Things to Know for Exam #4 Thermodynamics (approximately 40%-45% of exam) • Calculate the enthalpy change for a reaction given the standard ΔH of formation values. • Calculate the standard H of formation value for a substance in a reaction given the standard ΔH for the reaction and the ΔH of formation values of the other substances in the reaction. • Calculate the enthalpy change for a reaction using hessʼs law • Predict the sign on ΔSO for a reaction • Calculate ΔSO given standard entropy values • Calculate ΔGO given standard ΔG of formation values • Calculate ΔGO given information on ΔHO and ΔSO • Know how to tell when a reaction is spontaneous • Know how to calculate the boiling point for a substance using thermodynamic information • Calculate the equilibrium constant given thermodynamic quantities, or calculate ΔGO from an equilibrium constant. • First, second and third law of thermodynamics Electrochemistry (approximately 40%-45% of exam) • • • • • • • • • • • Given time and current determine mass of metal plated Given mass of metal plated out and the current determine the time required for the process Definitions – oxidizing agent, reducing agent, reduction, oxidation, salt bridge, anode, cathode, voltaic/galvanic cell, electrolytic cell Determine if a redox reaction is spontaneous as written Balance a redox reaction Calculate the standard cell potential given standard reduction potentials Convert short hand cell notation for a cell into a balanced redox equation Parts of a galvanic cell and direction of electron flow Calculate G from electrochemical information Use the Nernst equation to calculate the cell potential at nonstandard conditions Find the equilibrium constant from electrochemical information Nuclear Chemistry (approximately 5%-10% of exam) • • Types of Radioactive Decay Balancing nuclear reaction equations Coordination Chemistry (approximately 5%-10% of exam) • • • What are ligands? Recognizing different types of isomers Recognize the general structure of a porphyrin Electrochemistry Stuff Chapter 20 Part 1 1 Lightning and Batteries • The driving force for both lightning and the battery is the same. • Electrons flow away from negative charge and toward positive charge. • Most lightning occurs within the thundercloud itself or from one thundercloud to another. However, if the thundercloud gets close enough to the ground, the earth underneath the cloud develops a positive charge in response to the negative charge at the base of the cloud. 2 1 Lightning and Batteries • Cloud-to-ground lightning is visible and dramatic to observers on the ground. • A battery is composed of substances that are separated so that one end of the battery develops a positive charge and the other end develops a negative charge. • The charge separation exists until a conductive path connects the two ends, providing a path through which charge can flow. • As the electrons flow through the electrical load, they can do electrical work. 3 Electricity From Chemistry • The moving electrons in batteries constitute an electrical current. • Some batteries can recharge from solar cells, making it completely possible to live off the traditional electrical grid. • Today electricity is a fundamental form of energy, powering our entire economy. 4 2 Oxidation–Reduction • Reactions where electrons are transferred from one atom to another are called oxidation–reduction reactions. ü Redox reactions for short • Atoms that lose electrons are being oxidized; atoms that gain electrons are being reduced. • Increase in oxidation state is oxidation; decrease in oxidation state is reduction. 2 Na ( s ) + Cl2 ( g ) ® 2 Na +Cl– ( s ) Na ® Na + + 1 e – oxidation, Na is going from 0 to +1 oxidation state Cl2 + 2 e - ® 2Cl- reduction, Cl is going from 0 to –1 oxidation state 5 Half-Reactions • We generally split the redox reaction into two separate half-reactions—reactions involving just oxidation or reduction, as on the previous slide. ü The oxidation half-reaction has electrons as products. ü The reduction half-reaction has electrons as reactants. - 3 Cl2 + I + 3 H2O ® 6 Cl- + IO3 - + 6 H+ 0 -1 +1 -2 -1 +5 -2 +1 Oxidation: I- ® IO3 - + 6 e Reduction: Cl2 + 2 e - ® 2 Cl- 6 3 Balancing Redox Reactions by the HalfReaction Method • This method is a helpful way to balance complex redox reactions in solution. • The reaction is broken down into two half-reactions, one for oxidation and another for reduction. • Each half-reaction is balanced individually, for both mass and charge. • The two half-reactions are added back together to get the overall balanced equation. 7 Balancing Redox Reactions 1. Assign oxidation states. a) Determine the element oxidized and the element reduced. 2. Write oxidation and reduction half-reactions. 3. Balance the mass of half-reactions. a) First balance elements other than H and O. b) Balance O by adding H2O where O is needed. c) Balance H by adding H+ where H is needed. d) If the reaction is in a basic solution, neutralize H+ with OH−. 4. Balance charge of half-reactions by adding electrons. 5. Make the number of electrons in both half-reactions the same by multiplying one or both by a small whole number. 6. Add half-reactions, and cancel like terms. 7. Check by counting atoms and total charge. 