Question

1) Calculate ΔG∞ for the electrochemical cell Pb(s) | Pb²⁺(aq) || Fe³⁺(aq) | Fe²⁺(aq) | Pt(s).
A. –1.2 x 10² kJ/mol
B. –1.7 x 10² kJ/mol
C. 1.7 x 10² kJ/mol
D. –8.7 x 10¹ kJ/mol
E. –3.2 x 10⁵ kJ/mol

2)Determine the equilibrium constant (K_eq) at 25∞C for the reaction
Cl₂(g) + 2Br^– (aq)    2Cl^– (aq) + Br₂(l).
A. 1.5 x 10^–10
B. 6.3 x 10⁹
C. 1.3 x 10⁴¹
D. 8.1 x 10⁴
E. 9.8

Answer 1

Q1. Correct option is : (B.) -1.7 x 10^2 kJ/mol

Q2. Correct option is : (B.) 6.3 x 10^9

Explanation

Q1. Oxidation half-reaction : Pb (s) Pb2+ (aq) + 2e : Eo = 0.126 V

Reduction half-reaction : Fe3+ (aq) + e Fe2+ (aq) : Eo = 0.771 V

Overall reaction : Pb(s) + 2 Fe3+ (aq) Pb2+ (aq) + 2 Fe2+ (aq)

Standard cell potential, Eocell = (0.126 V) + (0.771 V)

Eocell = 0.897 V

Go = -(n) * (F) * (Eocell)

where n = number of electrons transferred = 2

F = Faraday's constant = 96485 J/V-mol

Go = -(2) * (96485 J/V-mol) * (0.897 V)

Go = -173094 J/mol

Go = -1.7 x 10^2 kJ/mol