Electrochemistry is a branch of chemistry, which is concerned, with the conversion of chemical energy to electrical energy and vise versa.

Transcription

1 Chemistry 12 Unit V Electrochemistry Notes V.1 - Introduction Electrochemistry is a branch of chemistry, which is concerned, with the conversion of chemical energy to electrical energy and vise versa. Let s look at the following reaction 2 Ag + (aq) + Cu (s) 2 Ag (s) + Cu 2+ (aq) The same reaction can be produced in the following experimental setup. The above set up is an example of a chemical system called an electrochemical cell a system that produces electrical energy. The voltmeter shows a flow of electrons from the copper strip to the silver strip. The reaction 2 Ag + (aq) + Cu (s) 2 Ag (s) + Cu 2+ (aq) only involves the loss and gain of electrons. Half of the electrochemical cell is called a half-cell and the reaction, which occurs in a halfcell, is called the half reaction (half-cell reaction). Half-reaction / half-cell reaction: is a reaction equation which shows only the gain or only the loss of electrons. Half reaction for the above equation (2 Ag + (aq) + Cu (s) 2 Ag (s) + Cu 2+ (aq)) are 2 Ag + (aq) + 2e - 2 Ag (s) and Cu (s) Cu +2 (aq) + 2e - (Reacting on the left side) (Reacting on the right side) Salt bridge: is an inverted u-tube filled with an electrolyte solution, such as Na 2 SO 4 or KCl, chosen so that it does not interfere with the operation of the cell. Regardless of how the cell is constructed, the solutions in each compartment remain more or less separated. Because ions can move into and out of the salt bridge, the solution will remain electrically neutral. If the flow of ions did not occur, a charge difference would build up between the compartments and the reaction would stop. If a salt bridge were removed, the reaction would stop. Simmer Down Inc. August

2 Oxidation reaction: half-reaction in which a species LOSES electrons. Reduction reaction: half-reaction in which a species GAINS electrons. Memory Aid: Every reduction reaction must be accompanied by an oxidation reaction. Reactions involving a loss and gain of electrons are called a reduction-oxidation or redox reaction. Let s look again at the above reaction Which is oxidized and which is reduced? Look at the half-reactions again 2 Ag + (aq) + Cu (s) 2 Ag (s) + Cu 2+ (aq) 2 Ag + (aq) + 2e - 2 Ag (s) Cu (s) Cu +2 (aq) + 2e Ag + gains electrons, thus it is being reduced. Cu is losing electrons, thus it is being oxidized. Ag + is said to be the agent with causes Cu to become oxidized. Ag + is called the oxidizing agent. Cu is the agent which causes Ag+ to become reduced, Cu is called the reducing agent. Oxidizing Agent (OA): a substance which causes another to be oxidized by itself becoming reduced. Reducing Agent (RA): a substance which causes another to be reduced by itself becoming oxidized. How can you tell if it is being reduced or oxidized? Species being oxidized become more positively charged. Species being reduced become more negatively charged. Example: Zn +2 + Mg Zn + Mg +2 Which is being oxidized, which is being reduced? What are the oxidizing and reducing agents? Simmer Down Inc. August

3 Answer: Break into half-reactions Zn +2 (aq) + 2e - Zn (s) Mg (s) Mg + 2e - Mg is losing electrons (becomes more positive) therefore it is being oxidized. Zn +2 is gaining electrons (becomes less positive) therefore it is being reduced. Mg is the reducing agent, Zn +2 is the oxidizing agent. Examples: 1. January June April 2000 Simmer Down Inc. August

4 4. April 2003 V.2 Oxidation Numbers Oxidation number is a real or apparent number which allows you to keep track or electrons in redox reactions. Oxidation number increases in oxidation (loss of electrons); decreases in reduction (gain of electrons). Here is the short and quick version of the rules what you will see most often: 1. Pure (neutral elements) = 0 2. Monatomic ion = ion charge 3. Oxygen in compounds = -2 (most of the time, except H 2 O 2 and O 2 ) 4. Hydrogen in compounds = +1 (most of the time except NaH metal hydrides) 5. Sum of the oxidation numbers in a neutral compound is zero (0). Here are the long winded rules a little advanced but I would rather tell you whole truth. a.) The oxidation number of an atom in a free element is always zero, regardless of the formula. For example, the oxidation number of oxygen in 0 2 and 0 3, phosphorus in P 4 and sulfur in S 8, is zero. b.) The oxidation number of a monatomic ion is equal to the charge on the ion. c.) Fluorine is the most electronegative element, so the oxidation number of fluorine in its compounds is always 1. d.) Oxygen, the second-most electronegative element, has an oxidation number of 2 except when it is bonded to Fluorine or itself. The oxidation number of oxygen is +2 in OF 2 (F-O-F), -1 in peroxides such as H (H-O-O-H), 0 in O 2 and 1/2 in the superoxide ion O 2 -. Simmer Down Inc. August