8 4 Example 20.1 Half-Reaction Method of Balancing Aqueous Redox Equations in Acidic Solution Balance the redox equation: Al(s) + Cu2+(aq) Al3+(aq) + Cu(s) How To: Balance Aqueous Redox Equations in Acidic Solution Using the Half-Reaction Method 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. Oxidation: Al(s) Reduction: Cu2+(aq) Al3+(aq) Cu(s) 9 Example 20.1 Half-Reaction Method of Balancing Aqueous Redox Equations in Acidic Solution Continued Step 3 Balance each half-reaction with respect to mass in the following order: • Balance all elements other than H and O. • Balance O by adding H2O. • Balance H by adding H+. All elements are balanced, so proceed to the next step. Step 4 Balance each half-reaction with respect to charge by adding electrons. (Make the sum of the charges on both sides of the equation equal by adding as many electrons as necessary.) Al(s) Al3+(aq) + 3 e– 2 e– + Cu2+(aq) Cu(s) Step 5 Make the number of electrons in both half-reactions equal by multiplying one or both half-reactions by a small whole number. 2[Al(s) Al3+(aq) + 3 e–] 2 Al(s) 2 Al3+(aq) + 6 e– 3[2 e– + Cu2+(aq) Cu(s)] 6 e– + 3 Cu2+(aq) 3 Cu(s) 10 5 Example 20.1 Half-Reaction Method of Balancing Aqueous Redox Equations in Acidic Solution Cont. Step 6 Add the two half-reactions together, canceling electrons and other species as necessary. Step 7 Verify that the reaction is balanced with respect to both mass and charge. For Practice 20.1 Balance the redox reaction in acidic solution: H+(aq) + Cr(s) H2(g) + Cr2+(aq) 11 Example 20.2 Balancing Redox Reactions Occurring in Basic Solution Balance the equation occurring in basic solution: I–(aq) + MnO4– (aq) I2(aq) + MnO2(s) Solution To balance redox reactions occurring in basic solution, follow the half-reaction method outlined in Examples 20.1 and 20.2, but add an extra step to neutralize the acid with OH- as shown in step 3. Step 1 Assign oxidation states. Step 2 Separate the overall reaction into two half-reactions. Oxidation: I– (aq) I2(aq) Reduction: MnO4–(aq) MnO2(s) 12 6 Example 20.2 Balancing Redox Reactions Occurring in Basic Solution Continued Step 3 Balance each half-reaction with respect to mass. • Balance all elements other than H and O. • Balance O by adding H2O. • Balance H by adding H+. • Neutralize H+ by adding enough OH– to neutralize each H+. Add the same number of OH– ions to each side of the equation. 13 Example 20.2 Balancing Redox Reactions Occurring in Basic Solution Step 4 Balance each half-reaction with respect to charge. 2 I–(aq) I2(aq) + 2 e– 4 H2O(l ) + MnO4–(aq) + 3 e– MnO2(s) + 2 H2O(l ) + 4 OH–(aq) Step 5 Make the number of electrons in both half-reactions equal. 3[2 I–(aq) I2(aq) + 2 e–] 6 I–(aq) 3 I2(aq) + 6 e– 2[4 H2O(l) + MnO4–(aq) + 3 e– 8 H2O(l) + 2 MnO4–(aq) + 6 e– MnO2(s) + 2 H2O(l ) + 4 OH–(aq)] 2 MnO2(s) + 4 H2O(l ) + 8 OH–(aq) Step 6 Add the half-reactions together. 14 7 Example 20.2 Balancing Redox Reactions Occurring in Basic Solution Continued Step 7 Verify that the reaction is balanced. For Practice 20.3 Balance the following redox reaction occurring in basic solution: ClO–(aq) + Cr(OH)4–(aq) CrO42–(aq) + Cl–(aq) 15 Electrical Current • Electrons flow through a conductor in response to an electrical potential difference similar to water flowing downhill in response to a difference in gravitational potential energy. • Electric current—the amount of electric charge that passes a point in a given period of time ü Whether as electrons flowing through a wire, or ions flowing through a solution 16 8 Electric Current Flowing Directly between Atoms 17 Electrical Current • Current is the number of electrons that flow through the system per second. ü Unit = ampere (or amp, A) • 1 A of current = 1 coulomb of charge flowing each second 18 ü1 A = 6.242 ´ 10 electrons per second 18 9 Potential Difference • The difference in potential energy between the reactants and products is the potential difference. ü Unit = volt • 1 V = 1 J of energy per coulomb of charge ü The voltage needed to drive electrons through the external circuit • The amount of force pushing the electrons through the wire is called the electromotive force, emf. 19 Cell Potential • The difference in potential energy between the anode and the cathode in a voltaic cell is called the cell potential. • The cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode. • The cell potential under standard conditions is called the standard emf, E °cell . ü 25 °C , 1 atm for gases, 1 M concentration of solution ü Sum of the cell potentials for the half-reactions elsius 20 10 Conceptual Connection Which statement best captures the difference between volts and amps? a. The volt is a unit that quantifies the difference in electrical potential energy, and the amp is a unit that quantifies the flow of electrical current. b. The amp is a unit that quantifies the difference in electrical potential energy, and the volt is a unit that quantifies the flow of electrical current. c. The volt and amp are two different units used to measure the same thing, the flow of electrical current. 21 Conceptual Connection Which statement best captures the difference between volts and amps? 22 11 Electrochemical Cells • Oxidation and reduction half-reactions are kept separate in half-cells. • Electron flow through a wire along with ion flow through a solution constitutes an electric circuit. • It requires a conductive solid electrode to allow the transfer of electrons. ü Through external circuit ü Metal or graphite • Requires ion exchange between the two half-cells of the system. ü Electrolyte 23 Electrodes and Salt Bridge • Anode ü Electrode where oxidation always occurs ü More negatively charged electrode in voltaic cell • Cathode ü Electrode where reduction always occurs ü More positively charged electrode in voltaic cell • Salt bridge is an inverted, U-shaped tube containing a strong electrolyte and connecting the two half-cells. 