Question

Calculate the emf of the following cell at 298 K:

Fe_(s) | Fe²⁺ (0.01 M) | | H⁺ (1 M) | H_2(g) (1 bar) Pt_(s)

Given \[\ce{E^0_{cell}}\] = 0.44 V.

Answer 1

Fe_(s) | Fe²⁺ (0.01 M) | | H⁺ (1 M) | H_2(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