Ch 18 Electrochemistry OIL-RIG Reactions Alessandro Volta’s Invention Modified by Dr. Cheng-Yu Lai Daily Electrochemistry Appliactions Electrochemistry: The area of chemistry that examines the transformations between chemical and electrical energy. Electrochemistry Terminology #1 Oxidation – A process in which an element attains a more positive oxidation state Na(s) Na+ + eReduction – A process in which an element attains a more negative oxidation state Cl2 + 2e- 2ClIn the following reaction, what is being oxidized, reduced? What is the oxidizing agent, the reducing agent? 2 2 FeS2 (s, pyrite) 7 O2 (g) 2H2O(l) 2 Fe 2 (aq) 4 SO 4 (aq) 4 H (aq) Electrochemistry Terminology #3 Oxidizing agent The substance that is reduced is the oxidizing agent Reducing agent The substance that is oxidized is the reducing agent You need to review the following terminology: oxidation, reduction, oxidizing agent, reducing agent and half-reaction (see CHEM 101). Electrochemistry Terminology #4 Anode The electrode where oxidation occurs Cathode The electrode where reduction occurs Memory device: Reduction at the Cathode Redox Examples 1. In the reaction 2Ca(s) + O2(g) ------> 2CaO(s), calcium is __________ a. Reduced b. Electrolyzed c. synthesized d. oxidized e. none of these 2. In the reaction SiO2(s) + 2C(s) -------> Si(s) + 2CO(g), which species is the oxidizing agent? a. Si b. C c. O d. SiO2 e. CO 3. In the reaction Zn + H2SO4 --------> ZnSO4 + H2, which element, if any, is oxidized? a. zinc b. hydrogen c. sulfur d. oxygen e. none is oxidized Example 18.1 Half-Reaction Method of Balancing Aqueous Redox Equations in Acidic Solution Balance the redox equation: General Procedure Step 1 Assign oxidation states to all atoms and identify the substances being oxidized and reduced. Step 2 Separate the overall reaction into two half-reactions: one for oxidation and one for reduction. Step 3 Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro © 2014 Pearson Education, Inc. Redox Reactions and Current • Redox reactions involve the transfer of electrons from one substance to another. • Therefore, redox reactions have the potential to generate an electric current. • To use that current, we need to separate the place where oxidation is occurring from the place where reduction is occurring. © 2014 Pearson Education, Inc. Example - Zn (s) + Cu2+ (aq) Voltaic Cells • Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s) – When run directly in a test tube • Cu metal plates out on surface of Zn metal • Zn metal enters solution as Zn2+ ions • Blue color of Cu2+ solution fades In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released. In order to harvest e /E, Voltaic Cells (aka Galvanic Cell) Redox reactions have the potential to generate an electric current. To use that current, we need to separate the place where oxidation is occurring from the place where reduction is occurring. A typical voltaic cell looks like this. The oxidation occurs at the anode. The reduction occurs at the cathode. • We can use that energy to do work if we make the electrons flow through an external device. Voltaic or Galvanic cell – chemical energy is used to produce electrical energy (DG<0) (i.e. a battery). Electrolytic cell – an external source of electrical energy is used to do work on a chemical system (i.e. charging a car battery with the alternator after the car has started). Cells Components : An electrochemical cell is a reaction system in which oxidation and reduction reactions occur in separate compartments (or 2 half cells) either consume or produce electrical energy. The cells are separated by a salt bridge or semi-permeable membrane that allows ions to migrate from one cell to the other. Electrons move from anode (oxd) to cathode (red). Voltaic Cells (+) charge (-) charge Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop. As the half reactions occur A surplus of positive ions builds up at the anode , (-) charge A surplus of negative ions builds up at the cathode, (+) charge Anions and cations must flow to balance charge Current is the number of electrons that flow through the system per second. Unit = ampere Salt Bridge • Therefore, we use a salt bridge, usually a Ushaped tube that contains a salt solution, to keep the charges balanced. Na +Cations move toward the cathode. The sign of this electrode is (+) NO3-Anions move toward the anode. The sign of this electrode is (-) Salt Bridges allows current to flow • The salt bridge