Question

Consider the following unbalanced equation.

H¹⁺(aq) + Fe(s) H₂(g) + Fe²⁺(aq)

(a) What are the following standard voltages? Include the sign. Change the sign as appropriate. Use the standard reduction potentials in these Reference Tables. Enter the number of decimal places allowed by the data in the table.

standard oxidation potential for the oxidation half-cell	V
reduction potential for the reduction half-cell	V
potential for the entire cell	V

(b) Select all that apply for the reaction under standard conditions.

H¹⁺ is the substance being oxidized.Fe is the substance being oxidized.H¹⁺ is the substance being reduced.Fe is the substance being reduced.The forward reaction is spontaneous. The reverse reaction is spontaneous.Electrons will flow from the H half-cell to the Fe half-cell.Electrons will flow from the Fe half-cell to the H half-cell.[H¹⁺] increases as the cell operates.[H¹⁺] decreases as the cell operates.[Fe²⁺] increases as the cell operates.[Fe²⁺] decreases as the cell operates.[H¹⁺] and [Fe²⁺] remain constant.

Answer 1

- 2H + (aq) + Fe (5) af Hene, ho Ht is being reduced H₂ (g) + Fe 2+ (aq) and te is being oxidized Oxidation hall cell reachion: Fe(s) Fe 2+ + ae Canode) The standard reduchons parenhicle of Fe 2is -0.45 V. standard rede oxidation potential = - Ened =_+0.45 v .:Standard oxidation potem nal fou the following oxidation reachon Fe(s) Festa e or + 0.45 Reduchon half cell reachion : att (aast ae- H₂(g) Ened = 0 (cathode) Ossidabon takes place at anode and reduction at cathode. Ecett . Bathode - Earnode (Here tº refers to standard I reduction potential y - O - (0.45 p) Ecell - + 0.45 V 11111

e 0 be fe is the substance being oxudized Ht is the substance being reduced The forw and greachan is spontaneous (as overall fett cell potentie Ecell is positive) Elections will flow from Fe half cell to the half cell (electrons floss from anode to cathode) #* decreases as the cell operating operates Fe2t increases as the cell operates