Calculate the emf of the following cell at 298 K:
Fe(s) | Fe2+ (0.001M)|| H+(1M) | H2(g) (1bar), Pt(s)
(Given E°cell = +0.44V)
At anode:
[Oxidation]
At cathode:
[Reduction]
Total electrons transferred (n) = 2.
Also, E°cell = +0.44V
From Nernst equation, we know,
Ecell = E0cell – (
)
[Coxid = concentration at the oxidation cell (0.001 M) and Cred = concentration at the reduction cell(1M)]
= 0.44 – (
)![]()
= 0.44 – (0.0295× -3)
= 0.44 + 0.08865 = 0.53V
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