e (reduction)

Transcription

1 OXIDATION-REDUCTION REACTIONS Some of the most important reaction in chemistry are oxidation-reduction (redox) reactions. In these reactions, electrons transfer from one reactant to the other. The rusting of iron, bleaching of hair and the production of electricity in batteries involve redox reactions. Synthesis, decomposition, single replacement and combustion reactions are all examples of redox reactions. In these reactions, one substance loses electrons (oxidized) while another substance gain electrons (reduced). Therefore, oxidation is defined as loss of electrons, while reduction is defined as gain of electrons. For example in reaction of magnesium metal and sulfur to form magnesium sulfide: 2e (oxidation) 0 +2 Mg + S MgS e (reduction) Mg : - is oxidized (loses electrons) S: - is reduced (gains electrons) - causes the reduction of S - causes the oxidation of Mg - called Reducing Agent - called Oxidizing Agent 2e Mg 0 + S 0 Mg 2+ S 2 Oxidation-Reduction reactions can be represented by half-reactions: Oxidation Half-Reaction Reduction Half-Reaction Mg Mg e S + 2e S 2 1

2 OXIDATION NUMBERS OR STATES Identifying the oxidation-reduction nature of reactions between metals and non-metals is straight forward because of ion formation. However, redox reactions that occur between two non-metals is more difficult to characterize since no ions are formed. Chemists have devised a scheme to track electrons before and after a reaction in order to simply this process. In this scheme, a number (oxidation state or number) is assigned to each element assuming that the shared electrons between two atoms belong to the one with the most attraction for these electrons. The oxidation number of an atom can be thought of as the charge the atom would have if all shared electrons were assigned to it. The following rules are used to assign oxidation numbers to atoms in elements and compounds. (Note: these rules are hierarchical. If any two rules conflict, follow the rule that is higher on the list) 1. The oxidation number of an atom in a free element is The oxidation number of a monatomic ion is equal to its charge. 3. The sum of the oxidation number of all atoms in: A neutral molecule or formula is equal to 0. An ion is equal to the charge of the ion. 4. In their compounds, metals have a positive oxidation number. Group 1A metals always have an oxidation number of +1. Group 2A metals always have an oxidation number of In their compounds non-metals are assigned oxidation numbers according to the table at right. Entries at the top of the table take precedence over entries at the bottom of the table. 2

3 OXIDIDATION-REDUCTION REACTIONS Oxidation numbers (or states) can be used to identify redox reactions and determine which substance is oxidized and which is reduced. To do so, assign oxidation numbers to all elements in the reactants and products, then track which elements change oxidation numbers from reactants to products. Elements that increase their oxidation numbers lose electrons, and are therefore oxidized. Elements that decrease their oxidation numbers gain electrons, and are therefore reduced. OXIDATION (LOSS OF ELECTRONS) REDUCTION (GAIN OF ELECTRONS) Examples: 1. Using oxidation numbers, identify the element is oxidized and the element that is reduced and the number of electrons transferred, in each of the following redox reactions: a) Sn (s) + 4 HNO 3 (aq) SnO 2 (s) + 4 NO 2 (g) + 2 H 2 O (g) b) Hg(NO 3 ) 2 (aq) + 2 KBr (aq) Hg 2 Br 2 (s) + 2 KNO 3 (aq) 3

4 HALF REACTIONS Redox Reactions are discussed (sometimes balanced) by writing two Half-Reactions: OXIDATION HALF-REACTION involves loss of electrons increase in Oxidation Number REDUCTION HALF-REACTION involves gain of electrons decrease in Oxidation Number Cu(s) Cu 2+ (aq) + 2 e 2Ag + (aq) + 2e 2Ag 0 (s) NOTE: Number of electrons lost in Oxidation reaction = Number of electrons gained in Reduction reaction CONCLUSION: Oxidation Reduction (Redox) Reactions are reactions in which the Oxidation Numbers of at least two elements change involve transfer of electrons: from : to: the element that is oxidized (called Reducing Agent) the element that is reduced (called Oxidizing Agent) Examples: 1. Determine the oxidized and reduced species and write half-reactions for the reaction of zinc and sulfur to form zinc sulfide: Zn (s) + S (s) ZnS (s) 4

