17.62 17.62 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 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
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