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EMF of Fe³⁺/Fe²⁺ and Sn²⁺/Sn cell

Short answer

Fe2+ is oxidized at anode, Sn2+ reduced at cathode. E°cell = 0.63 V, Q = [Fe3+]2/([Fe2+]2[Sn2+]) = 80. EMF = 0.63 – (0.0591/2) log(80) = 0.57 V.

Calculate the EMF of the following cell at 25°C. E° Fe³⁺/Fe²⁺ = −0.77 V, E° Sn²⁺/Sn = −0.14 V in which [Fe³⁺] = 0.4 M, [Fe²⁺] = 0.1 M, [Sn²⁺] = 0.2 M.

Reading of data: the E° values are used as given: Fe³⁺/Fe²⁺ = −0.77 V and Sn²⁺/Sn = −0.14 V, treated as electrode potentials of the form given in the question (as reduction potentials). Sn²⁺/Sn is the higher (less negative), so Sn²⁺ is reduced at the cathode and Fe²⁺ is oxidised to Fe³⁺ at the anode.

Anode   : 2Fe²⁺ → 2Fe³⁺ + 2e⁻
Cathode : Sn²⁺ + 2e⁻ → Sn
Cell    : 2Fe²⁺ + Sn²⁺ → 2Fe³⁺ + Sn      (n = 2)
Notation: Pt | Fe²⁺ (0.1 M), Fe³⁺ (0.4 M) || Sn²⁺ (0.2 M) | Sn

E°cell = E°(cathode) − E°(anode) = −0.14 − (−0.77) = 0.63 V
Q = [Fe³⁺]²/([Fe²⁺]²[Sn²⁺]) = (0.4)²/((0.1)² × 0.2) = 0.16/0.002 = 80
log Q = 1.903
E = E° − (0.0591/2) log Q
  = 0.63 − 0.02955 × 1.903
  = 0.63 − 0.0562 = 0.574 V

Answer: EMF ≈ 0.57 V.

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