5 e.) Hydrogen has an oxidation number of +1 in all compounds except metal hydrides such as CaH 2. The oxidation number of hydrogen in the hydride ion (H - ) is 1. f.) Some useful group oxidation numbers are: Group 1A: alkali metals: +1 Group 2A alkaline earth metals: +2 Group 7A halogens: -1, except when bonded to oxygen or to a more electronegative halogen. The oxidation number of chlorine, for example, is 1 in NaCl, 0 in Cl 2 and +1 in Cl 2 O and ClF. g.) The sum of the oxidation numbers in a neutral molecule is zero; the sum for an ion is equal to the charge on the ion. For example, the sum of the oxidation numbers is 0 in KBrO 3, +1 in NH 4 +, and 3 in PO 4-3. Examples: 1. What is the oxidation number of O in O 2? 2. What is the oxidation number of P in H 4 P 2 O 7? 3. What is the oxidation number of Cr in Cr +3? 4. What is the oxidation number of S in SO 4 2-? 5. Find the oxidation number of sulphur in Na 2 SO 4. Simmer Down Inc. August

6 Oxidation is a loss of electrons, reduction is a gain. Oxidation number of at least one atom increases during a redox reaction, while the other oxidation number decreases. The following example shows how oxidation numbers can be used to recognize a redox reaction. Example: Ethyl alcohol converts dichromate ion (Cr 2 O 7-2 ) to Chromium (III) ion (Cr +3 ). Is the dichromate ion oxidized or reduced in this reaction? Examples: 1. June April 2003 Simmer Down Inc. August

7 V.3 Predicting The Spontaneity of a Redox Reaction Look at the table of Standard Reduction Potentials and read what s below Please Simmer Down Inc. August

8 What you will notice 1. Left hand side of the table is increasing strength as an oxidizing agent Think Reduction!!! 2. Right hand side, increases going down the table, is increasing reducing agent Think Oxidation!!! 3. Most metals are located at the bottom right. (Except Cu, Ag, Hg and Au) 4. Halogens / oxyanions are located in the top left. 5. Some metals have more than one oxidation number. Therefore, there is more than one one-half reaction on the table. This can either be an oxidation or reduction. Example: Cu +, Sn 2+ and Fe 2+. Isolated half-reactions can go forward or backwards. Use equilibrium arrows. Example: Au e - If a half reaction is part of a redox reaction and MADE to go as either an oxidation or reduction, then use a one-way arrow. Simmer Down Inc. August Au (s) Oxidations, like bases in the previous section, run backwards on the table. How to predict if there is an oxidation or reduction ** If two half cells are joined, the higher half-reaction on the table will under go reduction and the lower will undergo oxidation.** Example: Half-reactions are Zn (s) in a Zn +2 solution Cu (s) in a Cu +2 solution Cu e - Zn e - Cu (s) Zn (s) Of the two species that are reduced Cu +2 and Zn +2, Cu +2 is higher on the left. Therefore, Cu +2 will be reduced. Of the two species that can oxidize, Cu (s) and Zn (s), the Zn is lower on the right, and thus will be oxidize. To write as a redox equation, the oxidation reaction must be written in reverse. Just remember to have electrons on both sides. Cu e - Zn e - Cu (s) Zn (s)

9 Cu +2 is reduced therefore the reaction stays the same Zn (s) is oxidized must be flipped Cu e - Cu (s) Zn (s) Zn e - Add the reactions together the electrons cancel and you are left with.. Cu +2 + Zn (s) Zn +2 + Cu (s) **If only one species in a half reaction is present, you CAN NOT assume the other is also present UNLESS you have been explicitly told.** Example: Cl 2 + 2e - 2Cl - Just because you have chlorine ions (Cl - ) does NOT mean you have chlorine gas think about a salt solution does it smell funny If you are given two potential reactants, rather than complete half-reactions, a reaction may or may not occur. The following procedure is used to determine Locate each reactant on the table If both reactants are found on the left OR right side of the table, then NO REACTION is possible. If one reactant is on the left and one is on the right, then there are two possible cases. 1. Reactant to be reduced (left side) is HIGHER on the table than the reactant to be oxidized (right side). In this case, the reaction will be SPONTANEOUS. 2. Reactant to be reduced (left side) is LOWER on the table than the reactant to be oxidized (right side). In the case, there will be NO REACTION. To sum it up a reaction will be spontaneous if and only if there is a reactant to be reduced (on the left side), which is ABOVE a reactant to be oxidized (on the right side). Examples: Predict what will happen 1. Reactants are Cu +2 and Zn (s) Simmer Down Inc. August

