Cell potentials and Reduction potentials

The light at the end of the tunnel Read 17.7 (pg. 716), do PE 7, 8 - use Ex (ignore Ex. 17.7). Also 17.62, (731) Note: you need to multiply equations so that e – cancel out. However, unlike Hesss 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 V = 1.37 V

PE 8 PE 8 - MnO 4 – has the greater reduction potential, thus it is reduced and Cr is oxidized … MnO 4 – + 8H + + 5e – Mn H 2 O Cr Cr e – Electrons cannot exist in isolation (they must cancel out), so first x 3 and second x 5 3MnO 4 – + 24H e – 3Mn H 2 O 5Cr 5Cr e – 3MnO 4 – + 24H + + 5Cr 3Mn H 2 O + 5Cr 3+ E°cell = E°reduced - E°oxidized = 1.49 V V = 2.23 V

Were 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 OE° = 0.96 V Fe 2+ Fe 3+ + e – E° = 0.77 V NO 3 – + 4H + + 3e – NO + 2H 2 OE° = 0.96 V 3Fe 2+ 3Fe e – E° = 0.77 V NO 3 – + 4H + 3Fe 2+ 3Fe 3+ + NO + 2H 2 O E°cell = E°reduced - E°oxidized = 0.96 V V = 0.19 V

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 V = 0.90 V

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