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Cell Potentials a Galvanic Cell Is a Combination of an Anode And

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Cell Potentials a Galvanic Cell Is a Combination of an Anode And Name ________________________ Cell Potentials A Galvanic cell is a combination of an anode and cathode, in which the oxidation and reduction, respectively, of a REDOX reaction occur in separate compartments in order to convert chemical energy into electrical energy. Shown in Figure 1, is a schematic diagram of a Galvanic cell that takes advantage of the oxidation of zinc and the reduction of copper (II) ions, eq. 1. The corresponding oxidation and reduction half- reactions are also shown in Figure 1. 2+ 2+ Zn (s) + Cu (aq) Zn (aq) + Cu (s) (1) Electrons flow from the anode (zinc) to the cathode (copper), which are positively and negatively charged, respectively. To complete the circuit, anions in the salt brigde must flow towards the solution containing the anode, because the zinc metal is being coverted into zinc (II) cations. Likewise, the cations in the salt bridge must flow towards the cathode to compensate for the copper (II) ions that are being removed from solution and being deposited on the copper electrode as copper metal. This flow of ions results in no net change in the amount of charged ions present in the solution during the reaction. As the cell discharges, the zinc electrode slowly dissolves, while the copper electrode slowly gets larger. A shorthand notation for representing a Galvanic cell has been developed in which the salt bridge is represented by a double slash, //, with the anode’s reactants and products occurring on the left side of the notation and the cathode’s reactants and products occurring on the right side of the notation. A single slash, /, is used whenever there is a phase difference between reactants and products of a half-reaction, i.e. a solid electrode and an ion in aqueous solution. The notation used to describe the Galvanic cell shown in Figure 1 would be Zn (s) / Zn2+ (aq) // Cu2+ (aq) / Cu (s) Figure 1. A schematic diagram (a) and a photgraph of an actual Galvanic cell based on eq. 1. Name ________________________ Cell Potentials o The voltage of a Galvanic cell, E cell, is the sum of the voltage produced by the reduction, o o E red, and oxidation, E ox, half-reactions, eq. 2. Shown in Table 1 are selected reduction potentials, which are for 1 M solutions and 1 atmosphere of pressure for gases. To obtain the corresponding oxidation half-reactions and their potentials, simply reverse the o o reduction half reaction and change the sign of the potential, E ox = -E red. o o o E cell = E red + E ox (2) Table 1. Reduction potentials for selected half-reactions. o Reduction half-reaction Potential, E red (Volts) + 2- PbO2 + 4 H + SO4 + 2 e- ---> PbSO4 (s) + 2 H2O +1.69 - + 2+ MnO4 + 8 H + 5 e- ---> Mn + 4 H2O +1.49 + 2+ PbO2 + 4 H + 2 e- ---> Pb + 2 H2O +1.46 + O2 + 4 H + 4 e- ---> 2 H2O +1.23 Ag+ + e- ---> Ag +0.80 Cu2+ + 2 e- ---> Cu +0.34 + 2 H + 2 e- ---> H2 0.00 Fe3+ + 3 e- ---> Fe -0.04 Pb2+ + 2 e- ---> Pb -0.13 Ni2+ + 2 e- ---> Ni -0.25 Co2+ + 2 e- ---> Co -0.29 2- PbSO4 + 2 e- ---> Pb + SO4 -0.359 Cd2+ + 2 e- ---> Cd -0.40 Fe2+ + 2 e- ---> Fe -0.41 Cr3+ + 3 e- ---> Cr -0.74 Zn2+ + 2 e- ---> Zn -0.76 Mn2+ + 2 e- ---> Mn -1.18 Al3+ + 3 e- ---> Al -1.66 Mg2+ + 2 e- ---> Mg -2.37 Write the appropriate oxidation half-reactions, reduction half-reactions, the balanced overall REDOX reaction and calculate the overall cell potential for the following. 1. Zn (s) / Zn2+ (aq) // Mg2+ (aq) / Mg (s) + 2+ 2. Pt (s) / H2 (g), H (aq) // Cu (aq) / Cu (s) 3. Mg (s) / Mg2+ (aq) // Ag+ (aq) / Ag (s) 4. Cr (s) / Cr3+ (aq) // Pb2+ (aq) / Pb (s) Name ________________________ Cell Potentials Answers 2+ - - 2+ 2+ 2+ 1. Zn (s) Zn (aq) + 2 e , 2 e + Mg (aq) Mg (s), Zn (s) + Mg (aq) Zn (aq) + Mg (s), Eo = -1.61 V. + - 2+ - 2+ + 2. H2 (g) 2 H (aq) + 2 e , Cu (aq) + 2 e Cu (s), H2 (g) + Cu (aq) 2 H (aq) + Cu (s), Eo = +0.34 V 3. Mg (s) Mg2+ (aq) + 2 e-, Ag+ (aq) + e- Ag (s), Mg (s) + 2 Ag+ (aq) Mg2+ (aq) + 2 Ag (s), Eo = +3.17 V 4. Cr (s) Cr3+ (aq) + 3 e-, Pb2+ (aq) + 2 e- Pb (s), 2 Cr (s) + 3 Pb2+ (aq) 2 Cr3+ (aq) + 3 Pb (s), Eo = +0.61 V .
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