5 Examples: 2. Determine the oxidized and reduced species and write half-reactions for the reaction of aluminum and iodine to form zinc aluminum iodine: Al (s) + I 2 (s) AlI 3 (s) 3. Determine the oxidized and reduced species and write half-reactions for the reaction shown below: Mg (s) + HCl (aq) MgCl 2 (aq) + H 2 (g) 5

6 BALANCING REDOX REACTIONS IN ACIDIC & BASIC SOLUTIONS When balancing redox reactions, the following must be observed: Atoms must balance Nr. of electrons lost = Nr. of electrons gained Net charges of ions on both sides of the equation must balance 1. Acidic Solution (H + ions and H 2 O molecules may be added as needed) Balance the following redox reaction that takes place in acidic solution: Zn(s) + NO 3 (aq) Zn 2+ (aq) + NH 4 + (aq) (acidic solution) Step 1: Assign oxidation numbers and determine the oxidized and reduced species and number of electrons involved Zn + NO 3 Zn 2+ + NH 4 + Step 2: Set up half-reactions. Oxidation loses electrons (write as product) Reduction gains electrons (write as reactant) Oxidation half-reaction Reduction half-reaction Zn Zn e NO e + NH 4 Step 3: Balance atoms; H 2 O molecules and H + ions may be added as needed Oxidation half-reaction Zn Zn e (balanced; no action needed) Reduction half-reaction NO e + NH 4 Balance O atoms by adding H 2 O molecules: NO e + NH H 2 O Balance H atoms by adding H + ions: NO H e NH H 2 O Check to see that atoms and charges are balanced for each half-reaction, 6

7 Step 4: Equalize loss and gain of electrons. Oxidation half-reaction Reduction half-reaction [ Zn Zn e ] x 4 [ NO H e + NH H 2 O ] x 1 4 Zn 4 Zn e NO H e + NH H 2 O Step 5: Add up half-reactions and cancel electrons. 4 Zn + NO H + + 8e 4 Zn e + NH H 2 O Check to see that atoms and charges are balanced for the overall reaction Examples: Balance the following redox reaction in acidic solution: 1. Br 2 (l) + SO 2 (g) Br (aq) + SO 4 2 (aq) 2. H 2 O 2 (aq) + MnO 4 (aq) Mn 2+ (aq) + O 2 (g) 7

8 2. Basic Solution (OH ions and H 2 O molecules may be added as needed) Basic Idea: O and H will be balanced as though the solution were acidic In a later step, the acidic solution will be converted to a basic solution Balance the following redox reaction that takes place in basic solution: MnO 4 (aq) + SO 3 2 (aq) MnO 2 (s) + SO 4 2 (aq) Step 1: Assign oxidation numbers and determine the oxidized and reduced species and number of electrons involved MnO 4 (aq) + SO 2 3 (aq) MnO 2 (s) + SO 2 4 (aq) Step 2: Set up half-reactions. Oxidation loses electrons (write as product) Reduction gains electrons (write as reactant) Oxidation half-reaction Reduction half-reaction SO 3 2 SO e MnO e MnO 2 Step 3: Balance atoms; H 2 O molecules and H + ions may be added as needed Oxidation half-reaction SO 3 2 SO e Balance O atoms by adding H 2 O molecules: SO H 2 O SO e Balance H atoms by adding H + ions: SO H 2 O SO e + 2 H + Reduction half-reaction MnO e MnO 2 Balance O atoms by adding H 2 O molecules: MnO e MnO H 2 O Balance H atoms by adding H + ions: MnO e + 4 H + MnO H 2 O Check to see that atoms and charges are balanced for each half-reaction, 8