is a gel-filled U-tube with a solution of a salt containing ions other than those involved in the redox reaction – KNO3 / NaNO3 is frequently used – Na+ Cations flow toward the cathode to neutralize the build-up of negative charge – NO3- Anions flow toward the anode to neutralize the build-up of positive charge Shorthand Cell Notation • Oxidation on the left • Reduction on the right • Single vertical line represents a phase boundary – Liquid-metal or liquid-gas, etc. The single line represents a phase boundary (electrode to electrolyte) • Double line is the salt bridge and the double line represents a Zn Zn 2 2 Cu Cu physical boundary (porous boundary) electrons flow The difference in potential energy between the reactants and products is the potential difference. Unit = volt The amount of force pushing the electrons through the wire is called the electromotive force, emf. Consider the reduction potential chart. Find the reduction equations for Ag+ Ag and Pb2+ Pb. 1. Which metal ion has the greater reduction potential? ____ see table 2. If these two metals (and their solutions) were used to create a galvanic cell, which metal would be the anode? ____ 3. Write the reaction at the anode: ___________________________ 4. Write the reaction at the cathode: ____________________________ 5. What is the overall reaction? ____________________________________ 6. Write the cell notation for the cell: ______|________||________|_______ 7. How many moles of electrons are involved in this reaction? Measuring Standard Electrode Potential Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts. Table of Reduction Potentials Measured against the Standard Hydrogen Electrode Standard Potentials • Once the hydrogen half cell has been assigned a voltage of 0.000 V, other half cells can be measured relative to it • Tables of standard potentials can be prepared – These are always reduction potentials, i.e., Ered – To obtain the oxidation potential, simply reverse the sign: Ered – Zn2+ (aq) + 2e- Zn (s) = -0.762V Eox – Zn (s) Zn2+ (aq) + 2e= +0.762V 25 °C, 1 atm for gases, 1 M concentration of solution • Standard voltages for oxidation and reduction are equal in magnitude and opposite in sign Oxidizing and Reducing Agents • The strongest oxidizers have the most positive reduction potentials. • The strongest reducers have the most negative reduction potentials. Predicting Spontaneity of Redox Reactions • A spontaneous reaction will take place when a reduction half-reaction is paired with an oxidation half-reaction lower on the table. • If paired the other way, the reverse reaction is spontaneous. Cu2+(aq) + 2 e− Cu(s) Zn2+(aq) + 2 e− Zn(s) Ered = +0.34 V Ered = −0.76 V Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Cu(s) + Zn2+(aq) Cu2+(aq) + Zn(s) © 2014 Pearson Education, Inc. spontaneous nonspontaneous Example 18.5 Predicting Spontaneous Redox Reactions and Sketching Electrochemical Cells ° Solution a. b. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro + 0.45 V © 2014 Pearson Education, Inc. A standard potential (Eo) is the electromotive force of a half-reaction written as a reduction reaction in which all reactants and products are in their standard states (see The standard cell potential (Eºcell) is the potential of a cell when all reactants and products are in their standard states, i.e. the pressure of all gases are 1 atm, and the concentration of dissolved species are 1 molar. Eºcell = Eºcathode – Eºanode or Eºcell = Eºreduction – Eºoxidation Because cell potential is based on the potential energy per unit of charge, it is an intensive property. ( not numbers ) Zn - Cu Galvanic Cell The less positive, or more negative reduction potential becomes the oxidation… Zn Zn2+ + 2eCu2+ + 2e- Cu E = +0.76V E = +0.34V Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V Calculating Cell Potentials under Standard Conditions • E°cell = E°oxidation + E°reduction • When adding E° values for the half-cells, do not multiply the half-cell E° values, even if you need to multiply the half-reactions to balance the equation. Oxidizing and Reducing Agents The greater the difference between the two, the greater the voltage of the cell. © 2014 Pearson Education, Inc. Ag+ (aq) + e Why 2Ag+ (aq) +2 e Why 3 Ag+ (aq) + 3e Ag Ered =+ 0.799 V Ag Ered =+ 0.799 V 3 Ag Ered =+ 0.799 V E=0v E =-1.201v Free Energy DG for a redox reaction can be found by using the equation DG = −nFE where n is the number of moles of electrons transferred, and F is a constant, the Faraday. 