10 2. Reactants are Zn +2 and Cu (s) 3. Reactants are Zn (s) and Cu (s) V.4 Balancing Half-Reactions Half reactions must be balanced for mass (atoms) and charge. Typically you will be given a skeleton equation containing the major atoms involved and it is up to you to balance the rest. Can be balanced in ACIDIC or BASIC solutions. Here are the steps involved when balancing half reactions 1. Balance MAJOR atoms first (Anything other than O and H) 2. Balance the OXYGEN atoms by adding water (H 2 O:) molecules. 3. Balance the HYDROGEN atoms by adding H +. (Used in acidic solutions. Also used in basic solutions, but changed in a later step) 4. Balance overall CHARGE by adding electrons. NEVER, EVER, NEVER, EVER vary the steps. Don t ask why, just do it. Basic solutions are the same steps except we use the water equilibrium, H + + OH - H 2 O to cancel out the H+. Use the following memory aid (corny, but it might help) Major Hydroxide (MAJOR OH - ) 1. Balance the MAJOR atoms first (MAJOR) 2. Balance the O atoms (O) 3. Balance the H atoms (H) 4. Balance the charge using e- (-) That s about all you need to know now sit back and let s do some examples. Simmer Down Inc. August

11 Examples: 1. Balance the half reaction RuO 2 Ru (in acidic solution). 2. Balance the half reaction Cr 2 O 7 2- Cr 3+ (In acidic solution) 3. Balance the half reaction Pb PbO 2 - (Occurs in basic solution) 4. Balance the half reaction N 2 H 4 N 2 (Occurs in basic solution) 5. Balance the following half reaction in acidic solution: H 2 **Anytime you find an isolated species made up of only H and/or O atoms just follow the steps as you normally would. Common examples H 2, O 2, O 3 and H 2 O 2 Simmer Down Inc. August

12 V.5 Balancing Redox Equations Using Half-Reactions **Note** There are two ways to balance redox equations: half reactions and using oxidation numbers. You DO NOT need to know both, just be able to balance the redox equations. Whatever method you use is up to you. We will be looking at the half-reaction method only Very similar to the previous section, just twice the work. Here are the steps 1. Break the equation into TWO half reactions. Usually the major atoms will be separated (One will be an oxidation, the other a reduction). 2. Balance each half reaction (assume all are acidic solution at this point) 3. Multiply the half reactions to make the electrons balance. This step is critical. Be very careful and take your time!!!! 4. Add the two half reactions and cancel out the electrons and any other species that are common (Usually H 2 O) If the reaction is in basic solution, use the water equilibrium and cancel out the H + as you did before. Do this AFTER you have added the two equations together. Watch out for Disproportionation reactions. This is a redox reaction in which the same species is both oxidized and reduced. The same species (reactant) appears in BOTH halfreactions. Examples 1. Balance Os + IO 3 - OsO 4 + I 2 in acidic solution Simmer Down Inc. August

13 2. Balance MnO C 2 O 4 2- MnO 2 + CO 2 in basic solution 3. Balance ClO 2 - ClO Cl - in basic solution 4. Balance Ca 3 (PO 4 ) 2 + SiO 2 + C P 4 + CaSiO 3 + CO Simmer Down Inc. August

14 V.7 Redox Titrations Very similar to acid-base titration s. Same calculations, same method just different overall reaction. Instead of a reaction based on neutralization you are using a redox equation. One of the most useful oxidizing agents is acidic KMnO 4. The half-reaction is: MnO H + + 5e - Mn H 2 O E o = 1.51 V Able to oxidize a large number of substances. (K+ is a spectator) Can be used to find the Fe 2+ concentration in a solution. Fe 2+ is easily oxidized to Fe 3+. Overall reaction is MnO H Fe 2+ Mn H 2 O + 5 Fe 3+ This is your balanced redox equation. Gives you the mole ratios very easy to calculate the concentration of Fe 2+ if you know how much KMnO 4 is used. What do you use as an indicator? As it turns out, one is already included. MnO 4 - is purple in solution and Mn 2+ is colourless. If MnO 4 - is added from the burette to a solution containing Fe 2+, the purple colour due to the MnO 4 - is continually destroyed. At the equivalence point, the last of the Fe 2+ is used up, so that the next drop of MnO 4 - does NOT react and a light purple colour remains in the solution. Endpoint of the titration is a change from colourless to a light purple colour. Example: 1. When ml of a solution containing an unknown concentration of Fe 2+ is titrated to an endpoint with acidic KMnO 4, the titration requires ml of acidified M KMnO 4. The equation is: MnO H + + 5Fe 2+ Mn H 2 O + 5Fe 3+ Simmer Down Inc. August