9 Step 4: Equalize loss and gain of electrons. Oxidation half-reaction Reduction half-reaction [SO H 2 O SO e + 2 H + ] x [MnO e + 4 H + MnO H 2 O] x SO H 2 O 3 SO e + 6 H + 2 MnO e + 8 H + 2 MnO H 2 O Step 5: Add up half-reactions and cancel electrons. 3 SO H 2 O + 2 MnO H + 3 SO H MnO H 2 O Note that the H + ions and H 2 O molecules may be canceled and the equation becomes: 3 SO MnO H + 3 SO MnO 2 + H 2 O Step 6: Add OH to cancel H + ion by forming H 2 O molecules 3 SO MnO H OH 3 SO MnO 2 + H 2 O + 2 OH 2 H 2 O 3 SO MnO H 2 O 3 SO MnO 2 + H 2 O + 2 OH Cancel one more time the H 2 O molecules: 3 SO MnO 4 + H 2 O 3 SO MnO OH Check to see that atoms and charges are balanced for each half-reaction, NOTE: For an alternate method of balancing basic solutions see Redox Review on my website 9

10 Examples: Balance the following redox reactions in basic solution: 1. Cr(OH) 4 (aq) + H 2 O 2 (aq) CrO H 2 O (l) 2. MnO 4 (aq) + Br (aq) MnO 2 (s) + BrO 3 (aq) 10

11 ELECTROCHEMICAL CELLS Electrochemical cells are systems consisting of electrodes that dip into an electrolyte and in which a chemical reaction either uses or generates an electric current There are of two types of electrochemical cells: 1. Voltaic Cells (galvanic cells, or battery cells) These are electrochemical cells in which a spontaneous reaction generates an electric current Example: An ordinary battery 2. Electrolytic Cells These are electrochemical cells in which an electric current drives a non-spontaneous reaction. Example: Electroplating of Metals Voltaic Cells Consist of two half-cells that are electrically connected Each half-cell is made from a metal strip that dips into a solution of its metal ion Zinc Zinc Ion Half-Cell (zinc electrode) Copper Copper Ion Half-Cell (copper electrode) Zinc strip Copper strip Zn 2+ Cu 2 11

12 VOLTAIC CELLS 12

13 VOLTAIC CELLS How does the voltaic cell function? Basic Idea: Zinc tends to lose electrons more readily than Copper (Zn is more active than Cu) Oxidation half-reaction Reduction half-reaction Zn (s) Zn 2+ (aq) + 2 e The electrons flow away from the zinc strip into the external circuit. The Zinc strip is used up. Cu 2+ (aq) + 2 e Cu (s) The electrons come from the external circuit and deposit onto the Cu strip. The Cu 2+ ions plate out as solid copper. Net Result: An electric current is generated through the external circuit. When sketching and labeling a voltaic cell, follow the following guidelines: The Electrodes: Anode: is the negative electrode is the more active metal Cathode: is the positive electrode is the less active metal Electron Flow: Away from the anode Towards the cathode Cell Reaction: Oxidation half-reaction Reduction half-reaction 13

14 SHORTHAND NOTATION FOR VOLTAIC CELLS Electrochemical cells are often represented by a compact notation as shown below: Zn (s) Zn 2+ (aq) Cu 2+ (aq) Cu (s) Anode terminal Phase boundary Anode electrolyte Cathode terminal Phase boundary Cathode electrolyte Salt bridge The Anode (Negative electrode) (the more active metal) The Oxidation half-cell The Cathode (Positive electrode) (the less active metal) The Reduction half-cell Examples: 1. Write the cell notation for the reaction shown below: 2 Al (s) + 3 Zn 2+ (a) 2 Al 3+ (aq) + 3 Zn (s) 2. Write the half-reactions and the overall reaction for the redox reaction that occurs in the voltaic cell shown below: Sc (s) Sc 3+ (aq) Ag + (aq) Ag (s) 14

