Chapter 18 Electrochemistry Homework 9 11 12 13
![Chapter 18 - Electrochemistry Homework: 9, 11, 12, 13, 14, 15, 17, 18, 19, Chapter 18 - Electrochemistry Homework: 9, 11, 12, 13, 14, 15, 17, 18, 19,](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-1.jpg)
![Oxidation # Redux n n Any ion will have an oxidation number equal to Oxidation # Redux n n Any ion will have an oxidation number equal to](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-2.jpg)
![18. 2 – Balancing Oxidation. Reduction Equations n How do we tell whether a 18. 2 – Balancing Oxidation. Reduction Equations n How do we tell whether a](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-3.jpg)
![Oxidized vs. Reduced n Oxidized n n An atom/element is oxidized when it loses Oxidized vs. Reduced n Oxidized n n An atom/element is oxidized when it loses](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-4.jpg)
![Changing Oxidation State n Zn(s) + 2 H+(aq) Zn 2+(aq) + H 2(g) n Changing Oxidation State n Zn(s) + 2 H+(aq) Zn 2+(aq) + H 2(g) n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-5.jpg)
![Changing Oxidation? We can write the previous reaction in terms of oxidation numbers n Changing Oxidation? We can write the previous reaction in terms of oxidation numbers n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-6.jpg)
![In some reactions there is no clear transfer of electrons, but oxidation states do In some reactions there is no clear transfer of electrons, but oxidation states do](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-7.jpg)
![n In the previous example n n n 2 H 2(g) + O 2(g) n In the previous example n n n 2 H 2(g) + O 2(g)](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-8.jpg)
![n In any redox reaction, both oxidation and reduction must occur n n In n In any redox reaction, both oxidation and reduction must occur n n In](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-9.jpg)
![n n Whenever we balance an equation, we MUST obey the law of conservation n n Whenever we balance an equation, we MUST obey the law of conservation](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-10.jpg)
![Half-Reactions n n Although oxidation and reduction take place at the same time, often Half-Reactions n n Although oxidation and reduction take place at the same time, often](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-11.jpg)
![n Equations that show either oxidation or reduction alone are called half-reactions n n n Equations that show either oxidation or reduction alone are called half-reactions n n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-12.jpg)
![18. 3 - Voltaic Cells** n The energy released in a spontaneous redox reaction 18. 3 - Voltaic Cells** n The energy released in a spontaneous redox reaction](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-13.jpg)
![n An example of this is when a strip of zinc is placed in n An example of this is when a strip of zinc is placed in](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-14.jpg)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-15.jpg)
![n Same basic principle as last n n n But this time the Zn(s) n Same basic principle as last n n n But this time the Zn(s)](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-16.jpg)
![How Does this Work? n The reduction of the Cu 2+ can occur only How Does this Work? n The reduction of the Cu 2+ can occur only](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-17.jpg)
![n The two solid metals that are connected by the external circuit are called n The two solid metals that are connected by the external circuit are called](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-18.jpg)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-19.jpg)
![Or This is my cat Matthew. • He is fat, and must go on Or This is my cat Matthew. • He is fat, and must go on](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-20.jpg)
![n As in this example, the electrodes can be made of materials that participate n As in this example, the electrodes can be made of materials that participate](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-21.jpg)
![n Each compartment in a voltaic cell is called a half-cell. n n One n Each compartment in a voltaic cell is called a half-cell. n n One](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-22.jpg)
![Zn(s) + Cu 2+(aq) Zn 2+(aq) + Cu(s) n Continuing with our present example. Zn(s) + Cu 2+(aq) Zn 2+(aq) + Cu(s) n Continuing with our present example.](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-23.jpg)
![n For a voltaic cell to work n n The solutions in the two n For a voltaic cell to work n n The solutions in the two](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-24.jpg)
![Similarly n The reduction of Cu 2+ at the cathode removes positive charge from Similarly n The reduction of Cu 2+ at the cathode removes positive charge from](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-25.jpg)
![How do we do this? n A salt bridge allows ions to move from How do we do this? n A salt bridge allows ions to move from](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-26.jpg)
![n As oxidation and reduction proceed at the electrodes, ions from the salt bridge n As oxidation and reduction proceed at the electrodes, ions from the salt bridge](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-27.jpg)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-28.jpg)
![n Note the direction of ions in the solution n Note the direction of ions in the solution](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-29.jpg)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-30.jpg)
![n Notice also that in any voltaic cell, the electrons will flow from the n Notice also that in any voltaic cell, the electrons will flow from the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-31.jpg)
![18. 4 - Cell EMF Under Standard Conditions n Why do electrons transfer spontaneously 18. 4 - Cell EMF Under Standard Conditions n Why do electrons transfer spontaneously](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-32.jpg)
![n The difference in potential energy per electrical charge (called potential difference) between two n The difference in potential energy per electrical charge (called potential difference) between two](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-33.jpg)
![n The potential difference between the two electrodes in a voltaic cell provides the n The potential difference between the two electrodes in a voltaic cell provides the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-34.jpg)
![n The actual emf of a particular cell depends on the n n n n The actual emf of a particular cell depends on the n n n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-35.jpg)
![At standard conditions n If we have 1 M concentrations for reactants and products At standard conditions n If we have 1 M concentrations for reactants and products](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-36.jpg)
![Standard Reduction (Half-Cell) Potentials n The emf of a voltaic cell depends on the Standard Reduction (Half-Cell) Potentials n The emf of a voltaic cell depends on the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-37.jpg)
![n The cell potential is the difference between the two electrode potentials n n n The cell potential is the difference between the two electrode potentials n n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-38.jpg)
![Calculating Eºcell n The cell potential (Eºcell) is given by the standard reduction potential Calculating Eºcell n The cell potential (Eºcell) is given by the standard reduction potential](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-39.jpg)
![n Since every voltaic cell involves two halfcells, it is impossible to measure the n Since every voltaic cell involves two halfcells, it is impossible to measure the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-40.jpg)
![n Various standard reduction potentials for half-reactions are found on pg. 873 (table 18. n Various standard reduction potentials for half-reactions are found on pg. 873 (table 18.](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-41.jpg)
![n Because electrical potential measures potential energy per electrical charge, standard reduction potentials are n Because electrical potential measures potential energy per electrical charge, standard reduction potentials are](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-42.jpg)
![Example n Given the standard reduction potential of Zn 2+ to Zn(s) is -0. Example n Given the standard reduction potential of Zn 2+ to Zn(s) is -0.](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-43.jpg)
![n Eºcell = Eºred(cathode) - Eºred(anode) n n Zn is oxidized, and is therefore n Eºcell = Eºred(cathode) - Eºred(anode) n n Zn is oxidized, and is therefore](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-44.jpg)
![Example 2 n n Using the standard reduction potentials provided, calculate the standard emf Example 2 n n Using the standard reduction potentials provided, calculate the standard emf](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-45.jpg)
![Cr 2 O 72 -(aq) + 14 H+(aq) + 6 I-(aq) 2 Cr 3+(aq) Cr 2 O 72 -(aq) + 14 H+(aq) + 6 I-(aq) 2 Cr 3+(aq)](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-46.jpg)
![Eºcell = Eºred(cathode) Eºred(anode) n Eºcell = 1. 33 V - 0. 54 V Eºcell = Eºred(cathode) Eºred(anode) n Eºcell = 1. 33 V - 0. 54 V](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-47.jpg)
![Analyzing the Equation n For each of the half-cells in the voltaic cell, the Analyzing the Equation n For each of the half-cells in the voltaic cell, the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-48.jpg)
![n We know that Eºcell is the difference between the standard reduction potential of n We know that Eºcell is the difference between the standard reduction potential of](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-49.jpg)
![Example n A voltaic cell is based on the following two half-reactions: n n Example n A voltaic cell is based on the following two half-reactions: n n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-50.jpg)
![Finding the Half-Reactions n Look up the Eºred for the two half-reactions n n Finding the Half-Reactions n Look up the Eºred for the two half-reactions n n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-51.jpg)
