Intro to Electrochemistry Unit 5 - Electrochemistry

Intro to Electrochemistry Unit 5 - Electrochemistry Chemistry 12 The branch of chemistry concerned with the conversion of chemical energy to electrical energy, and vice versa. Many of the concepts which we first learned in Acid-Base theory will apply to Electrochemistry. Consider the following reaction: 2 AgNO 3(aq) + Cu (s) 2 Ag (s) + Cu(NO 3 ) 2(aq) As a net ionic equation: 2 Ag + (aq) + Cu (s) 2 Ag (s) + Cu 2+ (aq) The previous reaction can be broken down into two halves: Reduction Reaction: A half-reaction in which a species GAINS electrons. Ag + (aq) + e - Ag (s) (Ag + (aq) is being reduced) Oxidation Reaction: A half-reaction in which a species LOSES electrons. Cu (s) Cu 2+ (aq) + 2 e - (Cu (s) is being oxidized) The overall balanced, net ionic equation is called a REDOX equation (REDuction + OXidation): OXIDATION: REDUCTION: REDOX: Cu (s) Cu 2+ (aq) + 2 e - 2 Ag + (aq) + 2 e - 2 Ag (s) 2 Ag + (aq) + Cu (s) 2 Ag (s) + Cu 2+ (aq) 1

REMEMBER: The number of electrons lost in one half reaction must equal the number of electrons gained in the other. Law of Conservation of Charge Here are two memory aids you can use: LEO the lion says GER LEO = Lose Electrons Oxidation GER = Gain Electrons Reduction OIL RIG Oxidation Involves Loss of electrons Reduction Involves Gain of electrons Going back to the previous example: 2Ag + (aq) + Cu (s) 2Ag (s) + Cu 2+ (aq) What causes Cu (s) to become oxidized? The Ag + (aq)!!! In this case, we call the Ag + (aq) the OXIDIZING AGENT. The oxidizing agent causes the other species to become oxidized, but is reduced itself. Every redox reaction will also have a REDUCING AGENT. The reducing agent causes the other species to become reduced, but is oxidized itself. For Example: What are the half reactions? Zn (s) + I 2(s) Zn 2+ (aq) + 2 I - (aq) What is the reducing agent and the oxidizing agent? 2

Homework: Oxidation Numbers Read: Pages 192-200 Do: #1, 2 The charge that an atom would possess if the species containing the atom were made up of ions. Somewhat fictitious, but will allow us to determine if a species was reduced or oxidized. Rules: The alkali metals (Li, Na, K, Rb, Cs) are ALWAYS +1. The alkaline earth metals (Be, Mg, Ca, Sr, Ba) are ALWAYS +2. The halogens (F, Cl, Br, I) are NORMALLY -1 (there are many exceptions). Oxygen is normally -2. Hydrogen is normally +1. The rule which allows us to calculate oxidation numbers is simply: The sum of the positive charges and the negative charges must equal the overall charge on the species. 3

For Example: What is the oxidation number of P in the molecule H 4 P 2 O 7? What is the oxidation number of P in P 4? We need to compare the oxidation number before and after the reaction has happened to see which species have been oxidized/reduced For Example: Consider the following unbalanced half-reaction: Is Se oxidized or reduced? H 2 SeO 3 Se What is the oxidation number of Cr in Cr 3+? What is the oxidation number of Mn in MnO 4-? H 2 SeO 3 Se ON = +4 ON = 0 ON = 0 4 = -4 Since the ON is negative, the species has gained electrons and is therefore REDUCED. To summarize: Examples: If the ON of a species increases (+ ON), then the species has been oxidized (loses e - ). If the ON of a species decreases (- ON), then the species has been reduced (gains e - ). 4

