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+ Given the following electrode potentials at 25°C Fe3+ e-- Fe2+ E° = 0.571 V 2e Fe(s) E° = -0.440 V Calculate the electrode
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Answer #1

Fe​​​​​3+ + e- ---> Fe​​​​​2+ --(i) (E°1 = 0.571 V)

∆G1 = -nFE°1 = -F(0.571) = -0.571F

Fe​​​​​2+ + 2e- ---> Fe ---(ii) (E°2 = -0.440 V)

∆G2 = -nFE°2 = -2F(-0.440) = 0.880F

(i) + (ii) we get,

Fe​​​​​3+ + 3e- ---> Fe

∆G3 = -nFE°3 = -3FE°3

=> ∆G1 + ∆G2 = ∆G3

=> 0.309F = -3FE°3

=> E°3 = -0.103 V

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