![Determine the standard cell potential n Eºcell = Eºred(cathode) - Eºred(anode) n Eºcell = Determine the standard cell potential n Eºcell = Eºred(cathode) - Eºred(anode) n Eºcell =](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-52.jpg)
![How to tell how easily something is reduced/oxidized n Generally, the more positive the How to tell how easily something is reduced/oxidized n Generally, the more positive the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-53.jpg)
![Easily reduced n Among the most easily reduced we find the halogens, O 2 Easily reduced n Among the most easily reduced we find the halogens, O 2](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-54.jpg)
![Easily oxidized n Like acid-base strength, the strongest oxidizers are the worst at being Easily oxidized n Like acid-base strength, the strongest oxidizers are the worst at being](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-55.jpg)
![n Easily oxidized things include: n n H 2, and active metals such as n Easily oxidized things include: n n H 2, and active metals such as](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-56.jpg)
![Table 18. 1, pg. 873 n The list orders the ability of substances to Table 18. 1, pg. 873 n The list orders the ability of substances to](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-57.jpg)
![18. 5 - Free Energy, Equilibrium and Redox Reactions n Voltaic cells use redox 18. 5 - Free Energy, Equilibrium and Redox Reactions n Voltaic cells use redox](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-58.jpg)
![n We start by applying a previous equation to all redox reactions, not just n We start by applying a previous equation to all redox reactions, not just](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-59.jpg)
![n Eº= Eºred(reducing process) - Eºred(oxidation process) n n A positive value of E n Eº= Eºred(reducing process) - Eºred(oxidation process) n n A positive value of E](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-60.jpg)
![Example n Using standard reduction potentials, determine whether or not the following reactions are Example n Using standard reduction potentials, determine whether or not the following reactions are](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-61.jpg)
![Cu(s) + 2 H+(aq) Cu 2+(aq) + H 2(g) n In this reaction, Cu Cu(s) + 2 H+(aq) Cu 2+(aq) + H 2(g) n In this reaction, Cu](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-62.jpg)
![Cl 2(g) + 2 I-(aq) 2 Cl-(aq) + I 2(s) n n Cl 2 Cl 2(g) + 2 I-(aq) 2 Cl-(aq) + I 2(s) n n Cl 2](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-63.jpg)
![Activity Series of Metals n Remember, when dealing with single and double replacement reactions Activity Series of Metals n Remember, when dealing with single and double replacement reactions](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-64.jpg)
![EMF and ΔG n The change in Gibbs free energy, ΔG measures the spontaneity EMF and ΔG n The change in Gibbs free energy, ΔG measures the spontaneity](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-65.jpg)
![Relationship n ΔG = -n. FE n n is a positive number without units Relationship n ΔG = -n. FE n n is a positive number without units](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-66.jpg)
![ΔG = -n. FE n Both n and F are positive numbers n So ΔG = -n. FE n Both n and F are positive numbers n So](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-67.jpg)
![Example n Use the standard reduction potentials to calculate the standard free-energy change, ΔGº Example n Use the standard reduction potentials to calculate the standard free-energy change, ΔGº](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-68.jpg)
![Find Eº First n To find Eº we need to break into halfreactions and Find Eº First n To find Eº we need to break into halfreactions and](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-69.jpg)
![n n Eº= 1. 23 V - 0. 80 V = 0. 43 V n n Eº= 1. 23 V - 0. 80 V = 0. 43 V](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-70.jpg)
![Now solve for K n n n ΔGº = -RT(ln. K) K = e-ΔGº/RT Now solve for K n n n ΔGº = -RT(ln. K) K = e-ΔGº/RT](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-71.jpg)
![Summary Triangle Summary Triangle](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-72.jpg)
![18. 6 - Cell Potential and Concentration n As a voltaic cell is discharged, 18. 6 - Cell Potential and Concentration n As a voltaic cell is discharged,](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-73.jpg)
![In General n If the concentrations of reactants increase relative to the concentrations of In General n If the concentrations of reactants increase relative to the concentrations of](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-74.jpg)
![n n n Note: X is copper V =. 47 V Overall reaction is: n n n Note: X is copper V =. 47 V Overall reaction is:](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-75.jpg)
![Example n n During a laboratory session, students set up the electrochemical cell shown. Example n n During a laboratory session, students set up the electrochemical cell shown.](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-76.jpg)
![n A student bumps the cell setup, resulting in the salt bridge losing contact n A student bumps the cell setup, resulting in the salt bridge losing contact](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-77.jpg)
![n A student spills a small amount of 0. 5 M Na 2 SO n A student spills a small amount of 0. 5 M Na 2 SO](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-78.jpg)
![n After the laboratory session is over, a student leaves the switch closed. The n After the laboratory session is over, a student leaves the switch closed. The](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-79.jpg)
![Concentration Cells n In each of the voltaic cells we have seen, the reactive Concentration Cells n In each of the voltaic cells we have seen, the reactive](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-80.jpg)
![Features of a Concentration Cell n Still requires two vessels, connected by a salt Features of a Concentration Cell n Still requires two vessels, connected by a salt](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-81.jpg)
![Overview n n In a concentration cell, the more dilute halfcell will seek to Overview n n In a concentration cell, the more dilute halfcell will seek to](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-82.jpg)
![n So with this cell n The. 01 M half-cell is more dilute. So n So with this cell n The. 01 M half-cell is more dilute. So](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-83.jpg)
![18. 8 - Electrolysis** n Voltaic cells are based on spontaneous redox reactions n 18. 8 - Electrolysis** n Voltaic cells are based on spontaneous redox reactions n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-84.jpg)
![Electrolytic Cell n An electrolytic cell is made of two electrodes in molten salt Electrolytic Cell n An electrolytic cell is made of two electrodes in molten salt](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-85.jpg)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-86.jpg)
![Things to notice n In a voltaic cell (or any other source of direct Things to notice n In a voltaic cell (or any other source of direct](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-87.jpg)
![Uses n Electroplating uses electrolysis to deposit a thin layer of one metal on Uses n Electroplating uses electrolysis to deposit a thin layer of one metal on](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-88.jpg)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-89.jpg)
![Electrical Work n Remember n A positive value of E is associated with a Electrical Work n Remember n A positive value of E is associated with a](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-90.jpg)
![n Since ΔG = -n. FE n n wmax = -n. FE The cell n Since ΔG = -n. FE n n wmax = -n. FE The cell](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-91.jpg)
![n In an electrolytic cell, we use external energy to bring about a nonspontaneous n In an electrolytic cell, we use external energy to bring about a nonspontaneous](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-92.jpg)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-93.jpg)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-94.jpg)
![A few final notes. . . n Electrical work is expressed in terms of A few final notes. . . n Electrical work is expressed in terms of](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-95.jpg)
![Electricity Misc. n Current n n The flow of the electrons is called current Electricity Misc. n Current n n The flow of the electrons is called current](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-96.jpg)
![Electrolysis Example n Calculate the number of grams of aluminum produced in 1. 00 Electrolysis Example n Calculate the number of grams of aluminum produced in 1. 00](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-97.jpg)
![Note: n n 10 A = 10 C/sec 1. 00 h = 3600 sec Note: n n 10 A = 10 C/sec 1. 00 h = 3600 sec](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-98.jpg)
![= 3. 36 g Al = 3. 36 g Al](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-99.jpg)
- Slides: 99
![Chapter 18 Electrochemistry Homework 9 11 12 13 14 15 17 18 19 Chapter 18 - Electrochemistry Homework: 9, 11, 12, 13, 14, 15, 17, 18, 19,](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-1.jpg)
Chapter 18 - Electrochemistry Homework: 9, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 31, 33, 34, 36, 37, 39, 41, 42, 43, 44, 45, 46, 47, 49, 51, 52, 53, 55, 57, 58, 60, 61, 63, 64, 83, 85, 87, 88, 89
![Oxidation Redux n n Any ion will have an oxidation number equal to Oxidation # Redux n n Any ion will have an oxidation number equal to](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-2.jpg)
Oxidation # Redux n n Any ion will have an oxidation number equal to its charge Oxygen in a compound (except in peroxide, then it is -1) will have an oxidation number of -2 Hydrogen will have an oxidation number of +1 or -1, depending on what it is bonded to Everything else s oxidation number is whatever is necessary to have the sum of the oxidation numbers of the compound add up to the charge on the compound.