24 12 Electrodes • Typically ü The anode is made of the metal that is oxidized. ü The cathode is made of the same metal as is produced by the reduction. • If the redox reaction involves the oxidation or reduction of an ion to a different oxidation state, or the oxidation or reduction of a gas, we may use an inert electrode. ü An inert electrode is one that not does participate in the reaction but just provides a surface on which the transfer of electrons can take place. 25 Redox Reactions & Current Batteries • Redox reactions involve the transfer of electrons from one substance to another • Therefore, redox reactions have the potential to generate an electric current • In order to use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring 26 26 13 Electric Current Flowing Indirectly Between Atoms 27 27 Conceptual Connection In a voltaic cell, in which direction do electrons flow? a. from higher potential energy to lower potential energy b. from the cathode to the anode c. from lower potential energy to higher potential energy 28 14 Conceptual Connection In a voltaic cell, in which direction do electrons flow? 29 Electrochemical Cells • electrochemistry is the study of redox reactions that produce or require an electric current • the conversion between chemical energy and electrical energy is carried out in an electrochemical cell • spontaneous redox reactions take place in a voltaic cell üaka galvanic cells • nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy 30 30 15 Cell Notation • Shorthand description of Voltaic cell • Electrode | electrolyte || electrolyte | electrode • Oxidation half-cell on left, reduction half-cell on the right • single | = phase barrier ü if multiple electrolytes in same phase, a comma is used rather than | ü often use an inert electrode • double line || = salt bridge 31 31 Electrochemical Cell Notation Because the half-reaction involves reducing the Mn oxidation state from +7 to +2, we use an electrode that will provide a surface for the electron transfer without reacting with the MnO 4−. Platinum works well because it is extremely nonreactive and conducts electricity. Fe ( s ) | Fe2+ ( aq ) || MnO 4 – ( aq ) , Mn2+ ( aq ) , H+ ( aq ) | Pt ( s ) 32 16 Standard Reduction Potential • We cannot measure the absolute tendency of a halfreaction; we can measure it only relative to another half-reaction. • We select as a standard halfreaction the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 v. ü Standard hydrogen electrode, SHE 33 Cell Potential • A half-reaction with a strong tendency to occur has a large positive half-cell potential. • When two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occur. 34 17 Which Way Will Electrons Flow? Under standard conditions, zinc has a stronger tendency to oxidize than copper. Zn ® Zn2+ + 2 e - E ° = +0.76 Cu ® Cu2+ + 2 e -E ° = -0.34 35 Measuring Electrode Potential 36 18 Half-Cell Potentials • S H E reduction potential is defined to be exactly 0 V. • Standard reduction potentials compare the tendency for a particular reduction half-reaction to occur relative to the reduction of H+ to H 2. ü Under standard conditions • Half-reactions with a stronger tendency toward reduction than ° . the S H E have a positive value for E red • Half-reactions with a stronger tendency toward oxidation than ° . the SHE have a negative value for E red ° ° • For an oxidation half-reaction, E oxidation = -E reduction . ° = E° ° • E call oxidation - E reduction . 37 Standard Electrode Potentials 38 19 Standard Electrode Potentials Table [continued] 39 Calculating Cell Potentials under Standard Conditions ° = E° ° • Ecell cathode - Eanode • When adding E° values for the half-cells, if you need to multiply the half-reactions to balance the equation, do not multiply the half-cell E° values. • E° values are intensive. 40 20 Tendencies from the Table of Standard Reduction Potentials • Higher on the table of standard reduction potentials, the stronger tendency for the reactant to be reduced • Lower on the table of standard reduction potentials, the stronger tendency for the product to be oxidized 41 21 Electrochemistry Stuff Chapter 20 Part 2 1 Predicting Spontaneity of Redox Reactions • Substances listed at the top of Table 20.1 tend to undergo reduction; they are good oxidizing agents. • Substances listed near the bottom of Table 20.1 tend to undergo oxidation; they are good reducing agents. • Any reduction reaction in Table 20.1 is spontaneous when paired with the reverse of any of the reactions listed below it on the table. ° Cu2+ ( aq ) + 2 e - ® Cu ( s ) E red = +0.34 V ° Zn2+ ( aq ) + 2 e - ® Zn ( s ) E red = -0.76 V Zn ( s ) + Cu2+ ( aq ) ® Zn2+ ( aq ) + Cu ( s ) spontaneous Cu ( s ) + Zn2+ ( aq ) ® Cu2+ ( aq ) + Zn ( s ) nonspontaneous 2 1 Predicting Whether a Metal Will Dissolve in Acid • Metals whose reduction half-reactions are listed below the reduction of H+ to H2 in Table 20.1 dissolve in acids. • Metals listed above H+ to H2 in Table 20.1 do not dissolve in acids. • Almost all metals will dissolve in HNO3. ü Having N reduced rather than H ü Au and Pt