Odpowiedź:
b. 0,32V
Wyjaśnienie:
E°(Zn l Zn2+) = -0.762V
E°(Fe l Fe2+) = -0.447V
SEM = E(katody) - E(anody)
SEM = -0.447V - (-0.762V) = 0.315 ≅ 0.32V
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Odpowiedź:
b. 0,32V
Wyjaśnienie:
E°(Zn l Zn2+) = -0.762V
E°(Fe l Fe2+) = -0.447V
SEM = E(katody) - E(anody)
SEM = -0.447V - (-0.762V) = 0.315 ≅ 0.32V