1 F = 96,485 C/mol = 96,485 J/V-mol Under standard conditions, DG = −nFE E°cell, DG°, and K • For a spontaneous reaction – one that proceeds in the forward direction with the chemicals in their standard states DG° < 1 (negative) – E° > 1 (positive) – K>1 • DG° = −RTlnK = −nFE°cell – n = the number of electrons – F = Faraday’s constant = 96,485 C/mol e− © 2014 Pearson Education, Inc. Calculating 0 DG for a Cell DG0 = -nFE0 n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e- Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V coulombs Joules DG (2 mol e )(96 485 )(1.10 ) mol e Coulomb 0 DG 212267 Joules 212 kJ 0 ° Example 18.6 Relating ΔG °and Ecell Use the tabulated electrode potentials to calculate ΔG° for the reaction. Is the reaction spontaneous? Sort You are given a redox reaction and asked to find ΔG°. Given: Find: ΔG° Strategize Refer to the values of electrode potentials in Table 18.1 to calculate °. ° . Then use Equation 18.3 to calculate G from Ecell Ecell Conceptual Plan Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro © 2014 Pearson Education, Inc. ° Example 18.6 Relating ΔG °and Ecell Continued Solve Separate the reaction into oxidation and reduction half-reactions and find the standard electrode potentials for ° by subtracting Ean from Ecat. each. Determine Ecell Solution - Calculate ΔG° from E°cell. The value of n (the number of moles of electrons) corresponds to the number of electrons that are canceled in the half-reactions. Remember that 1 V = 1 J/C. Since ΔG° is positive, the reaction is not spontaneous under standard conditions. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro © 2014 Pearson Education, Inc. Nernst Equation • Remember that Ch 17 DG = DG + RT ln Q • This means −nFE = −nFE + RT ln Q Nernst Equation • Remember that DG = DG + RT ln Q −nFE = −nFE + RT ln Q We can combine R (8.31 J/mol-K), T (25 ºC = 298 K), and F (96480 J/molV) to give 0.0257V Dividing both sides by −nF, we get the Nernst equation: E = E − E = E − RT ln Q nF 0.0257V ln Q n If concentration can gives voltage, then from voltage we can tell concentration Nernst Equation Nernst Equation Application- Concentration Cells • Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. would be 0, but Q would not. • For such a cell, Ecell • Therefore, as long as the concentrations are different, E will not be 0. Cell Potential When Ion Concentrations Are Not 1 M • We know there is a relationship between the reaction quotient, Q; the equilibrium constant, K; and the free energy change, DGº. • Changing the concentrations of the reactants and products so they are not 1 M will affect the standard free energy change, DGº. • Because DGº determines the cell potential, Ecell, the voltage for the cell will be different when the ion concentrations are not 1 M. © 2014 Pearson Education, Inc. Nernst Concentration Cells Calculations Zn2+ (1.0M) Zn2+ (0.10M) 0.0591 EE log(Q) n 0 E 0.0 Volts 0 n2 (0.10) Q (1.0) 0.0591 0.10 E 0.0 log( ) 0.030 Volts 2 1.0 Electrolytic Processes Electrolytic processes are NOT spontaneous. They have: A negative cell potential, (-E0) A positive free energy change, (+DG) Electrolysis • In electrolysis we use electrical energy to overcome the energy barrier of a nonspontaneous reaction, allowing it to occur. • The reaction that takes place is the opposite of the spontaneous process. 2 H2(g) + O2(g) 2 H2O(l) spontaneous 2 H2O(l) 2 H2(g) + O2(g) electrolysis • Some applications are (1) metal extraction from minerals and purification, (2) production of H2 for fuel cells, and (3) metal plating. © 2014 Pearson Education, Inc. Electrolysis • Running a galvanic cell backwards. • Put a voltage bigger than the potential and reverse the direction of the redox reaction. • Used for electroplating. Thin coatings of metals can be applied using electrolytic reactions. Electroplating of Silver Anode reaction: Ag Ag+ + eCathode reaction: Ag+ + e- Ag Electroplating requirements: 1. Solution of the plating metal 2. Anode made of the plating metal 3. Cathode with the object to be plated 4. Source of current Electrical Units • Charge – 1 mol electrons = 96,480 coulombs (of charge) • Current – 1 ampere = 1 coulomb/sec • Electrical energy – 1 joule = 1 C·V – 1 kWh = 3.600 X 106 J = 3.600 X 103 kJ Calculating plating • • • • • • Have to count charge. Measure current I (in amperes) 1 amp = 1 coulomb of charge per second q=Ixt q/nF = moles of metal Mass of plated metal