dissolve in HNO3 + HCl NO3 - ( aq ) + 4 H+ ( aq ) + 3 e - ® NO ( g ) + 2 H2O ( l ) 3 Example 20.3 Calculating Standard Potentials for Electrochemical Cells from Standard Electrode Potentials of the Half-Reactions Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25 °C. (The equation is balanced.) Al(s) + NO3–(aq) + 4 H+(aq) Al3+(aq) + NO(g) + 2 H2O(l ) Solution Begin by separating the reaction into oxidation and reduction half-reactions. (In this case, you can readily see that Al(s) is oxidized. In cases where it is not so apparent, you may want to assign oxidation states to determine the correct half-reactions.) Oxidation: Al(s) Al3+(aq) + 3 e– Reduction: NO3–(aq) + 4 H+(aq) + 3 e– NO(g) + 2 H2O(l ) 4 2 Example 20.3 Calculating Standard Potentials for Electrochemical Cells from Standard Electrode Potentials of the Half-Reactions Continued Look up the standard electrode potentials for each half-reaction in Table 20.1. Add the half-cell reactions together to obtain the overall redox equation. Calculate the standard cell potential by subtracting the electrode potential of the anode from the electrode potential of the cathode. E°cell = E°cat – E°an = 0.96 V – (–1.66 V) = 2.62 V For Practice 20.4 Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25 °C. (The equation is balanced.) 3 Pb2+(aq) + 2 Cr(s) 3 Pb(s) + 2 Cr3+(aq) 5 Eocell, DGo and K • for a spontaneous reaction üone the proceeds in the forward direction with the chemicals in their standard states üDG° < 1 (negative) üE° > 1 (positive) üK > 1 • DG° = −RTlnK = −nFEocell ün is the number of electrons üF = Faraday’s Constant = 96,485 C/mol e− 6 6 3 ° and K Relationship between Ecell • ΔG° = -RT ln K • -nFE °cell = -RT ln K RT • E °Cell = nF ln K • Substituting values for R, F, and 298.15K for T, and elvin converting to log instead of ln, we get • E °cell = 0.0592 V log K n 7 Conceptual Connection A redox reaction has an equilibrium constant of Which statement is true regarding K = 1.2 ´ 103. ° and Ecell ° for this DGrxn reaction ? ° ° is positive. a Ecell is positive and DGrxn . ° is negative and DGrxn ° is negative. b Ecell . ° is positive and DGrxn ° is negative. c Ecell . ° is negative and DGrxn ° is positive. d Ecell . 8 4 Conceptual Connection A redox reaction has an equilibrium constant of Which statement is true regarding K = 1.2 ´ 103. ° and Ecell ° for this DGrxn reaction? 9 Example 20.5 Relating 𝚫G° and E°cell Use the tabulated electrode potentials to calculate ΔG° for the reaction: I2(s) + 2 Br–(aq) 2 I–(aq) + Br2(l ) Is the reaction spontaneous? Sort You are given a redox reaction and asked to find ΔG°. Given: I2(s) + 2 Br–(aq) 2 I–(aq) + Br2(l ) Find: ΔG° Strategize Refer to the values of electrode potentials in Table 20.1 to calculate E°cell. Then use Equation 20.3 to calculate ΔG° from E°cell. Conceptual Plan Solve Separate the reaction into oxidation and reduction half-reactions and find the standard electrode potentials for each. Determine E°cell by subtracting Ean from Ecat. 10 5 Example 20.5 Relating 𝚫G° and E°cell Continued 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. 11 Example 20.5 Relating 𝚫G° and E°cell Cont. Check The answer is in the correct units (joules) and seems reasonable in magnitude (≈110 kJ). You have seen (in Chapter 19) that values of ΔG° typically range from plus or minus tens to hundreds of kilojoules. The sign is positive, as expected for a reaction in which E°cell is negative. For Practice 20.6 Use tabulated electrode potentials to calculate ΔG° for the reaction. 2 Na(s) + 2 H2O(l) H2(g) + 2 OH–(aq) + 2 Na+(aq) Is the reaction spontaneous? 12 6 Example 20.6 Relating E°cell and K Use the tabulated electrode potentials to calculate K for the oxidation of copper by H+ (at 25 °C): Cu(s) + 2 H+(aq) Cu2+(aq) + H2(g) Sort You are given a redox reaction and asked to find K. Given: Cu(s) + 2 H+(aq) Cu2+(aq) + H2(g) Find: K Strategize Refer to the values of electrode potentials in Table 20.1 to calculate E°cell. Then use Equation 20.6 to calculate K from E°cell. Conceptual Plan Solve Separate the reaction into oxidation and reduction half-reactions and find the standard electrode potentials for each. Find E°cell by subtracting Ean from Ecat. 13 Example 20.6 Relating E°cell and K Continued Solution Calculate K 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. 14 7 Example 20.6 Relating E°cell and K Cont. Check The answer has no units, as expected for an equilibrium constant. The magnitude of the answer is small, indicating that the reaction lies far to the left at equilibrium, as expected for a reaction in which E°cell is negative. For Practice 20.7 Use the tabulated electrode potentials to calculate K for the oxidation of iron by H+ (at 25 °C): 2 Fe(s) + 6 H+(aq) 2 Fe3+(aq) + 3 H2(g) 15 Cell Potential under Nonstandard Conditions ΔG = ΔG° + RT ln Q ° + RT ln Q -nFEcell = -nFEcell ° Ecell = Ecell RT ln Q nF ° Ecell = Ecell 0.0592 V log Q n This equation, known as the Nernst equation, helps us determine the cell potential at nonstandard conditions. 