![18 2 Balancing Oxidation Reduction Equations n How do we tell whether a 18. 2 – Balancing Oxidation. Reduction Equations n How do we tell whether a](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-3.jpg)
18. 2 – Balancing Oxidation. Reduction Equations n How do we tell whether a given reaction is an oxidation-reduction reaction? n n We do so by keeping track of the oxidation states/number of all the elements involved in the reaction This procedure tells us which elements (if any) are changing oxidation state
![Oxidized vs Reduced n Oxidized n n An atomelement is oxidized when it loses Oxidized vs. Reduced n Oxidized n n An atom/element is oxidized when it loses](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-4.jpg)
Oxidized vs. Reduced n Oxidized n n An atom/element is oxidized when it loses electrons Reduced n An atom/element is reduced when it gains electrons
![Changing Oxidation State n Zns 2 Haq Zn 2aq H 2g n Changing Oxidation State n Zn(s) + 2 H+(aq) Zn 2+(aq) + H 2(g) n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-5.jpg)
Changing Oxidation State n Zn(s) + 2 H+(aq) Zn 2+(aq) + H 2(g) n n We say that Zn is oxidized and H+ is reduced because electrons are transferred from zinc to hydrogen The transfer of electrons that occur produces energy in the form of heat n The reaction is thermodynamically favored, and proceeds spontaneously
![Changing Oxidation We can write the previous reaction in terms of oxidation numbers n Changing Oxidation? We can write the previous reaction in terms of oxidation numbers n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-6.jpg)
Changing Oxidation? We can write the previous reaction in terms of oxidation numbers n Zn(s) + 2 H+(aq) Zn 2+(aq) + H 2(g) 0 +1 +2 0 § By writing the oxidation state for each element above or below the equation, we can see the oxidation state changes that happen n
![In some reactions there is no clear transfer of electrons but oxidation states do In some reactions there is no clear transfer of electrons, but oxidation states do](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-7.jpg)
In some reactions there is no clear transfer of electrons, but oxidation states do change n Consider the combustion of hydrogen n 2 H 2(g) + O 2(g) 2 H 2 O(g) 0 0 +1 -2 § In this reaction, H 2 is oxidized from the 0 to +1 state and oxygen is reduced from the 0 to -2 state § Since oxidation states changes, this is an oxidationn reduction reaction
![n In the previous example n n n 2 H 2g O 2g n In the previous example n n n 2 H 2(g) + O 2(g)](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-8.jpg)
n In the previous example n n n 2 H 2(g) + O 2(g) 2 H 2 O(g) Water is not ionic, therefore there is not a complete transfer of electrons from hydrogen to oxygen Using oxidation states is a convenient bookkeeping method n n However, do not equate the oxidation state of an atom with its actual charge in a chemical compound DO think of the oxidation number as if we had an ionic compound, what would its charge be?
![n In any redox reaction both oxidation and reduction must occur n n In n In any redox reaction, both oxidation and reduction must occur n n In](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-9.jpg)
n In any redox reaction, both oxidation and reduction must occur n n In other words: If one substance is oxidized, then another must be reduced The electrons MUST go from one place to another
![n n Whenever we balance an equation we MUST obey the law of conservation n n Whenever we balance an equation, we MUST obey the law of conservation](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-10.jpg)
n n Whenever we balance an equation, we MUST obey the law of conservation of mass As we balance an oxidation-reduction reaction, we add another rule n n The gains and losses of electrons must be balanced This means if a substance loses a certain number of electrons during a reaction, some other substance must gain that same number of electrons Some redox reactions do this pretty much automatically Others do not.
![HalfReactions n n Although oxidation and reduction take place at the same time often Half-Reactions n n Although oxidation and reduction take place at the same time, often](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-11.jpg)
Half-Reactions n n Although oxidation and reduction take place at the same time, often easier to consider them separate processes Sn 2+(aq) + 2 Fe 3+(aq) Sn 4+(aq) + 2 Fe 2+(aq) n n n We can think of this as two separate process The oxidation of Sn 2+ and the reduction of Fe 3+ Oxidation: n n Reduction: n n Sn 2+(aq) Sn 4+(aq) + 2 e 2 Fe 3+(aq) + 2 e- 2 Fe 2+(aq) Note: During oxidation electrons are a product, while during reduction electrons are a reactant
![n Equations that show either oxidation or reduction alone are called halfreactions n n n Equations that show either oxidation or reduction alone are called half-reactions n n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-12.jpg)
n Equations that show either oxidation or reduction alone are called half-reactions n n In the overall redox reaction, the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction halfreaction When this condition is met and each half-reaction is balanced, the electrons on the two sides cancel when the two half-reactions are then added to give the overall, balanced equation
![18 3 Voltaic Cells n The energy released in a spontaneous redox reaction 18. 3 - Voltaic Cells** n The energy released in a spontaneous redox reaction](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-13.jpg)
18. 3 - Voltaic Cells** n The energy released in a spontaneous redox reaction can be used to perform electrical work n This is done through a voltaic (or galvanic) cell n n A device in which the transfer of electrons takes place through an external pathway Rather than directly between reactants
![n An example of this is when a strip of zinc is placed in n An example of this is when a strip of zinc is placed in](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-14.jpg)
n An example of this is when a strip of zinc is placed in contact with a solution containing Cu 2+ n n n As reaction proceeds, the blue color of the Cu 2+ ions fade and the copper metal deposits onto the zinc. At the same time, the zinc dissolves Zn(s) + Cu 2+(aq) Zn 2+(aq) + Cu(s)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-15.jpg)
![n Same basic principle as last n n n But this time the Zns n Same basic principle as last n n n But this time the Zn(s)](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-16.jpg)
n Same basic principle as last n n n But this time the Zn(s) and Cu 2+ are not in direct contact The Zn is placed in contact with Zn 2+ in one part of the cell The Cu is placed in contact with the Cu 2+ in another cell
![How Does this Work n The reduction of the Cu 2 can occur only How Does this Work? n The reduction of the Cu 2+ can occur only](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-17.jpg)
How Does this Work? n The reduction of the Cu 2+ can occur only by the flow of electrons through an external circuit n n The wire connecting the Zn and Cu strips By physically separating the reduction half of the reaction from the oxidation half n n We create a flow of electrons through an external circuit We have electricity!
