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Cell potentials and Reduction potentials. The light at the end of the tunnel . Read 17.7 (pg. 716), do PE 7, 8 - use Ex. 17.8 (ignore Ex. 17.7). Also 17.62, 17.64 (731) Note: you need to multiply equations so that e – cancel out. However, unlike Hess’s law problems, DO NOT also multiply E°
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The light at the end of the tunnel Read 17.7 (pg. 716), do PE 7, 8 - use Ex. 17.8 (ignore Ex. 17.7). Also 17.62, 17.64 (731) Note: you need to multiply equations so that e– cancel out. However, unlike Hess’s law problems, DO NOT also multiply E° PE 7 - NiO2 has the greater reduction potential, thus it is reduced and Fe is oxidized … NiO2 + 2H2O + 2e– Ni(OH)2 + 2OH– Fe + 2OH– Fe(OH)2 + 2e– NiO2 + 2H2O + Fe Ni(OH)2 + Fe(OH)2 E°cell = E°reduced - E°oxidized = 0.49 V - -0.88 V = 1.37 V
PE 8 PE 8 - MnO4– has the greater reduction potential, thus it is reduced and Cr is oxidized … MnO4– + 8H+ + 5e– Mn2+ + 4H2O Cr Cr3+ + 3e– Electrons cannot exist in isolation (they must cancel out), so first x 3 and second x 5 3MnO4– + 24H+ + 15e– 3Mn2+ + 12H2O 5Cr 5Cr3+ + 15e– 3MnO4– + 24H+ + 5Cr 3Mn2+ + 12H2O + 5Cr3+ E°cell = E°reduced - E°oxidized = 1.49 V - -0.74 V = 2.23 V
17.62 17.62 We’re looking for half cells that contain: NO3– … NO … and Fe2+ Fe3+ In table 17.1 we find: NO3– + 4H+ + 3e– NO + 2H2O E° = 0.96 V Fe2+ Fe3+ + e– E° = 0.77 V NO3– + 4H+ + 3e– NO + 2H2O E° = 0.96 V 3Fe2+ 3Fe3+ + 3e– E° = 0.77 V NO3– + 4H+3Fe2+ 3Fe3+ + NO + 2H2O E°cell = E°reduced - E°oxidized = 0.96 V - 0.77 V = 0.19 V
17.64 BrO3– has the greater reduction potential, thus it is reduced and I– is oxidized … BrO3– + 6H+ + 6e– Br– + 3H2O 2I– I2 + 2e– Electrons cannot exist in isolation (they must cancel out), so second x 3 BrO3– + 6H+ + 6e– Br– + 3H2O 6I– 3I2 + 6e– BrO3– + 6H+ + 6I– Br– + 3H2O + 3I2 E°cell = E°reduced - E°oxidized = 1.44 V - 0.54 V = 0.90 V
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