16 8 E° at Nonstandard Conditions 17 17 Example 20.7 Calculating Ecell under Nonstandard Conditions Determine the cell potential for an electrochemical cell based on the following two half-reactions: Oxidation: Cu(s) Cu2+(aq, 0.010 M) + 2 e– Reduction: MnO4–(aq, 2.0 M) + 4 H+(aq, 1.0 M) + 3 e– MnO2(s) + 2 H2O(l ) 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: [MnO4–] = 2.0 M; [H+] = 1.0 M; [Cu2+] = 0.010 M Find: Ecell Strategize Use the tabulated values of electrode potentials to calculate E°cell. Then use Equation 20.9 to calculate Ecell. Conceptual Plan Solve Write the oxidation and reduction half-reactions, multiplying by the appropriate coefficients to cancel the electrons. Find the standard electrode potentials for each half-reaction. Find E°cell. 18 9 Example 20.7 Calculating Ecell under Nonstandard Conditions Continued Solution Calculate Ecell from E°cell. The value of n (the number of moles of electrons) corresponds to the number of electrons (six 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 MnO2, and solid copper are omitted from the expression for Q.) 19 Example 20.7 Calculating Ecell under Nonstandard Conditions Cont. Check The answer has the correct units (V). The value of Ecell is larger than E°cell, as expected based on Le Châtelier’s 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 a greater cell potential. For Practice 20.8 Determine the cell potential of an electrochemical cell based on the following two half-reactions: Oxidation: Ni(s) Ni2+(aq, 2.0 M) + 2 e– Reduction: VO2+(aq, 0.010 M) + 2 H+(aq, 1.0 M) + e– VO2+(aq, 2.0 M) + H2O(l) 20 10 Driving Nonspontaneous Reactions • In all cells, whether voltaic or electrolytic, oxidation occurs at the anode, and reduction occurs at the cathode. • Voltaic cells—Spontaneous reaction generates electricity. ü Anode is the source of electrons and has a (−) charge. ü Cathode draws electrons and has a (+) charge. • Electrolytic cells—nonspontaneous reaction driven by external electrical current 21 Electrolysis • Electrolysis is the process of using electrical current to drive nonspontaneous reaction. • Electrolysis is carried out in an electrolytic cell. • Electrolytic cells can be used to separate compounds into their elements. 22 11 Voltaic Versus Electrolytic Cells 23 Conceptual Connection Which statement is true for both electrolytic and voltaic cells? a. The cell spontaneously produces a positive voltage. b. Electrons flow from the anode to the cathode. c. Oxidation occurs at the cathode. 24 12 Conceptual Connection Which statement is true for both electrolytic and voltaic cells? 25 Electrolytic Cells • Electrons are drawn away from the anode, which must be connected to the positive terminal of the external power source (anode +). • Electrons are forced to the cathode, which must be connected to the negative terminal of the power source (cathode −). • The reaction that takes place is the nonspontaneous process. 2 H2 ( g ) + O2 ( g ) ® 2 H2O ( l ) spontaneous 2 H2O ( l ) ® 2 H2 ( g ) + O2 ( g ) electrolysis 26 13 Electrolytic Cells • The electrical energy is supplied by a direct-current power supply, a battery or DC power supply. • Some electrolysis reactions require more voltage than Ecell predicts. This is called the overvoltage. 27 Stoichiometry of Electrolysis • In an electrolytic cell, the amount of product made is related to the number of electrons transferred. ü Essentially, the electrons are a reactant. • The number of moles of electrons that flow through the electrolytic cell depends on the current and length of time. ü 1 amp = 1 coulomb of charge/second ü 1 mole of e− = 96,485 coulombs of charge ØFaraday’s constant 28 14 Conceptual Connection Silver plating uses the reaction Ag+ ( aq ) + e - ® Ag ( s ) . How many moles of electrons must pass through an electrolytic cell for silver plating in order to plate 3 moles of Ag? a. 1 mole e− b. 2 mole e− c. 3 mole e− d. 4 mole e− 29 Conceptual Connection Silver plating uses the reaction Ag+ ( aq ) + e - ® Ag ( s ) . How many moles of electrons must pass through an electrolytic cell for silver plating in order to plate 3 moles of Ag? 30 15 Example 20.8 Stoichiometry of Electrolysis Gold can be plated out of a solution containing Au3+ according to the half-reaction: Au3+(aq) + 3 e– Au(s) 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, because current is a measure of charge per unit time, use the given current and the time to find the number of coulombs. 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. 31 Example 20.8 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 32 16 Example 20.8 Stoichiometry of Electrolysis Cont. Check The answer has the correct units (g Au). The magnitude of the answer is reasonable if you 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. For Practice 20.10 Silver can be plated out of a solution containing Ag+ according to the half-reaction: Ag+(aq) + e– Ag(s) How much time (in minutes) does it take to plate 12 g of silver using a current of 3.0 A? 33 17 4/30/21 Chapter 19 Radioactivity and Nuclear Chemistry 1 The Discovery of Radioactivity • Antoine-Henri Becquerel designed an experiment to determine if phosphorescent minerals also gave off X-rays 2 2 1 4/30/21 The Discovery of Radioactivity Cont. • Becquerel discovered that certain minerals were constantly producing penetrating energy rays he called uranic rays ü like X-rays ü but not related to fluorescence • Becquerel determined that ü all the minerals that produced these rays contained uranium ü the rays were produced even though the mineral was not exposed to outside energy • Energy apparently being produced from nothing?? 