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AP Topic 15: Electrochemistry

Goes with chapter 21: Silberberg’s Principles of General Chemistry Mrs. Laura Peck, 2013. AP Topic 15: Electrochemistry. Objectives/Study guide. Identify and compare the two types of electrochemical cells: galvanic and electrolytic

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AP Topic 15: Electrochemistry

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  1. Goes with chapter 21: Silberberg’s Principles of General Chemistry Mrs. Laura Peck, 2013 AP Topic 15: Electrochemistry

  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.

  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.

  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 Example #1: Give the correct line notation for the Galvanic cell pictured. Zn/Zn2+ II Cu2+/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.

  6. Standard reduction potentials • The cell potential, Ecell, 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.

  7. Example #2: consider a galvanic cell based on the reaction: Al + NI2+  Al3+ + Ni Give the balanced cell reaction and calculate the cell potential, E0cell 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, E0ox = -E0red Step 4: the cell potential for the overall Reaction is equal to the sum of the Reduction potential, E0red, and the Oxidation potential, E0ox. 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 E0 Oxidation: Al  Al3+ + 3e- Reduction: Ni2+ + 2e-  Ni Ni2+ + 2e-  Ni E0red = -0.23V Oxidation: Al Al3+ + 3e- E0ox= -E0red  -(-1.66V) = +1.66V E0cell= E0ox + E0red E0cell = -0.23V + 1.66V = 1.43V 3[Ni2+ + 2e-  Ni] E0red = -0.23V 2[Al  Al3+ + 3e-] E0ox = +1.66V 3Ni2+ + 2Al  3Ni + 2Al3+ E0cell = 1.43V

  8. Spontaneous Reactions • Gibbs free energy, ΔG°, can be calculated from the cell potential, E0cell. • ΔG° = -nFE0cell • 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 E0cell • 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 .

  9. Write half-reactions and Calculate E0cell • Example #3: Will 1M HCl dissolve silver metal and form Ag+ solution? 2H+ + 2e-  2H2 E0red = 0.00V 2Ag  2Ag+ + 2e- E0ox = -0.80V 2H+ + 2Ag  H2 + 2Ag+ E0cell = -0.80V The negative value for E0cell indicates that the reaction will not occur.

  10. Begin by writing the appropriate Half-reactions. Then calculate The cell potential for the overall Reaction. First, the reaction in which Br2 Oxidizes I- Then the reaction in which Br2 Oxidizes Cl- • Example #4: Bromine, Br2, can oxidize iodide, I-, to iodine, I2. However, Br2 cannot oxidize chloride, Cl-, to chlorine, Cl2. Explain why the first reaction occurs yet, the second one does not. Br2 + 2e-  2Br- E0red = 1.09V 2I-  I2 + 2e- E0ox = -0.54V Br2 + 2I-  2Br- + I2 E0cell = 0.55V This reaction occurs, E0cell is + Br2 + 2e-  2Br- E0red = 1.09V 2Cl-  Cl2 + 2e- E0ox = -1.36V Br2 + 2Cl-  2Br- + Cl2 E0cell = -0.27V This reaction does not occur, E0cell is negative

  11. Cell Dependence on Concentration • The galvanic cell represented by the reaction: • 3Ni2+ + 2Al  3Ni + 2Al3+ • Has a cell potential, E0cell, equal to 1.43V under standard conditions (all solutions are 1M) • Increasing the concentration of Ni2+ 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: • Ecell = E0cell – (0.0591/n)log Q • The cell potential, Ecell, is for nonstandard conditions. • The moles of electrons transferred are represented by n • The mass action quotient is represented by Q

  12. Example #5: Calculate the cell potential for the reaction: • 3Ni2+ + 2Al  3Ni + 2Al3+ • In which [Al3+] = 2.00M and [Ni2+] = 0.750M • (you already know the E0cell = 1.43V and the number of moles of electrons transferred, n, equals 6) Q = [Al3+]2 / [Ni2+]3 = (2.00)2 / (0.750)3 = 9.48 Ecell = 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) E0red, the stronger the oxidizing agent. • A reducing agent is oxidized; it is the reactant in the oxidation half-reaction. • The larger (more positive) the E0ox, the stronger the reducing agent.

  14. Oxidizing: Fe2+<Cu2+<I2<Br2 Reducing: Cl-<Fe2+<Mg • 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. • Br2, Mg, Fe2+, I2, Cl-, Cu2+ 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 Cu2+ (which can be reduced to Cu0 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 Cl0 and Mg2+. Fe2+ can Exist as Fe3+ or Fe0 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.

  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: • 3Ni2+ + 2Al  3Ni + 2Al3+ • The reverse of this reaction: 3Ni + 2Al3+  3Ni2+ + 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 Cu2+ + 2e-  Cu. • The anode can also be made of copper. • The oxidation of copper occurs at this anode.

  16. 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. Example #7: A solution of copper(II)sulfate is electrolyzed. Calculate the cell potential of the reaction, E0cell. possible reactions Cell potential, Eo(V) Cathode Cu2+ + 2e-  Cu 0.34 SO42- + 4H+ + 2e-  H2SO3 + H2O 0.20 2H2O + 2e-  H2 + 2OH- -0.83 anode Cu  Cu2+ + 2e- -0.34 2H2O  O2 + 4H+ + 4e- -1.23 For each electrode, the reaction with the more positive potential Will occur. At the cathode, Cu2+ will be reduced. At the anode, Cu Will be oxidized. Cu2+ + 2e-  Cu Cu  Cu2+ +2e-

  18. AP Tip • Frequently, the electrodes are inert for electrolysis. • For example, during the electrolysis of KI(aq) - K+, I-, and H2O are the only species present. • Only I- and H2O are present to be oxidized at the anode. • Note: In aqueous KI, there is no K(s) to be oxidized.

  19. 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. Example #8: A current of 10.0A is passed through a solution containing M2+ for 30.0 min. It produces 5.94 g of metal, M. Determine the identity of metal, M. 10.0A = 10.0C/s x 30.0 min x 60 s/min = 1.80x104 C 1.80x104 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

  21. 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. Comparison cont… 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.

  23. The end…

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