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Question
Calculate the emf of the following cell at 298 K:
Fe(s) | Fe2+ (0.01 M) | | H+ (1 M) | H2(g) (1 bar) Pt(s)
Given \[\ce{E^0_{cell}}\] = 0.44 V.
Solution
Fe(s) | Fe2+ (0.01 M) | | H+ (1 M) | H2(g) (1 bar) Pt(s)
\[\ce{E^0_{cell}}\] = 0.44 V
Reaction at anode: \[\ce{Fe -> Fe^{2+} + 2e^-}\]
Reaction at cathode: \[\ce{2H^+ + 2e^- -> H2}\]
Overall: \[\ce{Fe + 2H^+ -> Fe^{2+} + H2}\]
According to the Nernst equation,
E = `"E"_"cell"^0 - 0.0591/"n" log (["Fe"^(2+)])/(["H"^+]^2)`
As n = 2
E = `0.44 - 0.0591/2 log 0.01/1`
= `0.44 - 0.0591/2 (-2 log 10)`
= `0.44 + 2 xx 0.0591/2`
= 0.4991 V
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