Predicting Spontaneity Using the Standard Reduction Potentials of Half- Cells table on page 8 of the data booklet, we can predict the spontaneity of a redox reaction. For an isolated half-reaction, use a double arrow to indicate that the reaction may proceed in either the forward or backward direction: Au 3+ (aq) + 3 e - Au (s) When a half-reaction goes through oxidation or reduction as a result of being part of a redox reaction, use a one-way arrow. Au 3+ (aq) + 3 e - Au (s) Know the trends on the table!!! For Example: Considering the two species, Ag + is a stronger oxidizing agent and therefore has a greater tendency to reduce. The reduction reaction can be written as: Ag + (aq) + e - Ag (s) Consider two half-cells: In one half-cell, there is a piece of Ag (s) in a solution of Ag +. The second half-cell has a piece of Cu (s) in a solution of Cu +. Each half reaction can be written as: Since Cu + is lower on the table, it is the stronger reducing agent and has a greater tendency to oxidize. Think of the reaction as going in reverse: Ag + (aq) + e - Cu + (aq) + e - Ag (s) Cu (s) Cu + (aq) + e - Cu (s) When actually writing down the oxidation half-reaction, put Cu (s) on the reactants side. Cu (s) Cu + (aq) + e - 5

To find the overall redox reaction that will occur, add together the two half reactions: REDUCTION OXIDATION REDOX Ag + (aq) + e - Ag (s) Cu (s) Cu + (aq) + e - Ag + (aq) + Cu (s) Ag (s) + Cu + (aq) IF TWO HALF-CELLS ARE JOINED, THE HIGHER HALF-REACTION ON THE TABLE WILL UNDERGO REDUCTION, AND THE LOWER ONE WILL UNDERGO OXIDATION! How do we do this? 1. Locate both the reactants on the table. a) If both reactants only appear on the left, or both only appear on the right, then THERE IS NO REACTION! b) If one reactant appears on the left, and one reactant appears on the right, there are two possible cases. CASE 1: Case 2: When the reactant that will be reduced (left side) is higher on the table than the reactant that will be oxidized (right side) A SPONTANEOUS REACTION WILL OCCUR! When the reactant that will be reduced (left side) is lower on the table than the reactant that will be oxidized (right side) THERE IS NO REACTION! 6

Examples: NOTE: BE CAREFUL with half reactions that have H + as a reactant. H + must be treated as a reactant. These reactions can only occur in an ACIDIC environment. Homework: Balancing Half Reactions Read: Pages 193-210 Do: #3-18 must balance for mass and charge. Don t be sloppy about writing the charges on ions, or you will make many mistakes!!! You will be given a skeleton equation which contains the major atoms involved. 7

We have to complete the balancing by supplying other species as follows: Balance the MAJOR atoms by inspection. (The major atoms are any atoms other than O and H). Balance the OXYGEN atoms by adding H 2 O molecules. (The reactions will generally occur in water). Balance the HYDROGEN atoms by adding H +. (The reactions will initially be considered to occur in acidic solution). Balance the overall CHARGE by adding electrons. NEVER vary the order in which these steps are carried out!!! We can summarize the procedure with another memory aid: The officer s name was: MAJOR HYDROXIDE. MAJOR HYDROXIDE translates as MAJOR OH- 1. Balance the MAJOR species. (MAJOR) 2. Balance the O atoms. (O) 3. Balance the H atoms. (H) 4. Balance the charge, using electrons. (-) For Example: NH 3 N 2 H 4 (acidic) CrO 2-4 CrO - 2 (acidic) NO 3 - NO (acidic) If the solution is basic, we have to add another step: Using the self-ionization of water: H 2 O (l) we can cancel out all the H + s. H + (aq) + OH - (aq) Since the self-ionization equation can be written in either directions, write it in a way that will cancel out the H + s in the half-reaction. For Example: PbO 2 PbO (basic) 8

Balancing Redox Reactions Using Half Reactions For Example: Assume that the equation can be broken into separate reduction and oxidation half-reactions. After each is balanced, add the two reactions together to obtain the balanced redox equation. P 4 H 2 PO 2- + PH 3 (acidic) Homework: REDOX TITRATIONS Read: Pages 210-212 Do: #19 25 Study for your quiz!!! Redox titration calculations are the same as any other titration. They must be spontaneous reactions (higher on the left, lower on the right). 9

