Cell potentials Cell potentials and Reduction and Reduction

potentialspotentials

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 - 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)2E°cell = E°reduced - E°oxidized

= 0.49 V - -0.88 V = 1.37 V

PE 8 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

17.6217.6217.62 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

17.6417.64BrO3

– 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 3BrO3

– + 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

The End The End

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