![n The two solid metals that are connected by the external circuit are called n The two solid metals that are connected by the external circuit are called](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-18.jpg)
n The two solid metals that are connected by the external circuit are called electrodes n n The electrode at which oxidation occurs is called the anode The electrode at which reduction occurs is called the cathode n n Remember that oxidation and anode both start with a vowel While reduction and cathode both start with a consonant
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-19.jpg)
![Or This is my cat Matthew He is fat and must go on Or This is my cat Matthew. • He is fat, and must go on](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-20.jpg)
Or This is my cat Matthew. • He is fat, and must go on a diet. • Because cats must be reduced. • Cathode is where reduction takes place
![n As in this example the electrodes can be made of materials that participate n As in this example, the electrodes can be made of materials that participate](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-21.jpg)
n As in this example, the electrodes can be made of materials that participate in the reaction n n But typically the electrodes are made of a conducting material that does not gain or lose mass during the reaction Just acts as a surface at which electrons are transferred n Such as platinum or graphite
![n Each compartment in a voltaic cell is called a halfcell n n One n Each compartment in a voltaic cell is called a half-cell. n n One](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-22.jpg)
n Each compartment in a voltaic cell is called a half-cell. n n One half-cell is where the oxidation halfreaction takes place One half-cell is where the reduction halfreaction takes place
![Zns Cu 2aq Zn 2aq Cus n Continuing with our present example Zn(s) + Cu 2+(aq) Zn 2+(aq) + Cu(s) n Continuing with our present example.](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-23.jpg)
Zn(s) + Cu 2+(aq) Zn 2+(aq) + Cu(s) n Continuing with our present example. . . n n Anode: Zn(s) Zn 2+(aq) + 2 e. Cathode: Cu 2+(aq) + 2 e- Cu(s) Electrons become available as the zinc metal is oxidized at the anode n n n Zn is oxidized while Cu 2+ is reduced They flow through the external circuit to the cathode There they are consumed as Cu 2+(aq) is reduced Because Zn(s) is oxidized in the cell, the zinc electrode loses mass, and the concentration of the Zn 2+ solution increases n Similarly, the Cu electrode gains mass and the Cu 2+ solution becomes less concentrated as Cu 2+ is reduced to Cu(s)
![n For a voltaic cell to work n n The solutions in the two n For a voltaic cell to work n n The solutions in the two](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-24.jpg)
n For a voltaic cell to work n n The solutions in the two half-cells must remain electrically neutral As Zn is oxidized, Zn 2+ ions enter the solution n So we need a way to for the positive ions to leave the anode compartment or for negative ions to enter n To keep solution electrically neutral
![Similarly n The reduction of Cu 2 at the cathode removes positive charge from Similarly n The reduction of Cu 2+ at the cathode removes positive charge from](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-25.jpg)
Similarly n The reduction of Cu 2+ at the cathode removes positive charge from the solution n n Leaving an excess of negative charge in that half-cell Therefore, positive ions must migrate into the compartment or negative ions must leave
![How do we do this n A salt bridge allows ions to move from How do we do this? n A salt bridge allows ions to move from](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-26.jpg)
How do we do this? n A salt bridge allows ions to move from one solution to the other n n n Keeping each solution electrically neutral Maintaining the electron flow A salt bridge is simply a tube that contains an electrolyte that will not react with other ions in the cell or with the electrode material
![n As oxidation and reduction proceed at the electrodes ions from the salt bridge n As oxidation and reduction proceed at the electrodes, ions from the salt bridge](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-27.jpg)
n As oxidation and reduction proceed at the electrodes, ions from the salt bridge migrate to neutralize the charge in each cell n No matter how this works, anions will always migrate toward the anode and cations toward the cathode
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-28.jpg)
![n Note the direction of ions in the solution n Note the direction of ions in the solution](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-29.jpg)
n Note the direction of ions in the solution
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-30.jpg)
![n Notice also that in any voltaic cell the electrons will flow from the n Notice also that in any voltaic cell, the electrons will flow from the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-31.jpg)
n Notice also that in any voltaic cell, the electrons will flow from the anode through the external circuit to the cathode. n n Because the negative charged electrons flow from the anode to the cathode, the anode in a voltaic cell is labeled with a negative sign And the cathode is labeled with a positive sign n Think of the electrons as being attracted to the positive cathode from the negative anode through the external circuit
![18 4 Cell EMF Under Standard Conditions n Why do electrons transfer spontaneously 18. 4 - Cell EMF Under Standard Conditions n Why do electrons transfer spontaneously](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-32.jpg)
18. 4 - Cell EMF Under Standard Conditions n Why do electrons transfer spontaneously from a Zn atom to a Cu 2+ ion? n n In this section we will look at the main driving force that pushes the electrons through the external circuit in a voltaic cell The chemical processes that make up any voltaic cell are spontaneous n n Electrons flow from the anode of a voltaic cell to the cathode because of a difference in potential energy The potential energy of the electrons is higher in the anode than in the cathode n So they spontaneously move to lower their energy
![n The difference in potential energy per electrical charge called potential difference between two n The difference in potential energy per electrical charge (called potential difference) between two](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-33.jpg)
n The difference in potential energy per electrical charge (called potential difference) between two electrodes is measured in volts n One volt (V) is the potential energy required to impart 1 J of energy to a charge of 1 coulomb (C) n Remember that an electron has a charge of 1. 60 x 1019 C.
![n The potential difference between the two electrodes in a voltaic cell provides the n The potential difference between the two electrodes in a voltaic cell provides the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-34.jpg)
n The potential difference between the two electrodes in a voltaic cell provides the driving force to push the electrons through the external circuit n We will therefore call this potential difference the electromotive force, or emf n n electromotive means to “cause electron motion” The emf of a cell (Ecell) is also called the cell potential. n n Because Ecell is measured in volts, we often refer to this as the cell voltage. For any cell reaction that proceeds spontaneously (like in a voltaic cell), the cell potential will be positive.