Oxidizing Agents A very useful oxidizing agent is acidified KMnO 4. The reaction: MnO 4 - (aq) + 8 H + (aq) + 5 e - Mn 2+ (aq) + 4 H 2 O (l) has a strong tendency to reduce and is therefore able to oxidize many other substances (the K + in KMnO 4 is a spectator ion). Unlike an acid/base titration, the above reaction does not require the use of an indicator. The endpoint of this redox titration is the change from a clear solution to a faint purple one. Consider the titration of acidified MnO 4- with Fe 2+ (common application): MnO 4 - (aq) + 8 H + (aq) + 5 Fe 2+ (aq) Mn 2+ (aq) + 4 H 2 O (l) + 5 Fe 3+ (aq) As we add MnO 4- from a burette to a solution containing Fe 2+, the purple colouration of the MnO 4- is continually destroyed. At the equivalence point, the last of the Fe 2+ will have been used up, so that the next drop of MnO 4- will not react and a light purple colour will remain in the solution. Reducing Agents An example of a substance which is commonly used as a reducing agent is NaI or KI. Titrations involving this species usually occur in two steps: 1. The I - is oxidized to I 2 by the substance being reduced. 2. Next, the I 2 produced by the first step is reduced back to I - by a second reducing agent. An example of a reaction involving I - is the reduction of laundry bleach (NaOCl): NOTE: 2 I - I 2 + 2 e - 2 e - + 2 H + + OCl - Cl - + H 2 O 2 H+ + OCl - + 2 I - Cl - + H 2 O + I 2 All the OCl - must react, therefore, an excess of I - is added. 10

The second reaction involves reducing the I 2 formed in the first reaction back into I - by S 2 O 3 2- (usually Sodium thiosulphate, a strong reducing agent). I 2 + 2 e - 2 I - 2 S 2 O 3 2- S 4 O 6 2- + 2 e - 2 S 2 O 3 2- + I 2 S 4 O 6 2- + 2 I - When the S 2 O 3 2- reacts with the I 2 present, the brown colour of the I 2 will begin to become a pale yellow. At this point a starch solution is added. Homework: Read: Pages 215-232 Do: #26-33 The starch will turn the solution a dark blue due to its reaction with the I 2 in solution. The endpoint of the titration occurs when the blue colour has just disappeared. Electrochemical Cells An electrochemical cell is a device capable of deriving electrical energy from chemical reactions. Every electrochemical cell has two electrodes (a conductor at which a half-cell reaction occurs). Anode: The electrode at which OXIDATION occurs. The electrode toward which ANIONS travel. Cathode: The electrode at which REDUCTION occurs. The electrode toward which CATIONS travel. For Example: Two half-cells consisting of Ag (s) in an 1.0 M AgNO 3(aq) solution and Cu (s) in a1.0 M CuSO 4(aq) solution are connected to make an electrochemical cell. 11

Standard Reduction Potentials Voltage: The tendency of electrons to flow in an electrochemical cell. Measured in VOLTS. The voltage is always the work done per electron transferred. Electrons cannot flow in an isolated half-cell, so we cannot determine individual half-cell voltages. We can measure the difference in electrical potentials between two half-cells. To do this, we define the Hydrogen half-cell as a zeropoint on the voltage scale: Where: 2 H + (aq) + 2 e - H 2(g) E o = 0.00 V E o = the standard reduction potential, in Volts. (The o in E o implies that we are dealing with a standard state.) For Example: Cu 2+ + 2 e - Cu E o = +0.34 V The cell is at 25 o C. All gasses are at 101.3 kpa (1 atm). All elements are in their normal phase at 25 o C. All solutions in the half-cell have a concentration of 1.0 M. This reaction has a voltage which is 0.34 V more than that of the hydrogen half-cell. Zn 2+ + 2 e- Zn E o = -0.76 V This reaction has a voltage which is 0.76 V less than that of the hydrogen half-cell. 12

Something to remember: If we reverse a reduction reaction and cause it to become an oxidation reaction we find: Zn 2+ + 2 e - Zn E o = -0.76 V (as a reduction) Zn Zn 2+ + 2 e - and E o = +0.76 V (as an oxidation) If we reverse a half-cell, we reverse the sign of its E o value. Since we can add two half-cells together to give a redox equation, the voltages associated with the half-cells can also be added. For Example: Imagine we connect these two half-cells together: Fe 3+ (aq) + 3e - Fe (s) E o = +0.77 V Zn 2+ (aq) + 2e - Zn (s) E o = -0.76 V After we decide which species will be oxidized and reduced, we find the spontaneous reaction to be: (Fe 3+ (aq) + 3e - Fe (s) ) x2 (Zn (s) Zn 2+ (aq) + 2e - ) x3 E o (RED) = +0.77 V E o (OX) = +0.76 V 2 Fe 3+ (aq) + 3 Zn (s) 2 Fe (s) + 3 Zn 2+ (aq)e o (CELL) = + 1.53 V The value for E o CELL is the difference between the voltages for the reduction and oxidation reactions as written on the table of Standard Reduction Potentials of Half-Cells in your data booklet. As easier way to calculate this is: Our value of E o (CELL) also tells us whether the reaction will be spontaneous or not. If E o CELL is positive for a redox reaction, the reaction is expected to be SPONTANEOUS. If E o CELL is negative for a redox reaction, the reaction is expected to be NON-SPONTANEOUS. For Example: E o CELL = E o RED - E o OX Use the values for E o AS GIVEN on your table in your calculations for E o CELL. 13

