ? • (GER CPR) 2e− + Br2(l)  2Br−(aq) Eºred = +1.06 V • Plug and Chug… Eºcell = Eºred (cathode)−Eºred (anode) • 1.46 = 1.06 – [Eºred(anode) ] • Therefore… Eºred (anode) = – 0.40 V • Note: This reaction is spontaneous. Also, In+ was oxidized in this reaction, so you would expect the value for Eºred to be (−).

  22. Eºcell • Changing the coefficients of a half-reaction does not change Eºred. • Zn2+(aq), (1 Molar)  Zn(s)Eºred = – 0.76 V • 2Zn2+(aq), (1 Molar)  2Zn(s) Eºred = – 0.76 V • As you may suspect, the concentration will change Eºcell as we will see later on in the notes. • The cathode reaction (CPR) of a voltaic cell will always have the more positive Eºred than the reaction at the anode (ANO). • In essence, the greater driving force of the cathode half-reaction is used to force the anode reaction to occur “in reverse” as an oxidation. • The entire process ends up being spontaneous.

  23. Eºcell • Practice Problem: A voltaic cell is based on the following two standard half-reactions… • Cd2+(aq) + 2e− Cd(s) • Sn2+ (aq) + 2e− Sn(s) • Which reaction takes place at the cathode and which is at the anode? • Look up the Eºcell for each in Appendix E… • Cd2+… – 0.403 V Sn2+… – 0.136 V • Since Sn2+ is “more positive”, then it will be reduced at the cathode. • Therefore, Cd2+ will be oxidized at the anode. • What is the value for Eºcell? • Eºcell = Eºred(cathode) – Eºred(anode) • Eºcell= –0.136 V – (– 0.403) = +0.267 Volts • Note: The reaction had to be positive (and spontaneous) for a voltaic cell!!

  24. Oxidizing and Reducing Agents • We can use this table to determine the relative strengths of reducing (and oxidizing) agents. • The more positive Ered the stronger the oxidizing agent on the left. • The more negative Ered the stronger the reducing agent on the right. • We can use this to predict if one reactant can spontaneously oxidize another.

  25. Oxidizing and Reducing Agents

  26. EMF and Free Energy • How does emf relate to ∆G? • ∆G = – nFE • where... n = # of electrons transferred during the process • F = 96,500 C/mole • E = cell potential • Since n and F are positive, if E > 0 then G < 0 and the reaction would be spontaneous. • If the reactants and products are in their standard states, then… • ∆Gº = – nFEº

  27. Effect of Concentration on Cell EMF • The Nernst Equation • A voltaic cell will function 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… • RT lnQ • or expressed as log base 10… 2.303 RT • (See p.799 in the text for the derivation.) E = Eº – nF log Q E = Eº – nF

  28. The Nernst Equation • When T = 25º C, the equation is simplified to… • 0.0592 V • Let’s see how this equation can be used… • Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) • At standard conditions, the Zinc/Copper cell has an emf of +1.10 V. What is the voltage of the cell at 25º C when [Cu2+]= 5.0 M and [Zn2+] = .050 M? • Recall that Q = [products]/[reactants]. (Remember, pure solids and liquids do not appear in the expression for Q!) • Therefore… E= 1.10 V – 0.0592 V log [0.050]/[5.0] • Since Q < 1, then log Q < 0, and this will increase the voltage… • E = 1.16 Volts E = Eº – log Q n 2 e–

  29. Concentration Cells • The Nernst equation shows us that using the same species in the anode and cathode compartments of a cell but at different concentrations then a voltage will be generated. • For example… • If the concentrations of [Ni2+] are unequal, a voltage will be generated.

  30. Concentration Cells • At the anode, Ni(s)  Ni2+ increasing the concentration of nickel ions in this compartment. • At the cathode, Ni2+ Ni(s) and the concentration of nickel ions in this compartment is reduced. • Therefore… • Q = [Ni2+]dilute/[Ni2+]concentrated • At 25º C… • E = 0 – 0.0592 log [0.001]/[1] = 0.0888 V • When the compartments reach equilibrium, emf = zero. • The pH meter functions on this principle, and so does your heartbeat! 2 e−

  31. Cell EMF and Chemical Equilibrium • A voltaic cell is functional until E = 0 at which point equilibrium has been reached. The cell is then “dead.” • A system is at equilibrium when G = 0. • From the Nernst equation, at equilibrium and 298 K (E = 0 Volts and Q = Keq): • 0 = Eº – RT ln Keq • This equation can be simplified at 25º C to… • log Keq = nEº/0.0592 • So you can calculate Keq of a redox reaction by knowing Eº or vice versa! nF

  32. Batteries • A battery is a portable, self-contained electrochemical power source consisting of one or more voltaic cells. • The (+) end of the battery is called the cathode, and the (–) end is called the anode. • If you wanted to generate 9 V, you will have to link together multiple cells in series since no single redox reaction can produce that much voltage. • Primarycells cannot be recharged. • Secondarycells are rechargeable.

  33. Types of Batteries • Lead-Acid Battery (12 V car battery): • Cathode: lead dioxide plate • PbO2(s) + HSO4−(aq) 3H+(aq) + 2e− PbSO4(s) + 2H2O • Anode: (lead plate): • Pb(s) + HSO4−(aq) PbSO4(s) + H+(aq) + 2e− • The overall electrochemical reaction is • PbO2(s) + Pb(s) + 2SO42−(aq) + 4H+(aq) 2PbSO4(s) + 2H2O(l) • Ecell = Ered (cathode) – Ered (anode) • Ecell = (+1.685 V) – (– 0.356 V) • Ecell = +2.041 V. • Six of these cells are linked in series to make 12 V, & wood or glass-fiber spacers are used to prevent the electrodes from touching.