15 HALF-REACTIONS INVOLVING GASES An inert material (ex: Platinum) serves as: electrical terminal and electrode surface A Hydrogen electrode (catalyzes the half-reaction, but otherwise is not involved in it) Acidic Solution Cathode Reaction Half Reaction: 2 H + (aq) + 2 e H 2 (g) Anode Reaction Notation for the hydrogen electrode: As an Anode As a Cathode Pt H 2 (g) H + (aq) H + (aq) H 2 (g) Pt Anode Reaction: Cathode Reaction: H 2 (g) 2 H + (aq) + 2 e 2 H + (aq) + 2e H 2 (g) 15

16 Example: 1. Consider the voltaic cell shown below. (a) Identify and label the anode and the cathode, (b) determine the direction of the electron flow and (c) write a balanced equation for the overall reaction. 2. Write the notation for a cell in which the electrode reactions are: 2 H + (aq) + 2 e H 2 (g) Zn (s) Zn 2+ (aq) + 2 e 3. Give the overall cell reaction for the voltaic cell with the notation shown below: Cd (s) Cd 2+ (aq) H + (aq) H 2 (g) Pt 16

17 ELECTROMOTIVE FORCE Through the external wire, the electrons are forced to flow: From the negative electrons To the positive electrode (Anode) (Cathode) (high electric potential) (low electric potential) The Electrical Work needed to move an electrical charge (the electron) through a conductor depends on: the total electrical charge moved, and the potential difference This difference in electrical potential (voltage) is measured in volts, V (SI unit of potential difference) It follows: Electrical Work = Electrical Charge x Potential Difference Joules = coulombs x volts The magnitude of the electrical charge is conveniently measured as the electrical charge carried by 1 mole of electrons: Electrical charge carried by one electron = x coulombs/e Number of electrons in 1 mole of electrons = x electrons Electrical Charge coulombs carried by = x x x electrons 1 mole of electrons electron Electrical Charge carried by 9.65 x 10 4 coulombs 96,500 coulmbs 1 mole of electrons 17

18 The Electrical Charge carried by 1 mole of electrons: is equal to 9.65 x 10 4 coulombs (96,500 coulmbs) is referred to as the Faraday constant, F (in honor of Michael Faraday, who laid the foundations of Electrochemistry) Let: w = the electrical work done by the voltaic cell in moving 1 mole of electrons from one electrode to another w = Electrical Work = (Electrical Charge) x (Potential Difference) w = ( F ) (Potential Difference) negative sign indicates that the voltaic cell loses energy, as it does work on the surroundings Maximum Electrical Work = Maximum Voltage = Maximum Potential Difference obtainable Electromotice Force Maximum Electrical Maximum emf = Work = Maximum Voltage = Potential Difference E cell obtainable between the electrodes measured with a voltmeter In the normal operation of a voltaic cell: Conclusion: The Actual Voltage The Electromotive Force Reason: it takes energy to drive a current through the cell itself Maximum Work Obtainable = w max = n F E cell 18 Cell emf Faraday constant = 9.65 x 10 5 C Number of electrons transferred in the overall cell equation

19 Examples: 1. What is the maximum electrical work that can be obtained from 6.54 g of zinc metal that reacts in a Daniell cell (Zn as an anode and Copper as a cathode), whose emf is 1.10 V. The cell reaction is : Zn (s) + Cu 2+ aq) Zn 2+ (aq) + Cu (s) W max = n F E cell To determine n, write out the half-reactions: Anode ( ) half-reaction Cathode (+) half-reaction Zn (s) Zn 2+ (aq) + 2 e Cu 2+ (aq) + 2 e Cu (s) Note that: n = 2 w max = n F E cell = (2) (9.65 x 10 4 C) (1.10 V) = 2.12 x 10 5 J/mol The maximum work for 6.54 g of Zn(s) is: 1 mol Zn 2.12 x 10 5 J 6.54 g Zn x x = 2.12 x 10 4 J g Zn 1 mol Zn 2. Calculate the free energy change (G 0 ) for the reaction shown below with standard potential of 0.92 V at 25 C: Al (s) + Cr 3+ (aq) Al 3+ (aq) + Cr (s) (Hint: recall that G = maximum work) 19