A few final comments The surface area of an electrode has no effect on the cell potential. Since solids have a constant concentration, there is no shift in equilibrium by increasing or decreasing surface area. Omit the o symbol in E o when cells are not at standard conditions. Le Chatelier shifts occur with cells not at standard conditions. Consider the following: Cu 2+ (aq) + 2e - Shift left, E o (less work done by e - ) Cu (s) E o = +0.34 V Shift right, E o (more work done by e - ) Operating electrochemical cells ARE NOT at equilibrium, but tend towards it as they die out. Homework: Read: Pages 226-236 Do: #34 46 Study for your quiz!!! 14

Selecting Preferred Reactions Sometimes several different reactions appear to be possible simultaneously. When several different reduction half-reactions can occur, the half-reaction having the highest tendency to accept electrons (highest reduction potential) will occur preferentially. When several different oxidation half-reactions can occur, the half-reaction having the highest tendency to lose electrons (lowest reduction potential) will occur preferentially. What do we do? List all the species present. Break up all ionic compounds into ions. Starting in the upper left of the table (reduction side), look down the table until you find the first match with a species on the list. Starting in the bottom right of the table (oxidation side), look up the table until you find the first match with a species on the list. For Example: Applied Electrochemistry Examples include: (see Hebden pages 228 233) The Breathalyzer Lead-Acid Storage Battery Zinc-Carbon Battery Alkaline Dry Cell Fuel Cells 15

Corrosion of Metals Oxidizing (rusting of metals) is a spontaneous electrochemical reaction: The oxygen rich environment promotes reduction of water: ½ O 2(g) + H 2 O (l) + 2e - 2 OH - (aq) The electrons for the reduction come from the oxidation of the Iron metal: Fe (s) Fe 2+ (aq) + 2e - The yellow solid produced is a precipitate of Fe(OH) 2(s). Isolate the metal from it s environment. Apply plastic or paint on the surface. Oxygen can t reach the surface. Apply a metal which is corrosion resistant on the surface of the original metal. For Example: Chrome plated bumpers. Zinc coated metals (galvanized). Cathodic Protection The process in which a substance is protected from unwanted oxidation by connecting it to a substance having a higher tendency to oxidize. Consider steel underground gasoline storage tanks: Magnesium is a stronger reducing agent then Iron, and will therefore oxidize preferentially, sparing the Iron. Mg (s) H 2 O (l) O 2(g Fe (s) Fe 2+ (aq) + 2e - ) The iron tank will tend to rust in an Oxygen/water rich environment. If we use a substance with the greater tendency to oxidize than Fe (s), then the substance will corrode, but the iron will not. Mg (s) (sacrificial anode) is usually used for this task. 16

Another example: Ocean vessels are even more prone to oxidation. The salt water acts as an electrolyte, increasing the rate of oxidation. These vessels may be protected by using Zinc strips on the hull (same effect as the underground gas tank). The same effect can be achieved using a DC power source. Change the conditions in the chemical environment by removing O 2. Storing the metal in oil or another Oxygen free environment. Sodium is stored in oil to prevent its spontaneous oxidation with Oxygen. Homework: Electrolysis Read: Pages 237-241 Do: #47, 49-63 The process of supplying electrical energy to a molten ionic compound or to a solution containing ions so as to produce a chemical change. The process takes place in an ELECTROLYTIC CELL. Electrolysis reactions are non-spontaneous (E o CELL< 0). We have to supply the energy so that the reaction can occur. 17