15 Common reducing agent is NaI or KI. A large number of substances can oxidize I - to I 2. 2I - I 2 + 2e - Titration s involving I - generally involve two consecutive steps: 1. I - is oxidized to I 2 by the substance being reduced 2. I 2 produced in the first step is reduced back to I - by a second reducing agent, such as thiosulphate ion S 2 O Common reaction is the reduction of laundry bleach, NaOCl. The reaction between I- and OCl- proceeds according to: 2H + + OCl - + 2I - Cl - + H 2 O + I 2 (Note: I- is added in excess in order to ensure all OCl - has reacted.) The above reaction is just the initial reaction because the actual redox titration involves a second reaction between the I 2 produced and the reducing agent sodium thiosulphate, Na 2 S 2 O 3. The reaction is as follows How can you tell the endpoint? 2S 2 O I 2 S 4 O I - I 2 has a brown colour. When S 2 O 3 2- has reacted with most of the I 2, a pale yellow colour remains. At this stage, a starch solution is added producing a dark blue colour. (Blue colour due to starch and the remaining I 2 in solution). After the starch is added, the last of the S 2 O 3 2- is added, causing the blue colour to fade. The last bit of colour disappears at the equivalence point. Bottom line have to use both equations to figure out the concentration in the bleach. Same exact thing you have always done, just more work. Example: 1. A ml sample of bleach is reacted with excess KI according to the equation: 2H + + OCl - + 2I - Cl - + H 2 O + I 2 The I 2 produced requires exactly ml of M Na 2 S 2 O 3 to bring the titration to the endpoint according to the equation: 2S 2 O I 2 S 4 O I - using starch solution as an indicator. What is the OCl - concentration in the bleach? Simmer Down Inc. August

16 V.8 The electrochemical Cell Electrode - a conductor at which a half-reaction occurs (general term) Anode - electrode at which OXIDATION occurs. The electrode toward which anions travel. Cathode - electrode at which REDUCTION occurs. The electrode towards which cations travel Memory Aid: Oxidation at the Anode (both vowels) Reduction at the Cathode (both consonants) OR ***Please look at the electrochemical cell handout now*** The half reactions are: Ag + + e - Ag Cu e - Cu Look at the Standard Reduction Potential table, which half reaction is going to be the reduction and which is the oxidation? (Write it down below) Are the half-cells directly connected? Can they mix? Do you remember what the salt bridge does? Look it up After the half-cells are connected, the Cu has a greater tendency to oxidize than Ag, thus Copper will be oxidized. Because of this, there is an accumulation of electrons at the Cu electrode. This build up causes the electrons to flow from the Cu to the Ag electrode (through the wire). Because copper is oxidized, it becomes the ANODE. Simmer Down Inc. August

17 Electrons are supplied to the Ag electron (via the wire) and as a result the equilibrium Ag Ag + + e - is upset. According to Le Chatelier s Principles, the Ag + is forced to reduce. Because reduction occurs at the Ag electrode it is the CATHODE. ELECTRONS FLOWS FROM THE ANODE TO THE CATHODE Water and certain ions (Cu 2+ and NO 3 - in the hand out) are able to pass through the salt bridge, but free mixing of the solution is prevented. (This prevents direct transfer of the electrons from Cu to Ag +.) By separating the mixtures it forces the electrons to travel through the wire. (A voltmeter can be inserted along the wire). As Cu +2 ions are formed, they accumulate around the anode. The excess positive charge is depleted by simultaneous migration of: 1. Cu 2+ ions away from the anode by random movement. There is a greater probability that the ions will leave an area of high concentration rather than enter it. 2. Negative ions, such as SO 4 -, toward the anode. High concentration of positive charge towards one end of the half-cell will attract the negative ions towards the anode. As the concentration of Ag + is depleted around the cathode, the net amount of positive charge is decreased around the cathode. The resulting deficiency is corrected by the migration of: 1. Ag + ions towards the cathode by random movement. There is a higher probability that Ag + will move into a region of low Ag + concentration rather than away from it. 2. Negative ions, such as NO 3 -, away from the cathode. Greater amount of positive charge at the anode end of the cell will attract the negative ions toward the anode. **No electrons flow in solution, only in the wire** *Number of electrons in the oxidation reaction must equal the electrons in the reduction* Simmer Down Inc. August

18 Examples: 1. Assume two half-cells consisting of Pb (s) in a Pb(NO 3 ) 2 solution and Zn (s) in a ZnCl 2 solution are connected to make an electrochemical cell. During the reaction, more Pb (s) is formed. Draw and label the parts of the cell, write the equation for the individual half-reactions and overall reaction, and indicate the directions in which electrons and ions flow. (Have fun!!!) 2. June 2004 Simmer Down Inc. August

19 V.9 Standard Reduction Potentials Voltage / Electrical Potential Tendency of electrons to flow in an electrochemical cell. VOLTAGE is the WORK DONE PER ELECTRON TRANSFERRED Electrons cannot flow in an isolated half-cell; therefore individual half-cell voltages cannot be measured. The difference in electrical potentials between two half-cells can be measured. A zero point is arbitrarily defined on the voltage scale. The Hydrogen Half cell is defined to be 2H + + 2e - H 2 (g) E o = V Where E o = the standard Reduction potential, in volts and o implies is a standard state. Standard State implies o C 2. All gases at kpa (1 atm) 3. All elements in their standard state (normal phase at 25 o C) 4. Exists a 1 M concentration of ALL solutions involved in the half-cell (Both reactant and product sides) It uses a Platinum electrode. Platinum is used because it is an INERT ELECTRODE it doesn t react but conducts electrons and allows the reaction to occur. Inert electrodes (Platinum / carbon) are often used when all the materials in the halfreaction are ions, liquids, or gases. Voltages associated with each half-reactions are relative to the voltage for the hydrogen half-cell. Simmer Down Inc. August