20 STANDARD CELL Emf s AND STANDARD ELECTRODE POTENTIALS Emf is a measure of the driving force of the cell half-reactions Example: The Daniell Cell Anode half-reaction (Oxidation) Cathode half-reaction (Reduction) Zn (s) Zn 2+ (aq) + 2e Cu 2+ (aq) +2e Cu (s) Each of the two ions (Zn 2+ and Cu 2+ ) has a certain tendency to acquire electrons from the its respective electrode and become reduced Zn 2+ (aq) + 2e Zn (s) Cu 2+ (aq) + 2e Cu (s) Smaller tendency to acquire electrons Lower potential to undergo reduction Is forced to undergo Oxidation Greater tendency to acquire electrons Greater potential to undergo reduction Is forced to undergo Reduction Reduction Potential = the tendency of the metallic ion to become reduced E cell = the difference in the tendencies of the two ions to become reduced E cell Reduction Potential of the substance that actually undergoes reduction Cu 2+ /Cu Reduction Potential of the substance that is forced to undergo oxidation Zn 2+ /An E cell = E red (cathode) E red (anode) For the Daniell cell: E cell = E Cu E Zn Reduction Potential Reduction Potential of substance being of substance being reduced oxidized 20

21 An Oxidation half-reaction is the reverse of the corresponding Reduction half-reaction: Zn 2+ (aq) + 2e Reduction Zn(s) It follows that: Oxidation Oxidation Potential for a half-reaction = Reduction Potential for the reverse half-reaction E cell = Reduction Potential of the substance that actually undergoes reduction Cu 2+ /Cu + Oxidation Potential of the substance that is forced to undergo oxidation Zn 2+ /An Examples: 1. A voltaic cell is based on the following two half-reactions, shown below. Determine the standard potential for this cell. Cd 2+ (aq) + 2 e Cd (s) E = V Sn 2+ (aq) + 2 e Sn (s) E = V 2. Using standard cell potentials in your text, calculate the emf for a cell that operates on the following overall reaction: 2 Al (s) + 3 I 2 (s) 2 Al 3+ (aq) + 6 I (aq) 21

22 STANDARD ELECTRODE POTENTIALS E cell = standard emf = emf of a voltaic cell operating under standard-state conditions: solute concentrations are each 1M gas pressures are each 1 atm temperature is 25 C To find E cell a table of standard electrode potentials is needed. Experimentally it is not possible to measure the potential of a single electrode. However, it is possible to measure emf s of cell constructed from various electrodes connected to a chosen reference electrode The reference electrode chosen is arbitrarily assigned a Potential of Zero, under standard conditions This electrode is the standard Hydrogen electrode 1 atm pressure 2 H + (aq) + 2 e H 2 (g) Standard Potential = E = 0.00 V 25 C 1 M acidic solution Any substance that is more easily reduced than H + has a (+) value of E 0 : Cu 2+ (aq) + 2e Cu (s) E = V Any substance that is more difficult to reduce than H + has a (-) value of E 0 Zn 2+ (aq) + 2e Zn (s) E = 0.76 V To find a standard electrode potential, the electrode is connected to a standard Hydrogen electrode and the emf is measured with a voltmeter: 22

