Question

Devise a cell for the folkowing redox reaction H2(g) +12 (g) 2HI(aq) → Choose the tems below in correct order (trom 1 to 9) that ts consislont with the correct cell notation from Left to the Right Itis e easier to sketch the cel notation on a piece of peper first betore you choose the answers Hnt the product Hl(ag) can also written as "H dentical items, choose the one with lower number first aq) (aq where two ions are in the same solubon, therefore sall britge is not needed in this case (two vertical ines are not needed) Note: For twD 2(S) Ht(aq). r(aq) Pt Pt H2(g)

All are one question! They are separated by steps. Please answer all steps. Thanks!

Answer 1

a) The cell will comprise of the oxidation half cell (anode) and the reduction half cell (cathode). In a cell notation, the oxidation half is written followed by the reduction half separated by vertical bars (│). The vertical bars are used to denote different phases.

HI can be written as H⁺(aq)I^- (aq). H₂ is oxidized to H⁺ (hence forms the anode) while I₂ is reduced to I^- (hence forms the cathode). H₂ and I₂ are gases while HI is an aqueous species; hence, we shall need Pt electrodes to support the gaseous and aqueous species.

Therefore, the cell notation is

Pt│H₂ (g)│H⁺(aq), I^-(aq)│I₂ (g)│Pt

b) The half reactions are shown below.

Oxidation (left): H₂ (g) --------> 2 H⁺ (aq) + 2 e^-. E⁰_L = 0.00 V

Reduction (right): I₂ (s) ------> I₂ (g) + 2 e^- ---------> 2 I^- (aq); E⁰_R = +0.54 V

The total cell potential is given as

E⁰_cell = E⁰_R + E⁰_L = (+0.54 V) + (0.00 V) = 0.54 V (ans).

c) The equilibrium constant of the cell can be obtained by using the relation

n*F*E⁰_cell = R*T*ln K_eq where n = 2 (since 2 electrons are involved); F = 96485 J/V.mol; R = 8.314 J/mol.K and T = 298 K; therefore,

(2)*(96485 J/V.mol)*(0.54 V) = (8.314 J/mol.K)*(298 K)*ln K_eq

=====> ln K_eq = 42.0588

=====> K_eq = exp^(42.0588) = 1.8446*10¹⁸ ≈ 1.845E18 (ans).