20 Placing a piece of copper metal in a 1 M Cu 2+ solution, sets up the half-cell: Cu e - Cu ; E 0 = 0.34 V This half-cell has a voltage that is 0.34 V more than that of the hydrogen cell IF A HALF-REACTION IS REVERSED, THE SIGN OF ITS E 0 VALUE IS REVERSED. Let s see what happens when the following half-cells are joined. Hg e- Cu e- Hg ; E 0 = 0.85 V Cu ; E 0 = 0.34 V Which will be the reduction, which the oxidation? (Use your chart ) The experimentally observed voltages for the overall reaction is found to be the difference the voltages of the individual half-cells. That is: Hg e - Hg ; E 0 = 0.85 V Cu Cu e - ; E 0 = V Hg 2+ + Cu Hg + Cu 2+ ; E 0 cell = 0.51 V Conclusion: IF TWO-REACTIONS CAN BE ADDED TOGETHER TO GIVE A REDOX EQUATION, THE VOLTAGES ASSOCIATED WITH THE HALF-REACTIONS CAN ALSO BE ADDED. The cell potential can also be calculated E 0 cell = E 0 RED E 0 OX E 0 RED = the reduction potential of the reduction reaction E 0 OX = the reduction potential of the oxidation reaction Conclusion: THE POTENTIAL OF AN ELECTROCHEMICAL CELL IS JUST THE DIFFERENCE BETWEEN THE HALF-CELL VOLTAGES FOR THE REDUCTION REACTION AND THE OXIDATION REACTION. No matter which method you use to calculate the value of E 0 cell, all you are doing is finding the difference between the two half-reactions. Simmer Down Inc. August

21 Examples: 1. Calculate the potential of the cell: Ni 2+ + Fe Ni + Fe Calculate the potential of the cell: Ni + Fe 2+ Ni 2+ + Fe Conclusion: If E 0 cell is positive for a redox reaction, the reaction is expected to be spontaneous. If E 0 cell is negative for a redox reaction, the reaction is non-spontaneous. What happens to the E 0 value if you have to multiple the half-reaction to get the electrons to balance? DO NOT multiply the E 0 value Voltage is the work done per electron. If you need more work, than you need more electrons. The ratio is still the same. Calculate the cell potential of: 3Ag + + Al 3Ag + Al 3+ The surface area of an electrode has NO EFFECT on the cell potential. When the surface area of an electrode is increased, the reaction rate at the electrode increases. Meaning, the number of electrons transferred per second increases. This increased reaction rate DOES NOT increase the cell voltage. Simmer Down Inc. August

22 Increasing the amount of electrode material also increases the length of time the cell operates big batteries operate longer than small batteries made of identical materials. This increase in the amount of reactant does not increase the cell voltage. Half-cells not at standard conditions o Generally, if you decrease the concentration (less than 1M) than the reduction potential will decrease. If you increase the concentrations (more than 1M) than the reduction potential will increase. What happens when the cells reach equilibrium? Operating electrochemical cell are NOT in equilibrium. Let s look at the following cell: 2Ag + + Cu 2Ag + Cu 2+ Initially this reaction has a great tendency to form products. As the cell operates, using up reactants and making more products, two effects are found. 1. The Reduction Reaction: 2Ag + + 2e - 2Ag As the [Ag + ] decreases, the reduction potential decreases and the half-reaction goes lower on the chart (lower overall reduction potential). As a result, the tendency to form products decreases as the cell operates. 2. The Oxidation Reaction: Cu Cu e - As the [Cu 2+ ] increases the tendency for this reaction to undergo reduction increases (the reduction potential of Cu e - Cu increases). Therefore, the tendency to be oxidized is increasing opposed by the greater tendency to be reduced as the cell operates. Overall, the following occurs as the cell goes to equilibrium. Simmer Down Inc. August