23 23

24 STRENGTHS OF OXIDIZING AND REDUCING AGENTS reduction Oxidized Species + n e Reduced Species oxidation The higher the The higher the The stronger The larger (more +) tendency of the tendency of the Oxidizing Agent the electrode reduction species to gain species to be the species is potential (E) of the electrons reduced species Example: F 2 (g) + 2 e Strong Oxidizing Agent 2 F (aq) E = V The higher the The higher the The stronger The smaller (more -) tendency of the tendency of the Reducing Agent the electrode reduction species to lose species to be the species is potential (E) of the electrons oxidized species Example: Li + (aq) + e Li (s) Strong Reducing Agent E = 3.04 V 24

25 PREDICTING SPONTANEITY FOR OXIDATION-REDUCTION REACTIONS spontaneous Oxidized Species + n e Reduced Species nonspontaneous Stronger Oxidizing Weaker Oxidizing Agent (ex: F 2 ) Agent (ex: 2 F ) nonspontaneous Oxidized Species + n e Reduced Species spontaneous Weaker Reducing Agent (ex: Li + ) Stronger Reducing Agent (ex: Li) 25

26 Chemistry 102 Chapter 18 Examples: 1. Does the following reaction occur spontaneously in the direction indicated, under standard conditions? Cu 2+ (aq) + 2 I (aq) Cu(s) + I 2 (s) Write the half-reactions and look up the E values (Standard Reduction Potential) I 2 (s) + 2e 2I (aq) E = 0.54 V Cu 2+ (aq) + 2 e Cu (s) E = 0.34 V E(I 2 ) E(Cu 2+ ) I 2 has a higher tendency to be reduced than Cu 2+ I 2 is a stronger oxidizing agent than Cu 2+ nonspontaneous Cu 2+ (aq) + 2 I (aq) Cu(s) + I 2 (s) weaker spontaneous stronger oxidizing oxidizing agent agent 2. Which is the stronger oxidizing agent, NO 3 (aq) in acidic solution (to NO) or Ag + (aq)? First write the two appropriate reduction half-reactions: NO 3 (aq) + 4H + (aq) + 3 e NO(g) + 2 H 2 O(l) E = V Ag + (aq) + e Ag(s) E = V Compare the two reduction potentials: E(NO 3 ) E(Ag + ) NO 3 has a higher tendency to be reduced than Ag + NO 3 is a stronger oxidizing agent than Ag 26

27 Chemistry 102 EQUILBRIUM CONSTANTS AND EMF Chapter 18 The free energy change, G, for a reaction equals the maximum useful work of the reaction. G = w max For a voltaic cell, this work is the electrical work, nfe cell. G = nfe cell Also recall that, G = RTlnK Combining the last two equations give another method for obtaining the equilibrium constant for a reaction: nfe cell = RTlnK or RT RT E cell = ln K = log K nf nf Substituting values for constants R and F at 25C gives the equation shown below: E cell = log K (at 25 C) n Recall that free energy change, G, and standard free energy change, G, were related as shown below: Therefore, G = G + RT ln Q nfe cell = nfe cell + RT ln Q Rearrangement leads to the Nernst equation shown below: RT 2.303RT E cell = E cell - ln Q or E cell = E cell - log Q nf nf Substituting values for constants R and F at 25C gives the equation shown below: E cell = E cell - log Q (at 25 C) n 27

28 Chemistry 102 Chapter 18 Examples: 1. The standard emf for a Zn/Cu voltaic cell is 1.10 V. Calculate the equilibrium constant for the reaction shown below: Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) E = log K n = log K 2 log K = 37.2 Taking antilog of both sides K = 10 = 1.6x Calculate the equilibrium constant for the following reaction from the standard electrode potentials: Fe (s) + Sn 4+ (aq) Fe 2+ (aq) + Sn 2+ (aq) 3. What is the emf for the Zn/Cu voltaic cell in example 1 when [Zn 2+ ] = 1.00x10 5 M and [Cu 2+ ] = M 2+ [Zn ] 2+ [Cu ] Q = = E cell = E cell - log Q = n 4. What is the emf for the following voltaic cell? Zn Zn 2+ (0.200 M) Ag + ( M) Ag (s) 28