The Electrolysis of Molten NaCl This simple electrolytic cell only requires a source of ions and a source of electrical energy. Recall that NaCl is an ionic solid; when melted, the ions are mobile. There is no need for a salt bridge to keep the reactants separated because there is no spontaneous reaction which occurs between the reactants. From the above set up we have: Na + (l) + e - Na (s) E o RED = - 2.71 V 2 Cl - (l) Cl 2(g) + 2 e - E o OX = - 1.36 V 2 Na + (l) + 2 Cl - (l) 2 Na (s) + Cl 2(g) E o CELL = - 4.07 V In ALL electrolytic cells, the reduction half-reaction will be below the oxidation half-reaction on the table. To make this cell operate we must add at least +4.07 V. The Electrolysis of Aqueous NaI Since we are talking about a salt in an aqueous environment, we must now consider the water present. There are two possible reduction reactions: 2 H 2 O (l) + 2 e - H 2(g) + 2 OH - (aq) (10-7 M) AND Na + (aq) + e - Na (s) There are two possible oxidation reactions: H 2 O (l) 1/2 O 2(g) + 2 H + (aq)(10-7 M) + 2 e - AND 2 I - (aq) I 2(s) + 2 e - 18

How do we select the preferred half-reaction when we are dealing with electrolysis reactions? Consider this: Therefore: The preferred reaction will be the one requiring the least voltage input!! The preferred reaction (the one requiring the least voltage input) involves the higher of the two possible reductions and the lower of the two possible oxidations. During electrolysis, ALWAYS consider the possibility of water being reduced and/or oxidized. We get: According to the above voltage differences, we can get a reaction to occur if we add at least +0.95 V. 2 H 2 O (l) + 2 e - H 2(g) + 2 OH - (aq)(10-7 M) E o RED = - 0.41 V 2 I - (aq) I 2(s) + 2 e - E o OX = - 0.54 V 2 H 2 O (l) + 2 I - (aq) H 2(g) + I 2(s) + 2 OH - (aq)(10-7 M) E o CELL = - 0.95 V Water has a substantial overpotential. The Overpotential Effect The measured voltage we find when we connect two half-cells together will differ from the voltage expected from the table of Standard Reduction Potentials. This is a result of the activation energies required for the reactions to occur at the electrodes. For most substances, the overpotential is generally small or negligible (+0.02 V). On the table we see: The reduction of neutral water theoretically occurs at - 0.41 V, but actually occurs at around - 0.80 V. The oxidation of neutral water theoretically occurs at +0.82 V, but actually occurs a bit above +1.36 V. The following exceptions occur: OXIDATION REACTIONS: We will only consider Cl - and Br -. These ions will oxidize preferentially to the H 2 O. REDUCTION REACTIONS: We will only consider Zn 2+ and Cr 3+. These ions will reduce preferentially to the H 2 O. 19

Electrolysis Example: Homework: Read: Pages 243-246 Do: #64-72 Applications of Electrolysis For Example: Electroplating: A process in which we cause a metal to be reduced or plated at a cathode. The cathode is made out of the material which will receive the metal plating. The electroplating solution will contain the ions of the metal we want to plate onto the cathode. In some instances, but not always, the anode may be made of the same metal we want to have plated out on our cathode. 20

Electrorefining: The process of purifying a metal by electrolysis. Consider the following cell: The small amount of Zn or Pb present is preferentially oxidized as it is exposed at the surface. When any exposed Zn or Pb atoms have reacted, only the Cu atoms are available to be oxidized. Any Ag, Au, or Pt atoms present can t be oxidized because the anode is mostly copper, which is oxidized in preference to Au, etc. At the anode Apart from the small amounts of Pb and Zn, the Cu has the greatest tendency to oxidize. These metal particles simply drop off the anode and accumulate on the bottom of the electrolytic cell. At the cathode The Cu 2+ in solution is preferentially reduced at the cathode and none of the Pb 2+ and Zn 2+ can be reduced since Cu 2+ exists in larger concentrations and has a higher reduction potential than Zn 2+ and Pb 2+. The result of the electrorefining is that the PURE metal is deposited on the CATHODE. WE ARE DONE OUR NOTES!!!!!! Now just the exams Very pure metals (>99.99%) can be produced in this mannor. 21

Homework: Do: #73 80 Study for your quiz!!! Study for the final!!!!!! 22