Question

Consider the following cell reaction:
2Fe(s) O₂(g) 4H⁺(aq) 2Fe²⁺(aq) 2H₂O(l);E° = 1.67V
At [Fe²⁺] = 10⁻³ M, p(O₂) = 0.1 atm and pH = 3, the cell
potential at 25ºC is
(a) 1.47 V (b) 1.77 V
(c) 1.87 V (d) 1.57 V

Answer 1

the cell  potential at 25ºC is  1.57 V

- nerdst equation is given by :

E = E° - (0.059/n) log ([products]/[reactants])

- for the given reaction

n = 4

products are [Fe²⁺] and reactants are O₂ and H⁺

- given: p(O₂) = 0.1 atm

and              pH = 3

          - log [H⁺] = 3

                   [H⁺] = 10⁻³ M

- putting these in eqn, we get

E = 1.67 - (0.059/4) log {[Fe²⁺]²/p(O₂)[H⁺]⁴}

E = 1.67 - (0.059/4) log {(10⁻³)²/0.1 × (10⁻³)⁴}

E = 1.67 - (0.059/4) × 7

E = 1.67 - 0.103

E = 1.567 V = 1.57 V

- so the correct answer is option (d)