![n The actual emf of a particular cell depends on the n n n n The actual emf of a particular cell depends on the n n n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-35.jpg)
n The actual emf of a particular cell depends on the n n n specific reactions that take place at the cathode and anode concentrations of the reactants and products temperature (which we will generally assume to be 25ºC) unless told otherwise n We will focus on this temperature, and refer to this as standard conditions
![At standard conditions n If we have 1 M concentrations for reactants and products At standard conditions n If we have 1 M concentrations for reactants and products](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-36.jpg)
At standard conditions n If we have 1 M concentrations for reactants and products and 1 atm pressure for gases and 25ºC n Then the emf for these standard conditions is called the standard emf or the standard cell potential n n Eºcell For example, for the Zn-Cu voltaic cell seen earlier, the standard cell potential is +1. 10 V
![Standard Reduction HalfCell Potentials n The emf of a voltaic cell depends on the Standard Reduction (Half-Cell) Potentials n The emf of a voltaic cell depends on the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-37.jpg)
Standard Reduction (Half-Cell) Potentials n The emf of a voltaic cell depends on the particular cathode and anode half-cell involved n n n We COULD tabulate the standard potentials for all possible combinations However, that is not necessary We assign a standard potential to each individual half-cell, and use these half-cells to determine Eºcell
![n The cell potential is the difference between the two electrode potentials n n n The cell potential is the difference between the two electrode potentials n n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-38.jpg)
n The cell potential is the difference between the two electrode potentials n n One from the cathode, one from the anode Due to convention, the potential associated with each electrode is the potential for reduction to occur at that electrode n n Thus, standard electrode potentials are found for reduction reactions Called standard reduction potentials, Eºred
![Calculating Eºcell n The cell potential Eºcell is given by the standard reduction potential Calculating Eºcell n The cell potential (Eºcell) is given by the standard reduction potential](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-39.jpg)
Calculating Eºcell n The cell potential (Eºcell) is given by the standard reduction potential of the cathode, Eºred(cathode), minus the standard reduction potential of the anode, Eºred(anode) n Eºcell = Eºred(cathode) - Eºred(anode)
![n Since every voltaic cell involves two halfcells it is impossible to measure the n Since every voltaic cell involves two halfcells, it is impossible to measure the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-40.jpg)
n Since every voltaic cell involves two halfcells, it is impossible to measure the standard reduction potential directly n n All based off of a certain reference half-reaction The reference half-reaction is the reduction of H+(aq) to H 2(g) under standard conditions n n 2 H+(aq, 1 M) + 2 e- H 2(g, 1 atm) Eºred = 0 V The standard reduction potential of this reaction is defined to be 0 V
![n Various standard reduction potentials for halfreactions are found on pg 873 table 18 n Various standard reduction potentials for half-reactions are found on pg. 873 (table 18.](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-41.jpg)
n Various standard reduction potentials for half-reactions are found on pg. 873 (table 18. 1) or in Appendix II (Table D)
![n Because electrical potential measures potential energy per electrical charge standard reduction potentials are n Because electrical potential measures potential energy per electrical charge, standard reduction potentials are](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-42.jpg)
n Because electrical potential measures potential energy per electrical charge, standard reduction potentials are intensive properties n n n This means that if we increased the amount of chemicals in the redox reaction, we would increase both energy and charges involved But the ratio of the energy and charge would remain constant Therefore, changing the stoichiometric coefficients in a half-reaction does not affect the value of the standard reduction potential
![Example n Given the standard reduction potential of Zn 2 to Zns is 0 Example n Given the standard reduction potential of Zn 2+ to Zn(s) is -0.](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-43.jpg)
Example n Given the standard reduction potential of Zn 2+ to Zn(s) is -0. 76 V, calculate the Eºred for the reduction of Cu 2+ to Cu for the following voltaic cell n n Zn-Cu 2+ cell Zn(s) + Cu 2+(aq, 1 M) Zn 2+(aq, 1 M) + Cu(s) n n Eºcell = 1. 10 V Cu 2+(aq, 1 M) + 2 e- Cu(s)
![n Eºcell Eºredcathode Eºredanode n n Zn is oxidized and is therefore n Eºcell = Eºred(cathode) - Eºred(anode) n n Zn is oxidized, and is therefore](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-44.jpg)
n Eºcell = Eºred(cathode) - Eºred(anode) n n Zn is oxidized, and is therefore the anode We were also given the Eºcell 1. 10 V = Eºred(cathode) -(-0. 76) Eºred(cathode) = 1. 10 V - 0. 76 = 0. 34 V
![Example 2 n n Using the standard reduction potentials provided calculate the standard emf Example 2 n n Using the standard reduction potentials provided, calculate the standard emf](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-45.jpg)
Example 2 n n Using the standard reduction potentials provided, calculate the standard emf for the voltaic cell given by this reaction Cr 2 O 72 -(aq) + 14 H+(aq) + 6 I-(aq) 2 Cr 3+(aq) + 3 I 2(s) + 7 H 2 O(l)
![Cr 2 O 72 aq 14 Haq 6 Iaq 2 Cr 3aq Cr 2 O 72 -(aq) + 14 H+(aq) + 6 I-(aq) 2 Cr 3+(aq)](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-46.jpg)
Cr 2 O 72 -(aq) + 14 H+(aq) + 6 I-(aq) 2 Cr 3+(aq) + 3 I 2(s) + 7 H 2 O(l) n First we need to identify the half-reactions involved n Cr 2 O 72 -(aq) + 14 H+(aq) + 6 e- 2 Cr 3+(aq) + 7 H 2 O(l) n Cathode Eºred(cathode) = 1. 33 V 6 I-(aq) 3 I 2(s) + 6 en n n Anode n Eºred(anode) = 0. 54 V n Now we plug into equation n Eºcell = Eºred(cathode) - Eºred(anode)
![Eºcell Eºredcathode Eºredanode n Eºcell 1 33 V 0 54 V Eºcell = Eºred(cathode) Eºred(anode) n Eºcell = 1. 33 V - 0. 54 V](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-47.jpg)
Eºcell = Eºred(cathode) Eºred(anode) n Eºcell = 1. 33 V - 0. 54 V = 0. 79 V
![Analyzing the Equation n For each of the halfcells in the voltaic cell the Analyzing the Equation n For each of the half-cells in the voltaic cell, the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-48.jpg)
Analyzing the Equation n For each of the half-cells in the voltaic cell, the standard reduction potential gives a measure of the driving force for the reaction to occur n n This means that the more positive the value of Eºred, the greater the driving force for reduction In all voltaic cells the reaction at the cathode has a more positive value of Eºred than the reaction at the anode. n n So the greater driving force of the cathode half-reaction is used to force the anode reaction to occur in reverse (oxidation) Since we can think of oxidation as the reverse of reduction, while all values are given for oxidation
![n We know that Eºcell is the difference between the standard reduction potential of n We know that Eºcell is the difference between the standard reduction potential of](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-49.jpg)