23 Examples: 1. January June 2004 Simmer Down Inc. August

24 V.10 Selecting Preferred Reactions When a cell contains a mixture of reactions, several different reactions may be possible. If a reaction occurs in an acidic solution, the reduction of H+ (at 0.00 V) may be a possible reaction and must be considered along with any other possible reductions. 2H + + 2e - H 2(g) Similarly, if a reaction occurs in a neutral solution, the reduction of neutral water (at 0.41 V) may be a possible reaction and must be considered. 2H 2 O + 2e - H 2(g) + 2OH - (10-7 M) (You don t have to worry about basic solutions ) The oxidation of neutral water [at V shown below as a reduction of O 2(g) ] is another possibility: ½ O 2(g) + 2H + (10-7 M) + 2e - H 2 O Look / record all possible half-reactions. When several different reduction half-reactions are possible, the one with the highest tendency to accept electrons (highest on the left hand side of the chart, highest E 0 value) will occur preferentially. When several different oxidation half-reactions can occur, the half-reaction having the highest tendency to lose electrons (lowest on the right hand side, lowest E 0 value) will occur preferentially. Any ion capable of being reduced will be a spectator if there is another ion in the same solution that has a greater tendency to be reduced. Same principle applies to oxidation. The following steps will help 1. List all species present, making sure that all ionic compounds are broken up into ions. 2. Start at the upper left hand side of the reduction table and look down the table till you find the first species on the list. This will be the reduction. 3. Start at the lower right hand side of the reduction table and look up the table till you find the first match with a species on the list. This will be your oxidation. Simmer Down Inc. August

25 Let s look at some examples 1. Look at the diagram below. What are the preferred reactions? 2. An iron strip is placed in a mixture of Br 2(aq) and I 2(aq). What are the preferred reactions? 3. A beaker contains an iron nail wrapped with a piece of copper wire and a piece of magnesium ribbon, immersed in an aqueous solution containing CuSO 4 and some dissolved Cl 2(g). What are the preferred reactions. Simmer Down Inc. August

26 V.11 Applied Electrochemistry Breathalyser Oxidation of ethanol is represented by C 2 H 5 OH + K 2 Cr 2 O 7 + H 2 SO4 CH 3 COOH + Cr 2 (SO 4 ) 3 + K 2 SO 4 + H 2 O Determines alcohol consumed based on the colour change of the reaction. Dichromate ion (Cr 2 O 7-2 ) is orange-yellow. Chromium (III) ion (Cr +3 ) is dark green The alcohol is absorbed into the blood stream. Some of the alcohol passes through the blood and into the alveoli. The movement of alcohol back and forth in the blood stream is an equilibrium process. A lot of alcohol in the breath means lots of alcohol in the blood. Amount of alcohol in the exhaled air is called the breath alcohol content. Exhaled air is mixed with a solution containing K 2 Cr 2 O 7, H 2 SO 4 and a little Ag + to stabilize it. Batteries and Fuel Cells Placed in a spectrometer and analyzed for the green colouration (from Cr +3 ). Lead Acid storage battery Automotive battery Consists of alternating plates of Pb (S) and PbO 2(s) immersed in dilute H 2 SO 4. Anode Reaction: Pb(s) + HSO 4 - PbSO 4 (s) + H + (aq) + 2e - (Simply) Pb(s) Pb e- Cathode Reaction: PbO 2 (s) + HSO 4 - (aq) + 3H + +2e - (Simply) Pb e - Pb +2 PbSO 4 (s) + 2H 2 O(l) Overall Reaction: Pb(s) + PbO 2 (s) + 2H + +2HSO 4 - (aq) (Simply) Pb + Pb +4 Pb +2 + Pb +2 2PbSO 4 (s) + 2H 2 O This above reaction occurs when the battery is discharging, that is when reacting to produce electrical energy. PbSO 4 forms layers around Pb (s) and PbO 2(s) plates. (See picture below) Simmer Down Inc. August

27 When an external source of electrical energy is applied to the battery to recharge it. The discharging reaction is driven backwards. 2PbSO 4(s) + 2H 2 O (l) Pb (s) + PbO 2(s) + 2H + (aq) + 2HSO 4 - (aq) During the discharge / re-charge cycles, some PbSO 4 (s) flakes off. Thus, less Pb and PbO 2 can re-form and the cell deteriorates. During discharge, H 2 O is produced and H 2 SO 4 is used. This decrease in H 2 SO 4 concentration in the discharge cell lowers the density of the solution. Can measure the condition of the battery be using a hydrometer to check the density. Zinc-Carbon Battery Dry cell battery that is used in flashlights among other things. Cathode reaction: 2MnO 2(s) + 2NH 4 + (aq) +2e - (Simple) Mn +4 + e - Mn +2 Anode reaction: Zn (s) + 4NH 3(aq) Zn(NH 3 ) 4 +2 (aq) +2e - (Simple) Zn Zn e - 2MnO(OH) (s) + 2NH 3(aq) As Zn is used up, Zn(NH 3 ) 4 +2 accumulates around the anode and as a result, the battery wears out voltage decreases. Advantages: Cheap to make Disadvantages: Cannot be re-charged, short shelf life, useless after large current used. Alkaline Dry Cell Alkaline because the electrolyte used is basic. Very similar to Zinc-Carbon battery uses MnO 2 and Zinc as well. Major difference is that it operates under basic conditions. Cathode reaction: 2MnO 2(s) + H 2 O (l) + 2e - Mn 2 O 3(s) + 2OH - (aq) Anode reaction: Zn (s) + 2OH - (aq) ZnO (s) + H 2 O (l) + 2e - Advantages: greater current, more constant voltage. Simmer Down Inc. August