3 3 The Curies • Marie Curie used electroscope to detect uranic rays in samples • Discovered new elements by detecting their rays üradium named for its green phosphorescence üpolonium named for her homeland • Since these rays were no longer just a property of uranium, she renamed it radioactivity 4 4 2 4/30/21 Other Properties of Radioactivity • radioactive rays can ionize matter ücause uncharged matter to become charged übasis of Geiger Counter and electroscope • radioactive rays have high energy • radioactive rays can penetrate matter • radioactive rays cause phosphorescent chemicals to glow übasis of scintillation counter 5 5 Types of Radioactive Rays • Rutherford discovered there were three types of radioactivity • alpha rays (a) ühave a charge of +2 c.u. and a mass of 4 amu üwhat we now know to be helium nucleus • beta rays (b) ühave a charge of -1 c.u. and negligible mass üelectron-like • gamma rays (g) üform of light energy (not particle like a and b) 6 6 3 4/30/21 Penetrating Ability of Radioactive Rays a g b 0.01 mm 1 mm 100 mm Pieces of Lead 7 7 Facts About the Nucleus • Every atom of an element has the same number of protons üatomic number (Z) • Atoms of the same elements can have different numbers of neutrons üisotopes üdifferent atomic masses • Isotopes are identified by their mass number (A) ümass number = number of protons + neutrons 8 8 4 4/30/21 Facts About the Nucleus Cont. • The number of neutrons is calculated by subtracting the atomic number from the mass number • The nucleus of an isotope is called a nuclide üless than 10% of the known nuclides are nonradioactive, most are radionuclides • Each nuclide is identified by a symbol üElement -Mass Number = X-A mass number A Element = X atomic number Z 9 9 Radioactivity • Radioactive nuclei spontaneously decompose into smaller nuclei ü Radioactive decay ü We say that radioactive nuclei are unstable • The parent nuclide is the nucleus that is undergoing radioactive decay, the daughter nuclide is the new nucleus that is made • Decomposing involves the nuclide emitting a particle and/or energy • All nuclides with 84 or more protons are radioactive 10 10 5 4/30/21 Important Atomic Symbols Particle Symbol Nuclear Symbol proton p+ neutron n0 electron e- alpha a 4 2 beta b, b- 0 -1 positron b, b+ 0 +1 1 1 H 11p 1 0 0 -1 n e α 42 He β -01e β +01e 11 11 Transmutation • Rutherford discovered that during the radioactive process, atoms of one element are changed into atoms of a different element - transmutation ü Dalton’s Atomic Theory statement 3 bites the dust • in order for one element to change into another, the number of protons in the nucleus must change 12 12 6 4/30/21 Nuclear Equations • we describe nuclear processes with nuclear equations • use the symbol of the nuclide to represent the nucleus • atomic numbers and mass numbers are conserved üuse this fact to predict the daughter nuclide if you know parent and emitted particle 13 13 Alpha Emission • an a particle contains 2 protons and 4 2 neutrons ühelium nucleus • most ionizing, but least penetrating • loss of an alpha particle means üatomic number decreases by 2 ümass number decreases by 4 4 α 2 2 He 222 4 218 Ra ® 2 He + 86 Rn 88 14 14 7 4/30/21 Alpha Decay 15 15 Beta Emission • a b particle is like an electron ümoving much faster üproduced from the nucleus 0 -1 β -01e • when an atom loses a b particle its üatomic number increases by 1 ümass number remains the same • in beta decay, a neutron changes into a proton Th ® -01e + 234 91 Pa 234 90 16 16 8 4/30/21 Beta Decay 17 17 Gamma Emission 0 0 γ • gamma (g) rays are high energy photons of light • no loss of particles from the nucleus • no change in the composition of the nucleus ü Same atomic number and mass number • least ionizing, but most penetrating • generally occurs after the nucleus undergoes some other type of decay and the remaining particles rearrange 18 18 9 4/30/21 Positron Emission • positron has a charge of +1 c.u. and negligible mass 0 0 üanti-electron +1 +1 • when an atom loses a positron from the nucleus, its ümass number remains the same üatomic number decreases by 1 • positrons appear to result from a proton changing into a neutron β 22 11 e Na ® +01e + 22 10 Ne 19 19 Positron Emission Cont. 20 10 4/30/21 Electron Capture 0 -1 e • occurs when an inner orbital electron is pulled into the nucleus • no particle emission, but atom changes üsame result as positron emission • proton combines with the electron to make a neutron ümass number stays the same üatomic number decreases by one 92 0 92 44 Ru + -1e ® 43Tc 92 92 44 Ru ® 43Tc 21 21 Particle Changes • Beta Emission – neutron changing into a proton 1 1 0 n ® p + 0 1 -1b • Positron Emission – proton changing into a neutron 1 1 0 p ® n + 1 0 +1b • Electron Capture – proton changing into a neutron 1 0 1 p + e ® 1 -1 0n 22 22 11 4/30/21 23 23 Nuclear Equations Cont. • in the nuclear equation, mass numbers and atomic numbers are conserved • we can use this fact to determine the identity of a daughter nuclide if we know the parent and mode of decay 24 24 12 4/30/21 Write the Nuclear Equation for Positron Emission From K-40 (1 of 4) 1) Write the nuclide symbols for both the starting radionuclide and the particle K - 40 = 40 19 K 0 positron = +1e 25 25 Write the Nuclear Equation for Positron Emission From K-40 (2 of 4) 2) Set up the equation • emitted particles are products • captured particles are reactants 40 19 K ® +01e + AZ X 26 26 13 4/30/21 Write the Nuclear Equation for Positron Emission From K-40 (3 of 4) 3) Determine the mass number and atomic number of the missing nuclide • mass and