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cell-potential

# cell-potential - 17.62 17.62 17.62 We’re looking for half...

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Cell potentials and  Cell potentials and  Reduction potentials Reduction potentials

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The light at the end of the tunnel 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 - NiO 2 has the greater reduction potential, thus it is reduced and Fe is oxidized … NiO 2 + 2H 2 O + 2e Ni(OH) 2 + 2OH Fe + 2OH Fe(OH) 2 + 2e NiO 2 + 2H 2 O + 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 PE 8 - MnO 4 has the greater reduction potential, thus it is reduced and Cr is oxidized … MnO 4 + 8H + + 5e Mn 2+ + 4H 2 O Cr Cr 3+ + 3e Electrons cannot exist in isolation (they must cancel out), so first x 3 and second x 5 3MnO 4 + 24H + + 15e 3Mn 2+ + 12H 2 O 5Cr 5Cr 3+ + 15e 3MnO 4 + 24H + + 5Cr 3Mn 2+ + 12H 2 O + 5Cr 3+ E°cell = E°reduced - E°oxidized = 1.49 V - -0.74 V = 2.23 V

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Unformatted text preview: 17.62 17.62 17.62 We’re looking for half cells that contain: NO 3 – … ↔ NO … and Fe 2+ ↔ Fe 3+ In table 17.1 we find: NO 3 – + 4H + + 3e – ↔ NO + 2H 2 O E° = 0.96 V Fe 2+ ↔ Fe 3+ + e – E° = 0.77 V NO 3 – + 4H + + 3e – ↔ NO + 2H 2 O E° = 0.96 V 3Fe 2+ ↔ 3Fe 3+ + 3e – E° = 0.77 V NO 3 – + 4H + 3Fe 2+ ↔ 3Fe 3+ + NO + 2H 2 O E°cell = E°reduced - E°oxidized = 0.96 V - 0.77 V = 0.19 V 17.64 17.64 BrO 3 – has the greater reduction potential, thus it is reduced and I – is oxidized … BrO 3 – + 6H + + 6e – → Br – + 3H 2 O 2I – → I 2 + 2e – Electrons cannot exist in isolation (they must cancel out), so second x 3 BrO 3 – + 6H + + 6e – → Br – + 3H 2 O 6I – → 3I 2 + 6e – BrO 3 – + 6H + + 6I – → Br – + 3H 2 O + 3I 2 E°cell = E°reduced - E°oxidized = 1.44 V - 0.54 V = 0.90 V The End The End For more lessons, visit www.chalkbored.com...
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