28 Fuel Cell Fuel cell is a device into which a fuel is continuously supplied and from which electricity is continuously obtained. (Batteries contain all the chemical reactants within themselves) Advantages: Pollution free, operates silently, very efficient (70% - 80%) Disadvantages: Electrodes corrode quickly, requires constant maintenance, expensive to make and usually very large. Most common type is the hydrogen oxygen fuel cell. Examples: 1. June 2000 Simmer Down Inc. August

29 2. August 2000 V.12 Corrosion Of Metals: Causes and Prevention Most widely known form of corrosion is the rusting of iron. Rusting only applies to the oxidation of iron. Corrosion is the oxidation of other metals Major agents responsible for rusting are water and oxygen. When a drop of water rests on an iron surface, a spontaneous reaction occurs. Oxygen poor region, in the centre of the drop, iron oxidizes. This is the anode. Fe (s) Fe e - Fe +2 moves away to the cathode region (outside of the drop). As the Fe +2 moves away, more iron is exposed for oxidation. Simmer Down Inc. August

30 Cathode (O 2 rich area) ½ O 2 + H 2 O + 2e - 2OH - After reaching the outer region of the drop, the Fe +2 meets up with OH - and forms insoluble Fe(OH) 2(s) Precipitate forms at the outer edge of the drop. Character ring of rust is formed when the water evaporates. Anode area is O 2 poor because the reaction ½ O 2 + H 2 O + 2e - 2OH - uses up most of the O 2 at the OUTER most of the drop, so little O 2 makes it to the INTERIOR. Therefore, Fe +2 migrates a fair distance before rusts forms. Rust is Fe 2 O 3 XH 2 O, where X is a variable. Know why rust can be a different colour????? Answer: because of the variability of X. Different numbers of water molecules attached to the Fe 2 O 3 will change the colour. Metal may corrode if it touches a different type of metal in the presence of an electrolyte solution exposed to O 2. Example: Iron and Copper wire (in water) Fe (s) Preventing Corrosion Two ways to prevent corrosion Fe e - (Fe has a greater tendency to oxidize) Copper conducts the electrons away from the iron and makes the electrons available to the oxygen / water touching the wire ½ O 2 + H 2 O + 2e - 2OH - 1. Protect the metal from the corrosive environment. 2. Electrochemical properties to prevent corrosion even when the metal is exposed. A - Isolate the Metal from the environment 1. Protective layer. Coat the metal with paint or plastic. Prevents the water and oxygen. 2. Apply a metal that is corrosive resistant to the surface of the corrosive metal. Example: Steel cans are coated with an outer layer of tin. Outer layer of tin atoms is oxidizes quickly. Product is tin oxide. Helps to protect the tin from further oxidation. Simmer Down Inc. August

31 B Electrochemical Principles Cathodic protection: process of protecting a substance from unwanted oxidation by connecting it to a substance with a higher tendency to oxidize. Example: Iron and Magnesium. Mg has a higher tendency to oxidize. Magnesium acts as an anode. Fe is the cathode. Fe remains in reduced form. Another example: Zinc and boats. Zinc is bolted below the water line of iron-hulled ships. The zinc is sacrificed iron is kept from oxidizing. Change the conditions in the surroundings as to limit the tendency of the surroundings to reduce. Water + iron; the following reaction oxidizes the iron. ½ O 2 + 2H + + 2e - H 2 O; E = 0.82 V If you remove the O 2, E drops to the 0.41 Volts which just above the iron halfcell. 2H + + 2e - Fe (s) ; E = Fe e - Fe (s) ; E = If you remove O 2 and lower the H + concentration by adding [OH - ] the resulting half-cell is now much more lower. 2H 2 O + 2e - H 2(g) + 2OH - ; E = V As a result, piece of iron placed in a solution of NaOH will not rust. Examples: 1. August 2005 Simmer Down Inc. August

32 2. January 2001 V.13 - Electrolysis Electrolysis: supplying electrical energy to a molten ionic compound or a solution containing ions so as to produce a chemical change. Electrolytic cell or electrolysis cell is an apparatus which electrolysis can occur. Uses electrical energy to produce a chemical change that would otherwise not occur spontaneously. Electrolytic Cell: Electrolysis of a molten binary salt NaCl is electrolysed. NaCl is an ionic solid. When melted, the ions are mobile. There is no need for a salt bridge or barrier as the reaction is not spontaneous (that is why it is electrolytic). Both anode and cathode are made of inert materials (Platinum, carbon) Only ions present are Na + and Cl - Cl 2 + 2e - 2Cl - Na + + e - Na Anode: 2Cl- Cl 2 + 2e - E o = Cathode: Na + + e - Na E o = Overall reaction: 2Na + + 2Cl - 2Na + Cl 2 E cell = Reduction half-cell is below the oxidation half-cell. This is why the reaction is NOT spontaneous. To operate the above cell, you must put in 4.07 volts (at least) Some half-cells are not at standard state and thus may require more voltage. Na and Cl 2(g) react violently. This is averted as the molten sodium sinks to the bottom of the cell and the gas is able to escape. Simmer Down Inc. August