n We know that Eºcell is the difference between the standard reduction potential of the cathode reaction and the standard reduction potential of the anode n We can think of Eºcell as the net driving force that pushes the electrons through the external circuit
![Example n A voltaic cell is based on the following two halfreactions n n Example n A voltaic cell is based on the following two half-reactions: n n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-50.jpg)
Example n A voltaic cell is based on the following two half-reactions: n n Cd 2+(aq) + 2 e- Cd(s) Sn 2+(aq) + 2 e- Sn(s) Find the half-reactions that occur at the anode and cathode Find the standard cell potential
![Finding the HalfReactions n Look up the Eºred for the two halfreactions n n Finding the Half-Reactions n Look up the Eºred for the two half-reactions n n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-51.jpg)
Finding the Half-Reactions n Look up the Eºred for the two half-reactions n n Eºred(Cd 2+/Cd) = -0. 403 V Eºred(Sn 2+/Sn) = -0. 136 V n n n Since the standard reduction potential for Sn 2+ is more positive (less negative) the reduction of Sn 2+ occurs at the cathode Cathode: Sn 2+(aq) + 2 e- Sn(s) Anode: Cd(s) Cd 2+(aq) + 2 e-
![Determine the standard cell potential n Eºcell Eºredcathode Eºredanode n Eºcell Determine the standard cell potential n Eºcell = Eºred(cathode) - Eºred(anode) n Eºcell =](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-52.jpg)
Determine the standard cell potential n Eºcell = Eºred(cathode) - Eºred(anode) n Eºcell = (-0. 136 V) - (-0. 403 V) = 0. 267 V
![How to tell how easily something is reducedoxidized n Generally the more positive the How to tell how easily something is reduced/oxidized n Generally, the more positive the](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-53.jpg)
How to tell how easily something is reduced/oxidized n Generally, the more positive the Eºred value for a half-reaction, the greater the tendency for the reactant of the half-reaction to be reduced n Which means a greater chance of it to oxidize another species
![Easily reduced n Among the most easily reduced we find the halogens O 2 Easily reduced n Among the most easily reduced we find the halogens, O 2](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-54.jpg)
Easily reduced n Among the most easily reduced we find the halogens, O 2 and oxyanions n n Such as Mn. O 4 -, Cr 2 O 72 - and NO 3 All have a central atom with a large, positive oxidation state
![Easily oxidized n Like acidbase strength the strongest oxidizers are the worst at being Easily oxidized n Like acid-base strength, the strongest oxidizers are the worst at being](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-55.jpg)
Easily oxidized n Like acid-base strength, the strongest oxidizers are the worst at being reduced n n n So the half-reaction with the smallest reduction potential is the most easily reversed Means they most easily give up electrons to other species So they are easily oxidized
![n Easily oxidized things include n n H 2 and active metals such as n Easily oxidized things include: n n H 2, and active metals such as](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-56.jpg)
n Easily oxidized things include: n n H 2, and active metals such as the alkali metals and the alkaline earth metals Other metals whose cations have negative Eºred values are also used n n Such as Zn and Fe Often difficult to store solutions of these because of the presence of O 2 (a chemical which is easily reduced) in the air
![Table 18 1 pg 873 n The list orders the ability of substances to Table 18. 1, pg. 873 n The list orders the ability of substances to](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-57.jpg)
Table 18. 1, pg. 873 n The list orders the ability of substances to act as an oxidizing or reducing agent n Remember, the more positive the value of Eºred, the more it will tend to be reduced
![18 5 Free Energy Equilibrium and Redox Reactions n Voltaic cells use redox 18. 5 - Free Energy, Equilibrium and Redox Reactions n Voltaic cells use redox](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-58.jpg)
18. 5 - Free Energy, Equilibrium and Redox Reactions n Voltaic cells use redox reaction to proceed spontaneously n n Any reaction that occurs in a voltaic cell to produce a positive emf must be spontaneous So we can determine spontaneity of a redox reaction by deciding if it produces a positive emf
![n We start by applying a previous equation to all redox reactions not just n We start by applying a previous equation to all redox reactions, not just](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-59.jpg)
n We start by applying a previous equation to all redox reactions, not just limiting it to voltaic cells n Eºcell = Eºred(cathode) - Eºred(anode) n Becomes n Eº= Eºred(reducing process) - Eºred(oxidation process)
![n Eº Eºredreducing process Eºredoxidation process n n A positive value of E n Eº= Eºred(reducing process) - Eºred(oxidation process) n n A positive value of E](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-60.jpg)
n Eº= Eºred(reducing process) - Eºred(oxidation process) n n A positive value of E tells us this is a spontaneous process While a negative value of E indicates a nonspontaneous one Use E to represent emf under nonstandard conditions Use Eº to represent emf under standard conditions
![Example n Using standard reduction potentials determine whether or not the following reactions are Example n Using standard reduction potentials, determine whether or not the following reactions are](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-61.jpg)
Example n Using standard reduction potentials, determine whether or not the following reactions are spontaneous under standard conditions n n Cu(s) + 2 H+(aq) Cu 2+(aq) + H 2(g) Cl 2(g) + 2 I-(aq) 2 Cl-(aq) + I 2(s)
![Cus 2 Haq Cu 2aq H 2g n In this reaction Cu Cu(s) + 2 H+(aq) Cu 2+(aq) + H 2(g) n In this reaction, Cu](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-62.jpg)
Cu(s) + 2 H+(aq) Cu 2+(aq) + H 2(g) n In this reaction, Cu is oxidized to Cu 2+, while H+ is reduced to H 2 n n Half-reactions would be Reduction: n n Eºred = 0 V Oxidation: n n 2 H+(aq) + 2 e- H 2(g) Cu(s) Cu 2+(aq) + 2 e- Eºred = 0. 34 V Therefore, using Eº= Eºred(reducing process) Eºred(oxidation process) n n Eº= (0 V) - (0. 34 V) = -0. 34 V Because Eº is negative, the reaction is not spontaneous in this direction
![Cl 2g 2 Iaq 2 Claq I 2s n n Cl 2 Cl 2(g) + 2 I-(aq) 2 Cl-(aq) + I 2(s) n n Cl 2](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-63.jpg)
Cl 2(g) + 2 I-(aq) 2 Cl-(aq) + I 2(s) n n Cl 2 is reduced, while I-is oxidized Reduction: n n Eºred = 1. 36 V Oxidation: n n Cl 2(g) + 2 e- 2 Cl-(aq) 2 I-(aq) I 2(s) + 2 e- Eºred = 0. 54 V Eº = (1. 36 V) - (0. 54 V) = 0. 82 V n This reaction would be spontaneous, and could be used to build a voltaic cell
![Activity Series of Metals n Remember when dealing with single and double replacement reactions Activity Series of Metals n Remember, when dealing with single and double replacement reactions](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-64.jpg)
Activity Series of Metals n Remember, when dealing with single and double replacement reactions that you consult the activity series to determine if one metal will replace another n n If metal A is above metal B on the table, then A will replace B Any metal in the activity series will be oxidized by the ions of any metal below it n n n Activity series really shows the oxidation reactions of the metals, ordered from strongest reducing agent at the top to the weakest reducing agent at the bottom So we can use the relative strength of oxidizers to predict the results of displacement reactions with metals In other words, activity series is the OPPOSITE of the reduction potential series.
