Chem 1B Dr. White 1 WS 18 Key Galvanic Cells and Strength of Oxidizing/Reducing Agents (You may have to look at a table of standard reduction potentials to solve many of these problems!) 1. A galvanic cell is to be constructed using the Ag/Ag+ half cell and the Pb/Pb+2 half-cell. Measurement shows that the silver electrode is positive. a) Write balanced half-reactions and the overall spontaneous reaction, without using a table of half-cell potentials. Since the silver electrode is positive, that means it is the cathode in this voltaic cell, and so Ag+ is reduced, thus the reactions must be 2Ag+ (aq) + 2 e- Pb (s) 2 Ag (s) Pb+2 (aq) + 2e- 2Ag+ (aq) + Pb (s) Pb+2 (aq) + 2 Ag (s) a) Diagram the cell, labeling electrodes as anode and cathode, labeling the salt bridge, showing what ions are in solution, and showing the direction of electron flow in the circuit and of cation and anion flow in the salt bridge. e- V cations anions Salt Bridge Pb Pb+2 Ag Ag+ c) Pb (s) ⏐Pb2+ (aq)⏐⏐Ag= (aq)⏐Ag (s) d) Using a table of half-cell potentials, calculate the standard cell potential for the cell you have diagrammed above. Did you choose your electrodes correctly? 2Ag+ (aq) + 2 ePb (s) 2 Ag (s) Pb+2 (aq) + 2e- 2Ag+ (aq) + Pb (s) Pb+2 (aq) + 2 Ag (s) E˚cell = E˚cathode - E˚anode = 0.80 V – (- 0.13 V) = + 0.93 V Chem 1B Dr. White 2 Yes, we have chosen the electrodes correctly, as we ended up with a + E˚cell, as we should for a voltaic cell 2. In principle, a battery (galvanic cell) could be made from aluminum metal and chlorine gas. a) Write a balanced equation for the reaction that would occur in a battery using Al (s) /Al+3(aq) and Cl2(g)/Cl- (aq) half-reactions. A battery, just like a voltaic cell, must produce a +E˚cell, so use table of reduction potentials to see which way half-reactions would be used. Al has a more negative reduction potential than Cl2, so it will be oxidized. 2 Al (s) + 3 Cl 2 (g) ----> 2 Al +3 (aq) + 6 Cl - (aq) b) Tell which half-reaction occurs at the anode and which at the cathode. What are the polarities of these electrodes? The oxidation of Al (s) occurs at anode, negative polarity: Al (s) ----> Al +3 (aq) + 3eThe reduction of Cl 2 (g) occurs at the cathode, positive polarity : Cl 2 (g) + 2e- ----> 2 Cl - (aq) c) Calculate the standard potential, E˚cell for this battery. E˚cell = E˚cathode - E˚anode = + 1.36 – (-1.66) = 3.02 V d) What other metal besides Al could conceivably be used in conjunction with the Cl2(g)/Cl-(aq) half-reaction to produce a battery with a larger voltage output? Leave the Cl2/Cl- as whatever half-reaction you decided it should be above. Show calculation of E˚cell to justify your choice. Could use any metal that has a more negative reduction potential than Al on the table of reduction potentials, this will ensure a larger voltage output. For example, Mg (s) Mg(s) ----> Mg+2 (aq) + 2 eCl 2 (g) + 2e- ----> 2 Cl - (aq) E˚cell = E˚cathode – E˚anode = 1.36 – (-2.37) = 3.73 V Chem 1B Dr. White 3 3. Cathodic protection is the protection of a metal from being oxidized by another metal. Name two metals you could you use to cathodically protect Sn metal from being oxidized, and explain your answer. You could use any number of metals to cathodically protect Sn – any metal that has a more negative reduction potential than Sn will be preferentially oxidized and keep Sn from oxidizing. For example: Sn+2 + 2eZn+2 + 2e- Sn Zn E˚ = -0.14 V E˚ = -0.76 V So Zn will be oxidized preferentially in presence of Sn 4. Look at the table of standard reduction potentials and suggest a reason why dentists use metals like gold or mercury for fillings in your cavities rather than something like copper or zinc. I don’t know if this is really why dentists use gold or mercury for fillings, but it makes sense that since these metals have very high reduction potentials compared to copper and zinc, they won’t oxidize in your mouth, You wouldn’t want your fillings to corrode, would you?