33 Electrolysis of aqueous NaI Again, inert electrodes are used. **Have to consider water (aqueous)** Species in solution are Na +, I - and H 2 O. Solution is neutral because no acid is shown / was used. During the electrolysis of aqueous solutions, you must consider the possibility that H 2 O may oxidize or reduce. Two possible reductions Na + e - Na (s) 2H 2 O + 2e - H 2 g) + 2OH - Two possible oxidations H 2 O ½ O 2 + 2H + (10-7 M) + 2e - 2I - I e - Which one will happen? Look at the diagram below Preferred reaction will be the one that requires the least voltage input. Remember this highest reduction and the lower of the oxidations Two that are the closest together. Half-reactions having the greatest tendency to reduce and greatest tendency to oxidize are preferred. Concentration of the cell will NOT be a concern. As along as there is sufficient material in the cell, you can assume it will proceed as predicted. Assuming a neutral aqueous solution is present, there are two water equations that must be considered. Simmer Down Inc. August

34 Oxidation: H 2 O ½ O 2(g) + 2H + (10-7 ) + 2e - ; E o = V Reduction: 2H 2 O + 2e - H 2(g) + 2OH - (10-7 ) ; E o = V In acidic solution there are also two water equations which must be considered Oxidation: H 2 O ½ O 2(g) + 2H + + 2e - ; E o = V Reduction: 2H + + 2e - H 2(g) ; E o = 0.00 V You do not need to worry about basic conditions Overpotential effect When dilute neutral aqueous solutions containing Cl - or Br - are electrolysed, we expect that water will oxide before either Cl - or Br - since water is lower on the table. 2Cl - Cl 2 + 2e - ; E o = V H 2 O ½ O 2(g) + 2H + (10-7 ) + 2e - ; E o = V O 2 is NOT produced, Cl 2 or Br 2 is. You don t need to know why, just remember this exception to the rule. Electrolysis of aqueous solutions containing Cl - and Br - will produce Cl 2 and Br 2 at the anode. Examples: 1. What are the products formed at the anode and cathode and what is the overall reaction when a solution containing NiSO 4(aq) is electrolysed using inert electrodes? 2. What is the overall reaction which occurs when a 1 M solution of HCl (aq) is electrolyzed using carbon electrodes? Simmer Down Inc. August

35 3. August April June 2000 Simmer Down Inc. August

36 Electroplating Electroplating is an electrolytic process in which a metal is reduced or plated out at the CATHODE. Cathode is made out of the material that will receive the metal plating. Electroplating solution contains ions of the metal, which is to be plated onto the cathode. Anode: usually made out of the same metal, which is to be plated out onto the cathode. ***We will only be concerned with neutral solutions*** Example: Design a cell to electroplate a copper medallion with nickel metal. Include in the design: the ions present in solution, the direction of ion flow, the substances used for the anode and cathode and the direction of electron flow when the cell is connected to a DC power source. Recap. Cathode Ions in solution Substances used at the anode and cathode Anode Direction of ion flow Direction of electron flow Simmer Down Inc. August

37 Examples: 1. August 2005 Electrorefining Process of purifying a metal by electrolysis Example: Purification of Pig copper. Simmer Down Inc. August

38 Anode: Apart from the small amounts of Pb and Zn, Cu has the greatest tendency to oxidize. When exposed Pb/Zn atoms have oxidized and gone into solution as ions, only Cu atoms are available to be oxidized. Au / Ag / Pt can t be oxidized because Cu preferentially is oxidized. Particles of Au / Ag / Pt drop off and form an anode sludge which is very valuable. Why? Cathode: Cu +2 in solution preferentially reduced at the cathode. Pb +2 and Zn +2 CANNOT be reduced as copper in larger amounts and has a higher reduction potential. Overall: No metals above Cu can oxidize and go into solution and no metal below Cu can be reduced and go back out of solution. Therefore, only copper is involved in both the oxidation and reduction Anode: Cu (impure) Cu e - ; E o = V Cathode: Cu e - Cu (pure) ; E o = 0.34 V Overall: Cu (impure) + Cu 2+ Cu (pure) + Cu 2+ ; E o = 0.00 V Although it says 0.00V, several volts must be supplied to get the reaction started. Examples: 1. April 2004 Simmer Down Inc. August

39 2. August April 2003 Simmer Down Inc. August