Question

A voltaic cell is made by placing a solid iron electrode in a compartment in which the Fe²⁺ concentration is 2.0 x 10 ^-5 M and by placing a Pt electrode in the other compartment, in which the H⁺ concentration is 3.4 M and PH ₂ = 0.700 atm. (This is a Hydrogen Electrode)
What is emf (E) for the cell at the concentrations given?
Fe(s) + 2H⁺(aq) --> Fe²⁺(aq) + H₂(g)
Hint, you do not have cell diagram, so you have to decide which is the anode, it is the species that gets oxidized (loses electrons)

Answer 1

which 2 2+ Reduction (Cathod) • A voltic cell is made by placing a solidiron electrode in a com compartment in the Feat concentration is 2.0X1055m and by placing a pt electrode in the other compartment, in which the li concentration is 3-ym and PH 2 20.700 atm. What is the emf (E) for the cell at the concentrations given? Fe(59+ 2H+ (aq) → Festag) + H2CH), - The reun is oxidahon (amodos Fe (3).+ 2H+ (aq) → Fé"caq) + 172(8) E'Fearcsas/Pecay = – 0-44V. E Ecell = (Ered cathod - (Efes ) anode And given, Paul, = 0.700 am Fett Efett? Fels [H+] = 3-4M 0.00 - (-0.94) v [ Fe 2+] = 2.0x105 m Ecun Oyuv electron involved in the redox reaction (M) is = 2 Now applying permit equation to the above redox rex" we get- [ Fezt].PH log (at 25°c) Ht | H₂ 0.0ov, And no. of Ecelu Feu - 0.059 n TH+72 0.44 -0.059 2 -log (2.0x105) x 0.700 (304)? = 10-40-(-01791] Ecen - 0.614 v Answer SO, the emf (E) for the cell is 0.614 v.