1. Goes with chapter 21: Silberberg’s Principles of General Chemistry Mrs. Laura Peck, 2013 1

2. Objectives/Study guide  Identify and compare the two types of electrochemical cells: galvanic and electrolytic  Draw and label a galvanic cell, including labeling the electrodes, the flow of electrons, and the flow of ions  Write half-reactions and determine which reaction occurs at the anode and which reaction occurs at the cathode.  Give the line notation for a galvanic cell or write a balanced redox reaction from the given line notation.  Calculate the cell potential for a galvanic cell and an electrolytic cell.  Determine if a reaction is spontaneous from its cell potential.  Calculate the cell potential under nonstandard conditions when the solutions are not 1M. This involves the use of the Nernst equation.  Determine the strengths of oxidizing agents and reducing agents.  Draw and label an electrolytic cell.  Determine the reactions which occur at the anode and the cathode during electrolysis.  Perform stoichiometric calculations involving electrolysis. 2

3. Basic differences in cells.  Electrochemistry is the study of the interchange of electrical and chemical energy.  There are two types of electrochemical cells, galvanic and electrolytic.  In galvanic cells, spontaneous redox reactions generate electric current.  In electrolytic cells, a nonspontaneous chemical reaction occurs with the application of an electric current. 3

4. Line Notation  A galvanic cell can be abbreviated with line notation.  Reactant/product II reactant/product (anode reaction) (cathode reaction)  The salt bridge is indicated by the symbol II 4 Example #1: Give the correct line notation for the Galvanic cell pictured. Zn/Zn 2+ II Cu 2+ /Cu

5. AP tips:  Here are some mnemonic devices to help you remember some facts about electrochemistry and redox reactions.  ‘LEO’ goes ‘GER’ means Loss of Electrons is Oxidation and Gain of Electrons is Reduction.  To recall what happens at the anode and the cathode: RedCat and AnOx means Reduction Occurs at the cathode and oxidation occurs at the anode.  To know the migration of ions toward the electrodes for both types of cells, ‘Cat’ ions move to the ‘Cat’ode and ‘An’ ions move to the ‘An’ode.  You will be provided with a table of standard reduction potentials on the AP test.  You should be able to sketch a galvanic cell and label the electrodes, the flow of electrons, and the flow of ions. 5

6. Standard reduction potentials  The cell potential, E cell, is the potential of the cell to do work on its surroundings by driving an electric current through a wire.  By definition, a potential of 1 volt is produced when 1 joule of energy moves 1 coulomb of electric charge across a potential.  The magnitude of the cell potential is a measure of the driving force behind an electrochemical reaction.  Sometimes it is referred to as the electromotive force or emf.  Tables of reduction potentials give standard voltages for reduction half-reactions measured at standard conditions of 1 atm, 1 molar solution, and 25*C  The reaction occurring in a galvanic cell can be broken down into an oxidation half-reaction and a reduction half-reaction.  Using the table of standard reduction potentials in your text, you can calculate the cell potential of the overall reaction. 6

7. 7 Example #2: consider a galvanic cell based on the reaction: Al + NI 2+  Al 3+ + Ni Give the balanced cell reaction and calculate the cell potential, E 0 cell For the reaction. Step 1: Write the oxidation & reduction Half-reactions. Step 2: For the reduction half-reaction Look up the potential in your book. Step 3: For the oxidation half-reaction, E 0 ox = -E 0 red Step 4: the cell potential for the overall Reaction is equal to the sum of the Reduction potential, E 0 red, and the Oxidation potential, E 0 ox. Step 5: to obtain the balanced cell Reaction, you must make sure that the Electrons lost equal the electrons gained. When multiplying the half-reactions through By a coefficient, do not change the value of E 0 Oxidation: Al  Al 3+ + 3e- Reduction: Ni 2+ + 2e-  Ni Ni 2+ + 2e-  Ni E 0 red = -0.23V Oxidation: Al  Al 3+ + 3e- E 0 ox = -E 0 red  -(-1.66V) = +1.66V E 0 cell = E 0 ox + E 0red E 0 cell = -0.23V + 1.66V = 1.43V 3[Ni 2+ + 2e-  Ni] E 0 red = -0.23V 2[Al  Al 3+ + 3e-] E 0 ox = +1.66V 3Ni 2+ + 2Al  3Ni + 2Al 3+ E 0 cell = 1.43V