![EMF and ΔG n The change in Gibbs free energy ΔG measures the spontaneity EMF and ΔG n The change in Gibbs free energy, ΔG measures the spontaneity](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-65.jpg)
EMF and ΔG n The change in Gibbs free energy, ΔG measures the spontaneity of a process at constant temperature and pressure n Because EMF, E, of a redox reaction is spontaneous, there is a relationship between these two terms
![Relationship n ΔG n FE n n is a positive number without units Relationship n ΔG = -n. FE n n is a positive number without units](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-66.jpg)
Relationship n ΔG = -n. FE n n is a positive number without units n n F is called Faraday’s constant n n Represents the number of electrons transferred in the reaction Equal to 96, 485 J/V-mol The units of ΔG is J/mol n The /mol means per mole of reaction as written
![ΔG n FE n Both n and F are positive numbers n So ΔG = -n. FE n Both n and F are positive numbers n So](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-67.jpg)
ΔG = -n. FE n Both n and F are positive numbers n So a positive value of E means a negative ΔG n n So a positive value of E and a negative value for ΔG both indicate a spontaneous reaction Now, since ΔGº = -RT(ln. K) n We can now relate the standard emf to the equilibrium constant for the reaction
![Example n Use the standard reduction potentials to calculate the standard freeenergy change ΔGº Example n Use the standard reduction potentials to calculate the standard free-energy change, ΔGº](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-68.jpg)
Example n Use the standard reduction potentials to calculate the standard free-energy change, ΔGº and the equilibrium constant, K, at room temperature (T = 298 K) for the reaction n 4 Ag(s) + O 2(g) + 4 H+(aq) 4 Ag+(aq) + 2 H 2 O(l)
![Find Eº First n To find Eº we need to break into halfreactions and Find Eº First n To find Eº we need to break into halfreactions and](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-69.jpg)
Find Eº First n To find Eº we need to break into halfreactions and obtain Eºred values n Reduction: n n n O 2(g) + 4 H+(aq) + 4 e- 2 H 2 O(l) Eºred = +1. 23 V Oxidation: n n 4 Ag(s) 4 Ag+(aq) + 4 e. Eºred = +0. 80 V
![n n Eº 1 23 V 0 80 V 0 43 V n n Eº= 1. 23 V - 0. 80 V = 0. 43 V](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-70.jpg)
n n Eº= 1. 23 V - 0. 80 V = 0. 43 V Now we use ΔGº = -n. FEº n n ΔGº = -(4)(96485)(0. 43) = -1. 7 x 105 J/mol = -170 k. J/mol
![Now solve for K n n n ΔGº RTln K K eΔGºRT Now solve for K n n n ΔGº = -RT(ln. K) K = e-ΔGº/RT](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-71.jpg)
Now solve for K n n n ΔGº = -RT(ln. K) K = e-ΔGº/RT = e-1. 7 x 10^5/(8. 314 x 298) K = 9 x 1029 n n n Why would we do this? Such a large K value very hard to measure However, the voltage measured is easy to measure.
![Summary Triangle Summary Triangle](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-72.jpg)
Summary Triangle
![18 6 Cell Potential and Concentration n As a voltaic cell is discharged 18. 6 - Cell Potential and Concentration n As a voltaic cell is discharged,](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-73.jpg)
18. 6 - Cell Potential and Concentration n As a voltaic cell is discharged, the reactants are consumed and products generated n n Which changes the concentrations The emf keeps dropping until E = 0, at which point we say the cell is “dead” n n n At this point the concentrations of products and reactants will cease to change They are at equilibrium This section will look at how emf changes with concentration, which will show nonstandard conditions
![In General n If the concentrations of reactants increase relative to the concentrations of In General n If the concentrations of reactants increase relative to the concentrations of](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-74.jpg)
In General n If the concentrations of reactants increase relative to the concentrations of the products, the emf increases n Whereas if the concentrations of products increase relative to the reactants, the emf decreases n n Which is what happens when a batter goes dead! Think of this as this: The more reactants you have, the stronger the cell. The more product you have, the weaker.
![n n n Note X is copper V 47 V Overall reaction is n n n Note: X is copper V =. 47 V Overall reaction is:](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-75.jpg)
n n n Note: X is copper V =. 47 V Overall reaction is: Cu 2+ + Pb Cu + Pb 2+
![Example n n During a laboratory session students set up the electrochemical cell shown Example n n During a laboratory session, students set up the electrochemical cell shown.](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-76.jpg)
Example n n During a laboratory session, students set up the electrochemical cell shown. For each of the following three scenarios, choose the correct value of the cell voltage and justify your choice. Ecell =0. 47 V
![n A student bumps the cell setup resulting in the salt bridge losing contact n A student bumps the cell setup, resulting in the salt bridge losing contact](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-77.jpg)
n A student bumps the cell setup, resulting in the salt bridge losing contact with the solution in the cathode compartment. Is V equal to 0. 47, less than 0. 47 or greater than 0. 47?
![n A student spills a small amount of 0 5 M Na 2 SO n A student spills a small amount of 0. 5 M Na 2 SO](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-78.jpg)
n A student spills a small amount of 0. 5 M Na 2 SO 4(aq) into the compartment with the Pb electrode, resulting in the formation of a precipitate. Is V less than 0. 47 or is V greater than 0. 47 or is V equal to 0. 47?
![n After the laboratory session is over a student leaves the switch closed The n After the laboratory session is over, a student leaves the switch closed. The](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-79.jpg)
n After the laboratory session is over, a student leaves the switch closed. The next day, the student opens the switch and reads the voltmeter. Is V less than 0. 47, great than. 47 or is V equal to 0. 47 ?