atomic numbers are conserved 40 19 K ® +01e + 40 18 X 27 27 Write the Nuclear Equation for Positron Emission From K-40 (4 of 4) 4) Determine the element from the atomic number 40 19 K ® +01e + 40 18 Ar 28 28 14 4/30/21 Practice - Write a nuclear equation for each of the following • alpha emission from U-238 238 4 234 92 U ® 2 He + 90Th • beta emission from Ne-24 24 0 24 10 Ne®-1 e + 11Na • positron emission from N-13 • 13 0 13 7 N ® +1 e + 6 C electron capture by Be-7 7 0 7 4 Be + -1e ® 3 Li 29 29 What Causes Nuclei to Break Down? • the particles in the nucleus are held together by a very strong attractive force only found in the nucleus called the strong force üacts only over very short distances • the neutrons play an important role in stabilizing the nucleus, as they add to the strong force, but don’t repel each other like the protons do 30 30 15 4/30/21 N/Z Ratio • the ratio of neutrons : protons is an important measure of the stability of the nucleus • if the N/Z ratio is too high – neutrons are converted to protons via b decay • if the N/Z ratio is too low – protons are converted to neutrons via positron emission or electron capture üor via a decay – though not as efficient 31 31 Valley of Stability for Z = 1 Þ 20, stable N/Z ≈ 1 for Z = 20 Þ 40, stable N/Z approaches 1.25 for Z = 40 Þ 80, stable N/Z approaches 1.5 for Z > 83, there are no stable nuclei 32 32 16 4/30/21 Ex 19.3b Determine the kind of radioactive decay that Mg-22 undergoes • Mg-22 ü Z = 12 ü N = 22 – 12 = 10 • N/Z = 10/12 = 0.83 • from Z = 1 Þ 20, stable nuclei have N/Z ≈ 1 • since Mg-22 N/Z is low, it should convert p+ into n0, therefore it will undergo positron emission or electron capture 33 33 Magic Numbers • besides the N/Z ratio, the actual numbers of protons and neutrons • • • effects stability most stable nuclei have even numbers of protons and neutrons only a few have odd numbers of protons and neutrons if the total number of nucleons adds to a magic number, the nucleus is more stable ü same idea as the electrons in the noble gas resulting in a more stable electron configuration ü most stable when N or Z = 2, 8, 20, 28, 50, 82; or N = 126 34 34 17 4/30/21 Rate of Radioactivity • it was discovered that the rate of change in the amount of radioactivity was constant and different for each radioactive “isotope” ü change in radioactivity measured with Geiger counter Øcounts per minute ü each radionuclide had a particular length of time it required to lose half its radioactivity Øa constant half-life ü we know that processes with a constant half-life follow first order kinetic rate laws • rate of change not affected by temperature ü means that radioactivity is not a chemical reaction! 35 35 Kinetics of Radioactive Decay • Rate = kN üN = number of radioactive nuclei • t1/2 = 0.693/k • the shorter the half-life, the more nuclei decay every second – we say the sample is hotter ln Nt rate t = -kt = ln N0 rate0 36 36 18 4/30/21 Half-Lives of Various Nuclides Half-Life Type of Decay Th-232 1.4 x 1010 yr alpha U-238 4.5 x 109 yr alpha C-14 5730 yr beta Rn-220 55.6 sec alpha Th-219 1.05 x 10–6 sec alpha Nuclide 37 37 If you have a 1.35 mg sample of Pu-236, calculate the mass that will remain after 5.00 years. The half life of Pu-236 is 2.86 yr. Given: mass Pu-236 = 1.35 mg, t = 5.00 yr, t1/2 = 2.86 yr Find: mass, mg Concept Plan: Relationships: t1/2 k 1 Solve: + 0.693 t = k 2 m0, t mt N ln t = -kt N0 N 0.693 lnt =t = -kt N0 k 1 2 - kt 0.693 )e - (0.2423 yr )(-51.00 yr ) Nk t ==0N.693 0 e == (1.35 mg = 0.2423 yr -1 t N t = 0.402 mg2.86 yr 1 2 Check: units are correct, the magnitude makes sense since it is less than ½ the original mass for longer than 1 half-life 38 38 19 4/30/21 Object Dating • mineral (geological) ücompare the amount of U-238 to Pb-206 ücompare amount of K-40 to Ar-40 • archaeological (once living materials) ücompare the amount of C-14 to C-12 üC-14 radioactive with half-life = 5730 yrs. üwhile substance living, C-14/C-12 fairly constant ØCO2 in air ultimate source of all C in organism Øatmospheric chemistry keeps producing C-14 at the nearly the same rate it decays üonce dies C-14/C-12 ratio decreases ülimit up to 50,000 years 39 39 An ancient skull gives 4.50 dis/min∙gC. If a living organism gives 15.3 dis/min∙gC, how old is the skull? Given: ratet = 4.50 dis/min∙gC, ratet = 15.3 dis/min∙gC Find: time, yr Concept Plan: Relationships: t1/2 k 1 Solve: + rate0, ratet 0.693 t = k t rate t ln = -kt rate0 2 rate t 0.693 t ln=rate0 = -kt k 4.50 dis min gC rate 0.693t 0.693 ln ln k = rate = = 1dis .2min 09 gC ´ 10 - 4 yr -1 4 15.3 0 - t =5 t 730 = - yr = 1.0 ´ 10 yr -4 -1 1 2 • • 1 k2 1.209 ´ 10 yr Check: units are correct, the magnitude makes sense since it is less than 2 half-lives 40 40 20 4/30/21 Nonradioactive Nuclear Changes • a few nuclei are so unstable that if their nucleus is hit just right by a neutron, the large nucleus splits into two smaller nuclei - this is called fission • small nuclei can be accelerated to such a degree that they overcome their charge repulsion and smash together to make a larger nucleus - this is called fusion • both fission and fusion release enormous amounts of energy ü fusion releases more energy per gram than fission Lise Meitner 41 41 21 Free Energy and Thermodynamics Chapter 19 1 What you Know from CHEM101 • First Law of Thermodynamics üEnergy can neither be created or destroyed but can be turned from one form into another. Ønonspontaneous processes require energy input to go • Internal Energy üE = q + w • State Functions • Thermochemical Equations • Hess’s Law 2 2 1 Thermodynamics and Spontaneity • Thermodynamics predicts whether a process will proceed under the given conditions üspontaneous process Ønonspontaneous processes require energy input to go • The tendency for a process to advance to equilibrium without external influence. • Something that happens naturally is spontaneous. 