8. Spontaneous Reactions  Gibbs free energy, ΔG°, can be calculated from the cell potential, E 0 cell.  ΔG° = -nFE 0 cell  Faraday’s constant, F, has a value of 96,485 C/mol e-  The number of moles of electrons transferred in a redox reaction is represented by n  A spontaneous reaction is one that has a negative value for ΔG° or a positive value for E 0 cell  You may be asked if an element or ionic species is capable of reducing another element or ion.  To determine if the reaction will occur, write the half-reactions and calculate the cell potential. 8

9.  Example #3: Will 1M HCl dissolve silver metal and form Ag + solution? 9 Write half-reactions and Calculate E 0 cell 2H + + 2e-  2H 2 E 0 red = 0.00V 2Ag  2Ag + + 2e- E 0 ox = -0.80V 2H + + 2Ag  H 2 + 2Ag + E 0 cell = -0.80V The negative value for E 0 cell indicates that the reaction will not occur.

10.  Example #4: Bromine, Br 2, can oxidize iodide, I-, to iodine, I 2. However, Br 2 cannot oxidize chloride, Cl-, to chlorine, Cl 2. Explain why the first reaction occurs yet, the second one does not. 10 Begin by writing the appropriate Half-reactions. Then calculate The cell potential for the overall Reaction. First, the reaction in which Br 2 Oxidizes I - Then the reaction in which Br 2 Oxidizes Cl - Br 2 + 2e-  2Br- E 0 red = 1.09V 2I-  I 2 + 2e- E 0 ox = -0.54V Br 2 + 2I-  2Br- + I 2 E 0 cell = 0.55V This reaction occurs, E 0 cell is + Br 2 + 2e-  2Br- E 0 red = 1.09V 2Cl-  Cl 2 + 2e- E 0 ox = -1.36V Br 2 + 2Cl-  2Br- + Cl 2 E 0 cell = -0.27V This reaction does not occur, E 0 cell is negative

11. Cell Dependence on Concentration  The galvanic cell represented by the reaction:  3Ni 2+ + 2Al  3Ni + 2Al 3+  Has a cell potential, E 0 cell, equal to 1.43V under standard conditions (all solutions are 1M)  Increasing the concentration of Ni 2+ will shift the reaction to the right by Le Chatelier’s principle, increasing the driving force on the electrons and increasing the cell potential.  The relationship between the cell potential and concentrations at 25°C is given by the Nernst equation:  E cell = E 0 cell – (0.0591/n)log Q  The cell potential, E cell, is for nonstandard conditions.  The moles of electrons transferred are represented by n  The mass action quotient is represented by Q 11

12.  Example #5: Calculate the cell potential for the reaction:  3Ni 2+ + 2Al  3Ni + 2Al 3+  In which [Al 3+ ] = 2.00M and [Ni 2+ ] = 0.750M  (you already know the E 0 cell = 1.43V and the number of moles of electrons transferred, n, equals 6) 12 Q = [Al 3+ ] 2 / [Ni 2+ ] 3 = (2.00) 2 / (0.750) 3 = 9.48 E cell = 1.43V – (0.0591/6) log 9.48 = 1.33V

13. Determining the Strength of Oxidizing and Reducing Agents.  You may be asked to list atoms or ions in order of increasing strength as reducing agents or oxidizing agents.  For a substance to be oxidized, it must lose electrons and another substance must gain electrons because oxidation and reduction always occur together.  The substance that causes another substance to be oxidized is called an oxidizing agent.  An oxidizing agent is reduced; it is the reactant in the reduction half- reaction.  The larger (more positive) E 0 red, the stronger the oxidizing agent.  A reducing agent is oxidized; it is the reactant in the oxidation half- reaction.  The larger (more positive) the E 0 ox, the stronger the reducing agent. 13

14.  Example #6: classify each of the following as an oxidizing agent, reducing agent, or both. Within each list, arrange in order of increasing strength as oxidizing agents and reducing agents.  Br 2, Mg, Fe 2+, I 2, Cl-, Cu 2+ 14 To be an oxidizing agent, a substance must be capable of gaining Electrons or being reduced. Of the species listed, Mg and Cl- are The only ones listed that cannot have a lower oxidation state. For The oxidizing agents listed above, the respective reduction potentials Are -0.44V, 0.16V, or 0.34V for Cu 2+ (which can be reduced to Cu 0 Or Cu + ), 0.54V, and 1.09V. The more positive the cell potential, the Stronger the oxidizing agent. Reducing agents must be capable of being oxidized to a higher Oxidation state. Cl- and Mg can go the Cl 0 and Mg 2+. Fe 2+ can Exist as Fe 3+ or Fe 0 so it can act as an oxidizing agent or reducing Agent. For the reducing agents listed above, the corresponding Oxidation potentials are -1.36V, -0.77V, + 2.37V. The more positive The cell potential, the stronger the reducing agent. Oxidizing: Fe 2+ <Cu 2+ <I 2 <Br 2 Reducing: Cl-<Fe 2+ <Mg