![Concentration Cells n In each of the voltaic cells we have seen the reactive Concentration Cells n In each of the voltaic cells we have seen, the reactive](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-80.jpg)
Concentration Cells n In each of the voltaic cells we have seen, the reactive species at the anode has been difference from the one at the cathode n n Cell emf depends on concentration Which means that a voltaic cell can be made using the same species at both the anode and cathode n n As long a the concentrations are different A cell based only on the emf generated because of difference in concentration is called a concentration cell
![Features of a Concentration Cell n Still requires two vessels connected by a salt Features of a Concentration Cell n Still requires two vessels, connected by a salt](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-81.jpg)
Features of a Concentration Cell n Still requires two vessels, connected by a salt bridge, and an external wire
![Overview n n In a concentration cell the more dilute halfcell will seek to Overview n n In a concentration cell, the more dilute halfcell will seek to](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-82.jpg)
Overview n n In a concentration cell, the more dilute halfcell will seek to be more concentrated (increasing # of ions) The more concentrated half cell will seek to become more dilute (decreasing # of ions)
![n So with this cell n The 01 M halfcell is more dilute So n So with this cell n The. 01 M half-cell is more dilute. So](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-83.jpg)
n So with this cell n The. 01 M half-cell is more dilute. So it will create ions. n n n Cu 2++2 e. Oxidation taking place in this cell The 2 M half-cell is more concentrated. So it will lose ions. n n Cu 2++2 e- Cu Reduction taking place in this cell
![18 8 Electrolysis n Voltaic cells are based on spontaneous redox reactions n 18. 8 - Electrolysis** n Voltaic cells are based on spontaneous redox reactions n](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-84.jpg)
18. 8 - Electrolysis** n Voltaic cells are based on spontaneous redox reactions n It is possible to use electrical energy to cause nonspontaneous redox reactions to occur n n For example, we can use electricity to decompose molten sodium chloride into its component elements 2 Na. Cl(l) 2 Na(l) + Cl 2(g) A nonspontaneous process driven by outside electrical energy is called an electrolysis reaction These reactions take place is electrolytic cells
![Electrolytic Cell n An electrolytic cell is made of two electrodes in molten salt Electrolytic Cell n An electrolytic cell is made of two electrodes in molten salt](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-85.jpg)
Electrolytic Cell n An electrolytic cell is made of two electrodes in molten salt or a solution n A battery or some other source of direct electrical current acts as an electron pump n n Pushing electrons into one electrode and pulling them from the other Just like in a voltaic cell, the electrode at which reduction occurs is called the cathode (and where oxidation occurs is the anode)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-86.jpg)
![Things to notice n In a voltaic cell or any other source of direct Things to notice n In a voltaic cell (or any other source of direct](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-87.jpg)
Things to notice n In a voltaic cell (or any other source of direct current) n n So the electrode of the electrolytic cell that is connected to the negative terminal of the voltage source is the cathode n n electrons move from the negative terminal It receives the electrons that are used to reduce the substance The electrons that are removed during the oxidation process at the anode travel to the positive terminal of the voltage source
![Uses n Electroplating uses electrolysis to deposit a thin layer of one metal on Uses n Electroplating uses electrolysis to deposit a thin layer of one metal on](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-88.jpg)
Uses n Electroplating uses electrolysis to deposit a thin layer of one metal on another metal n n Metal in solution becomes deposited onto cathode Only requires a small voltage source to accomplish
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-89.jpg)
![Electrical Work n Remember n A positive value of E is associated with a Electrical Work n Remember n A positive value of E is associated with a](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-90.jpg)
Electrical Work n Remember n A positive value of E is associated with a negative value of free-energy change n n And thus spontaneous processes We also know that for any spontaneous process ΔG measures the maximum useful work, wmax for the process n ΔG = wmax
![n Since ΔG n FE n n wmax n FE The cell n Since ΔG = -n. FE n n wmax = -n. FE The cell](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-91.jpg)
n Since ΔG = -n. FE n n wmax = -n. FE The cell emf for a voltaic cell is positive, so wmax will be negative n Negative work means work is being done by the system on the surroundings
![n In an electrolytic cell we use external energy to bring about a nonspontaneous n In an electrolytic cell, we use external energy to bring about a nonspontaneous](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-92.jpg)
n In an electrolytic cell, we use external energy to bring about a nonspontaneous electrochemical reaction n Therefore ΔG is positive and Ecell is negative To force the process to occur, we need to apply an external potential, Eext, whose magnitude is larger than Ecell When Eext is applied to a cell, the surroundings are doing work on the system, and therefore n n w = n. FEext So we can now calculate maximum work from a voltaic cell, and the minimum work needed for electrolysis
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-93.jpg)
![](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-94.jpg)
![A few final notes n Electrical work is expressed in terms of A few final notes. . . n Electrical work is expressed in terms of](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-95.jpg)
A few final notes. . . n Electrical work is expressed in terms of watts time n n n 1 W = 1 J/s Therefore, a watt-second = 1 J The unit power companies use is the kilowatthour (k. Wh), which is equal to 3. 6 x 106 J
![Electricity Misc n Current n n The flow of the electrons is called current Electricity Misc. n Current n n The flow of the electrons is called current](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-96.jpg)
Electricity Misc. n Current n n The flow of the electrons is called current Measured in the unit of the ampere (A) n n Is really Coulomb/second Faraday’s constant n F = 96500 coulombs per mole of electrons n Which means that 1 mole of electrons has a charge of 96500 C
![Electrolysis Example n Calculate the number of grams of aluminum produced in 1 00 Electrolysis Example n Calculate the number of grams of aluminum produced in 1. 00](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-97.jpg)
Electrolysis Example n Calculate the number of grams of aluminum produced in 1. 00 h by the electrolysis of molten Al. Cl 3 if the electrical current is 10. 0 A. n Note: Al 3+ + 3 e- Al
![Note n n 10 A 10 Csec 1 00 h 3600 sec Note: n n 10 A = 10 C/sec 1. 00 h = 3600 sec](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-98.jpg)
Note: n n 10 A = 10 C/sec 1. 00 h = 3600 sec There are 3 moles of electrons per mole Al Faradays constant brings this together (96000 C/mol e)
![3 36 g Al = 3. 36 g Al](https://slidetodoc.com/presentation_image/52ba901c83a89e9eec7d6d692645e7da/image-99.jpg)
= 3. 36 g Al
Homework oh homework i hate you you stink
Homework oh homework i hate you you stink
Jack prelutsky homework oh homework
Jack prelutsky homework oh homework
Alitteration definition
Literal and figurative language
Ap chemistry chapter 18 electrochemistry test
Introduction to electrochemistry
Ap chem electrochemistry review
Cell chapter 21
Applications of electrolytic cell
Chapter 20 electrochemistry
Faraday's constant
Transport number chemistry
Junction potential
Equilibrium constant formula electrochemistry
What is electrochemistry in chemistry
Electrochemistry balancing equations
Electrochemistry stoichiometry
Balancing redox reactions khan academy
Aee cd20f
Balancing complex redox reactions
Ap chemistry electrochemistry
What are redox reactions
E cell formula
Interfacial method in electrochemistry
Electrolysis table
Fundamentals of electrochemistry
What is electrochemistry
What is electrochemistry in chemistry
Line notation galvanic cell
Diagonal rule electrochemistry
Polarization in electrochemistry
Electrochemistry lesson
Branches of electrochemistry
Chemistry
Mass transport electrochemistry
Electrochemistry formulas
Fundamentals of electrochemistry
Electrochemistry
Diagonal rule electrochemistry
Ir drop in electrochemistry
Electrochemistry balancing equations
Analytical electrochemistry
Red cat electrochemistry
The cell reaction for the zn-h2 cell is
Oxidation or reduction
Electrochemistry
Geometry homework answers
Two-column journal
Yesterday's homework
I did my homework last night
Homework 3 multiplying binomials and trinomials
Find the measure of arc mk
Unit 8 rational functions homework 1
Who tf invented homework
Unit 6 radical functions homework 3
Unit 5: polynomial functions
Congruence theorem
Unit 3 parallel, perpendicular lines answer key
Types of home work
Two way tables homework 5
The homework machine shel silverstein poems
Main idea 2 being a clown answer key
Take out your homework
Spelling words for grade 2
Simplifying radicals homework
Macbeth homework tasks
Tangents secants and angle measures
Special right triangles radians
How often do you do your homework
The teacher assigns homework after 3/4 of the lessons
Two events are independent
Chemistry online homework
Central tendency and spread homework
Master asl unit 7
What is the perimeter
Second grade homework
Signing naturally - homework 5:8
Signing naturally unit 5
Signing naturally page 144
Signing naturally homework 3:10
Signing naturally homework 4:1
Signing naturally homework 5:7
The elevator incident by melinda answers
Write an email to a teacher
My maths co uk
Homework
Homework
Unsullied definition to kill a mockingbird
Irrational roots
Homework 122
Reading homework ideas
Alternate interior angles conjecture
Drama homework
Two roads diverged in a yellow wood figure of speech
What is literal language
Homework 259
Straight line motion revisited homework
Homework è countable or uncountable