3 3 Reversibility of Process • Any process will be spontaneous in one direction. • The reverse process is non-spontaneous. • If work needs to be done, it is not spontaneous. ü A rock naturally rolls down a hill – spontaneous ü It must be pushed back up – non-spontaneous ü A hot object naturally cools - spontaneous 4 4 2 Other Examples • Matter disperses – gas fills a container, two liquids mix • Heat disperses – hot object cools on a cold surface • Motion disperses – a ball stops bouncing • The reverses of these three well known process never occur spontaneously. 5 5 Indicators of Spontaneity • What is the indicator of spontaneity? • Heat evolved? • But … endothermic reactions occur spontaneously as well (ice melting, salt dissolving) • Enthalpy is not an indicator of spontaneity, although most spontaneous process are exothermic – energy is conserved not created. • The amount of energy does not change in any process – but it is redistributed. 6 6 3 Spontaneity and Speed • The speed of a reaction is not an indicator of its spontaneity. • Spontaneity is determined by the relative positions of the initial and final states (thermodynamic state functions.) • Speed is determined by the pathway (kinetics). • Two independent regimes. 7 7 8 Diamond → Graphite Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous. But don’t worry: It’s so slow that your ring won’t turn into pencil lead in your lifetime (or through many of your generations). 8 4 Thermodynamics vs. Kinetics 9 9 Factors Affecting Whether a Reaction Is Spontaneous • The two factors that determine the thermodynamic favorability are the enthalpy and the entropy. • The enthalpy is a comparison of the bond energy of the reactants to the products. übond energy = amount of energy needed to break a bond. ü DH 10 10 5 Entropy and Probability • Ordered states are less likely because there are fewer ways to obtain them. üDo our socks become matched spontaneously? üNo – only one of many possible arrangements. • With only a few molecules the ordered state becomes massively less probable than a disordered state. 11 11 Entropy • entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S • S generally J/mol • S = k ln W ü k = Boltzmann Constant = 1.38 x 10-23 J/K ü W is the number of possible arrangements of the state. • The entropy is proportional to the natural log of the number of arrangements of the state. • An ordered arrangement has W = 1, so S = 0 (this would happen at 0K.) 12 12 6 13 Macrostates and Microstates • Macrostate: the state defined by a given set of conditions (P, V, and T ) • Microstate: the exact internal energy distribution among the particles at any one instant ü W is the number of possible microstates of a system. 13 14 Microstates 14 7 W Energetically Equivalent States for the Expansion of a Gas 15 15 Macrostates → Microstates These microstates all have the same macrostate So there are 6 different particle arrangements that result in the same macrostate 16 16 8 Macrostates and Probability There is only one possible arrangement that gives State A and one that gives State B There are 6 possible arrangements that give State C Therefore State C has higher entropy than either State A or State B The macrostate with the highest entropy also has the greatest dispersal of energy 17 17 Changes in Entropy, DS • The change in entropy is the heat supplied divided by the Kelvin temperature • DSsys = q/T üClausius definition 18 18 9 Changes in Entropy, DS Cont. • entropy change is favorable when the result is a more random system. ü DS is positive • Some changes that increase the entropy are: üreactions whose products are in a more disordered state. Ø(solid < liquid < gas) üreactions which have larger numbers of product molecules than reactant molecules. üincrease in temperature üsolids dissociating into ions upon dissolving 19 19 Increases in Entropy 20 20 10 The 2nd Law of Thermodynamics • Second Law – In any spontaneous process the total entropy of a system and its surroundings increases. ü for reversible process DSuniv = 0, ü for irreversible (spontaneous) process DSuniv > 0 • DSuniverse = DSsystem + DSsurroundings • If the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount. ü when DSsystem is negative, DSsurroundings is positive • The increase in DSsurroundings often comes from the heat released in an exothermic reaction. 21 21 22 Entropy Change in the System and Surroundings When the entropy change in a system is unfavorable (negative), the entropy change in the surroundings must be favorable (positive) and large in order to allow the process to be spontaneous. 22 11 23 Entropy of System, Surroundings, and Universe 23 Entropy and Temperature • Increasing temperature causes an increase in entropy through molecular motion (rotational, vibrational and translational), and changes of state. • Disorder and motion üGreater motion corresponds to a greater number of microstates. üEntropy increases with temperature. 24 24 12