15. Electrolytic cells  In an electrolytic cell, a nonspontaneous reaction is made to occur by forcing an electric current through the cell.  In an earlier example, it was shown that the following reaction is spontaneous:  3Ni 2+ + 2Al  3Ni + 2Al 3+  The reverse of this reaction: 3Ni + 2Al 3+  3Ni 2+ + 2Al is nonspontaneous and can be made to occur by the addition of an external power source.  This electrolytic cell can be set up with two compartments just like the galvanic cell, with the replacement of a power supply for the voltmeter.  In the process of electroplating, the electrolytic cell can also be set up using only one compartment  For example, if an object is to be plated with copper, make it the cathode and immerse it into a copper(II)sulfate solution.  At the cathode, the reaction that will occur and deposit copper onto the object is Cu 2+ + 2e-  Cu.  The anode can also be made of copper.  The oxidation of copper occurs at this anode. 15

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17. Reactions that occur in an electrolytic cell  To determine which reaction occurs at the anode and the cathode during electrolysis, you must consider all possible reactions and their reduction and oxidation potentials.  If the reaction takes place in an aqueous solution, the oxidation and reduction of water must be considered. 17

18.  Example #7: A solution of copper(II)sulfate is electrolyzed. Calculate the cell potential of the reaction, E 0 cell. 18 possible reactions Cell potential, E o (V) Cathode Cu 2+ + 2e-  Cu 0.34 SO 4 2- + 4H + + 2e-  H 2 SO 3 + H 2 O 0.20 2H 2 O + 2e-  H 2 + 2OH- -0.83 anode Cu  Cu 2+ + 2e- -0.34 2H 2 O  O 2 + 4H + + 4e- -1.23 For each electrode, the reaction with the more positive potential Will occur. At the cathode, Cu 2+ will be reduced. At the anode, Cu Will be oxidized. Cu 2+ + 2e-  Cu Cu  Cu 2+ +2e-

19. AP Tip  Frequently, the electrodes are inert for electrolysis.  For example, during the electrolysis of KI (aq) - K +, I -, and H 2 O are the only species present.  Only I - and H 2 O are present to be oxidized at the anode.  Note: In aqueous KI, there is no K (s) to be oxidized. 19

20. Stoichiometry of electrolytic processes.  Lets review how much chemical change occurs with the flow of a given current for a specified time.  You might be asked how much metal was plated (formed) or how long an electroplating process will take or how much current is required to produce a specified amount of metal over a period of time.  Some units to be familiar with include A, amperes; 1A = 1C/s; coulombs, C; Faraday’s constant is 96,4895 C = 1mol e- 20

21.  Example #8: A current of 10.0A is passed through a solution containing M 2+ for 30.0 min. It produces 5.94 g of metal, M. Determine the identity of metal, M. 21 10.0A = 10.0C/s x 30.0 min x 60 s/min = 1.80x10 4 C 1.80x10 4 C x 1 mol e- /96,485C x 1 mol M/2 mol e- = 9.33x10 -2 mol M Molar mass of M = g M/mol M = 5.94gM / 9.33x10 -2 mol M = 63.7g/mol 63.7 g/mol is the molar mass of Cu. You can also do this is one step 5.94g M x 2 mol e- x 96485C x 1 s x 1 min x 1 = 63.7g/mol 1 mol M 10.0C mole e- 60 s 30.0 min

22. Comparison of Galvanic and Electrolytic cells  Galvanic and electrolytic cells have few features in common.  For both types of cells, reduction always occurs at the cathode and oxidation at the anode.  In an electrolytic cell, electrons travel from the battery to the cathode.  In both cases, electrons travel in the wire, but you wouldn’t say the electrons travel from the anode to the cathode in an electrolytic cell.  Positive ions are always attracted to the cathode whether the cell is electrolytic or galvanic. 22

23. Comparison cont… GalvanicElectrolytic Sign of the cathode+- Sign of the anode-+ Ions attracted to the cathode Cations (+) Ions attracted to the anode Anions (-) Sign of E0cell+- SpontaneitySpontaneousnonspontaneous 23 In both types of cells, the + ions or cations move toward the cathode Because there is an excess of negative ions at the cathode caused by The reduction of + ions in solution. Likewise, oxidation at the anode produces + ions, so negative ions or anions in the salt bridge must move to the anode to maintain electrical Neutrality.

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