Determine E° for the half-reaction Fe³⁺(aq) + e⁻ → Fe²⁺(aq)

Question

Fe^3+(aq) + H2(g) -> 2H+(aq) + Fe^2+(aq)

E = +0.77 V

(17/16)

Answer 1

Fe³⁺ + 3e- ==> Fe(s) Eº = +0.77 V (from a table of standard reduction potentials)

2H⁺ + 2e- ==> H2(g) Eº = 0.00 V

Fe3+ will be reduced (cathode) and H2 will be oxidized (anode)

Eºcell = cathode - anode = 0.77 - 0 = +0.77 V (answer)