  34. Lead Acid Batteries

  35. Lead-Acid Batteries • Lead-Acid batteries can be recharged. • [H2SO4] decreases as the battery discharges and increases as the cell is charged back up while the reactions run in reverse. • H2SO4 is much more dense than H2O, so the condition of the battery can be monitored by testing the specific gravity, (density), of the solution… • * specific gravity = 1.25 to 1.30 charged up and ready to go! • * specific gravity < 1.20 needs charging • These batteries are called “wet cells” because the electrolyte is a liquid.

  36. Alkaline Batteries (Dry Cells) • The most common nonrechargeable battery is the alkaline battery. • • Powdered zinc metal is immobilized in a gel in contact with a concentrated solution of KOH. • The electrolyte is a base, thus these batteries are alkaline. • The electrolyte is a paste, hence the name “dry cell”. • • The reaction at the anode is: • Zn(s) + 2OH−(aq) Zn(OH)2(s) + 2e− • • The reaction at the cathode is the reduction of MnO2: • 2MnO2(s) + 2H2O(l) + 2e− 2MnO(OH)(s) + 2OH−(aq) • • The cell potential of these batteries is 1.55 V at room temperature.

  37. Alkaline Batteries (Dry Cells)

  38. Other Batteries • A common rechargeable battery is the nickel–cadmium (NiCad) battery. • The reaction at the cathode is: • 2NiO(OH)(s) + 2H2O(l) + 2e− 2Ni(OH)2(s) +2OH−(aq) • The reaction at the anode is: • Cd(s) + 2OH−(aq) Cd(OH)2(s) + 2e− • The cell potential of this battery is about 1.30 V at room temp. • Cadmium is a toxic heavy metal. • There are environmental concerns to be addressed with respect to disposal of such batteries. • • Other rechargeable batteries have been developed. • - NiMH batteries (nickel–metal–hydride). • - Li–ion batteries (lithium–ion batteries)…(greater energy density)

  39. Corrosion • An example of an undesirable redox reaction is the corrosion of metals. • • A metal is attacked by a substance in the environment and converted to an unwanted compound. • • Consider the corrosion (rusting) of iron: • Since Eºred(O2) > Eºred(Fe2+), iron can be oxidized by O2. • Cathode: • O2(g) + 4H+(aq) + 4e− 2H2O(l) Eºred = +1.23 V. • Anode: • Fe(s) Fe2+(aq) + 2e−Eºred = −0.44 V.

  40. Corrosion •Dissolved oxygen in water usually causes the oxidation of iron. • The Fe2+ initially formed can be further oxidized to Fe3+, which forms rust, Fe2O3·xH2O(s). • Oxidation occurs at the site with the greatest concentration of O2. • Other factors to consider are the pH, presence of salts, stress on the iron, and contact with other metals.

  41. Corrosion of Iron

  42. Preventing the Corrosion of Iron • • Corrosion can be prevented by coating the iron with paint or another metal to keep it away from O2 and water. • Another way to protect the iron is by a process called cathodic protection. • Coating iron with a layer of zinc, “galvanizediron”, can protect the iron from corrosion even after the surface coat is broken. • Zn2+(aq) + 2e− Zn(s)Eºred = − 0.76 V • Fe2+(aq) + 2e− Fe(s)Eºred = − 0.44 V • The standard reduction potentials indicate that Zinc is easier to oxidize than iron, so the Zinc will be oxidized! • Zinc is the “sacrificial anode” and is slowly destroyed. Iron acts as the cathode where the oxygen is reduced. • • We can use something similar to protect underground pipelines. • Often, Mg is used as a sacrificial anode…(Eºred = − 2.37 V)

  43. Preventing the Corrosion of Iron

  44. Preventing the Corrosion of Underground Pipes

  45. Electrolysis • It is possible to cause a nonspontaneous redox reaction to occur. Such processes are driven by an outside source of electrical energy. • These reactions are called “electrolysis reactions” and take place in “electrolytic cells”. • In voltaic and electrolytic cells, reduction occurs at the cathode, and oxidation occurs at the anode. • However, in electrolytic cells, electrons are forced to flow from the anode to the cathode. • In electrolytic cells the anode is positive and the cathode is negative. • (In voltaic cells the anode is negative and the cathode is positive.) • Here’s what an electrolytic cell looks like…

  46. Electrolytic Cell

  47. Electrolysis of Molten NaCl • Example of Electrolysis: • Cathode (GER): 2Na+(l)+ 2e− 2Na(l) • Anode (LEO): 2Cl−(l) Cl2(g) + 2e− • A battery or some other source of direct current acts as an electrical pump pushing electrons to the cathode and pulling them from the anode. • As Na(s) is produced at the cathode (–) , additional Na+ migrates in. • As Cl2(g) is produced at the anode (+), additional Cl− migrates in. • Industrially, electrolysis is used to produce metals like aluminum and gases like chlorine.

  48. Electrolysis of Aqueous Solutions • Do we get the same products if we electrolyze an aqueous solution of the salt? • No! Water complicates the issue. • Example: Consider the electrolysis of NaF(aq): • Cathode (GER): Na+(aq) + e− Na(s)Eºred = − 2.71 V • Cathode (GER): 2H2O(l) + 2e− H2(g) + 2OH−(aq)Eºred = − 0.83 V • Thus water is more easily reduced than the sodium ion! • Anode (LEO): F−(aq) F2(g) + 2e−Eºred = +2.87 V • Anode (LEO): 2H2O(l) O2(g) + 4H+(aq) + 4e− Eºred = +1.23 V • Thus it is easier to oxidize water than the fluoride ion. • What you end up doing is electrolyzing water into H2(g) and O2(g)!

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