Solving an Electroplating Problem Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO3 using a 20.0 Ampere current? Ag+ + e- Ag 5.0 g 1 mol Ag 1 mol e- 96 485 C 107.87 g 1 mol e- 1 mol Ag 1s 20.0 C = 2.2 x 102 s Electrolysis 1.0 M aqueous solutions of AgNO3, Cu(NO3)2 and Au(NO3) 3 are electrolyzed in the apparatus shown, so the same amount of electricity passes through each solution. If 0.10 moles of solid Cu are formed how many moles of Ag and Au are formed? Example 18.10 Stoichiometry of Electrolysis Gold can be plated out of a solution containing Au 3+ according to the half-reaction: What mass of gold (in grams) is plated by a 25-minute flow of 5.5 A current? Sort You are given the half-reaction for the plating of gold, which shows the stoichiometric relationship between moles of electrons and moles of gold. You are also given the current and duration. You must find the mass of gold that will be deposited in that time. Given: 3 mol e– : 1 mol Au 5.5 amps 25 min Find: g Au Strategize You need to find the amount of gold, which is related stoichiometrically to the number of electrons that have flowed through the cell. Begin with time in minutes and convert to seconds. Then, since current is a measure of charge per unit time, use the given current and the time to find the number of coulombs. You can use Faraday’s constant to calculate the number of moles of electrons and the stoichiometry of the reaction to find the number of moles of gold. Finally, use the molar mass of gold to convert to mass of gold. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro © 2014 Pearson Education, Inc. Example 18.10 Stoichiometry of Electrolysis Continued Conceptual Plan Solve Follow the conceptual plan to solve the problem, canceling units to arrive at the mass of gold. Solution Check The answer has the correct units (g Au). The magnitude of the answer is reasonable if we consider that 10 amps of current for 1 hour is the equivalent of about 1/3 mol of electrons (check for yourself), which would produce 1/9 mol (or about 20 g) of gold. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro © 2014 Pearson Education, Inc. Other uses • Electroysis of water. • Seperating mixtures of ions. • More positive reduction potential means the reaction proceeds forward. • We want the reverse. • Most negative reduction potential is easiest to plate out of solution. Example 18.8 Calculating Ecell under Nonstandard Conditions Determine the cell potential for an electrochemical cell based on the following two half-reactions: Sort You are given the half-reactions for a redox reaction and the concentrations of the aqueous reactants and products. You are asked to find the cell potential. Given: Find: Ecell Strategize ° . Then use Equation 18.9 to calculate Ecell. Use the tabulated values of electrode potentials to calculate Ecell Conceptual Plan Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro © 2014 Pearson Education, Inc. Example 18.8 Calculating Ecell under Nonstandard Conditions Continued Solve Write the oxidation and reduction half-reactions, multiplying by the appropriate coefficients to cancel the °. electrons. Find the standard electrode potentials for each. Find Ecell Solution ° . The value of n (the number of moles of electrons) corresponds to the number of electrons Calculate Ecell from Ecell (6 in this case) canceled in the half-reactions. Determine Q based on the overall balanced equation and the given concentrations of the reactants and products. (Note that pure liquid water, solid MnO 2, and solid copper are omitted from the expression for Q.) Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro © 2014 Pearson Education, Inc. Example 18.8 Calculating Ecell under Nonstandard Conditions Continued Check ° , as expected based on Le Châtelier’s The answer has the correct units (V). The value of Ecell is larger than Ecell principle because one of the aqueous reactants has a concentration greater than standard conditions and the one aqueous product has a concentration less than standard conditions. Therefore, the reaction has a greater tendency to proceed toward products and has a greater cell potential. For Practice 18.8 Determine the cell potential of an electrochemical cell based on the following two half-reactions: Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro © 2014 Pearson Education, Inc. Electrolysis Electrolysis Electrolysis Electrolysis

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