Calculate $\Delta$G^$\circ$ for the electrochemical cell Fe(s) | Fe²⁺(aq) || Sn⁴⁺(aq) | Sn²⁺(aq) | Pt(s).

Question

Quizplus · Accepted Answer

We can use the following equations and values:
Fe(s) + FeS1U1P12+S1S1P0(aq) ⇌ 2FeS1U1P12+(aq) + 2e-     E° = +0.07 V
SnS1U1P14+S1S1P0(aq) + 2e- ⇌ SnS1U1P12+S1S1P0(aq)        E° = +0.14 V
Adding these two equations, we get:
Fe(s) + SnS1U1P14+S1S1P0(aq) ⇌ 2FeS1U1P12+(aq) + SnS1U1P12+S1S1P0(aq)    E°cell = +0.21 V
Using the Nernst equation:
Ecell = E°cell - (RT/nF)ln(Q)
where Q = [FeS1U1P12+]^2/[SnS1U1P14+][FeS1U1P12+S1S1P0]^2
At 298 K, we have:
R = 8.314 J mol^-1 K^-1
F = 96,485 C mol^-1
n = 2 (from the balanced cell reaction)
Q = (FeS1U1P12+)^2 / (SnS1U1P14+)(FeS1U1P12+S1S1P0)^2
Q = (1 x 10^-10)^2 / (1 x 10^-4)(1 x 10^-10)^2
Q = 1 x 10^-18
Substituting in the Nernst equation, we get:
Ecell = 0.21 - (8.314*298/(2*96,485))*ln(1 x 10^-18)
Ecell = -1.1 x 10^2 kJ/mol = -1.1 x 10^4 J/mol = -11,000 J/mol = -11 kJ/mol
Therefore, the answer is B, -1.1 x 10^2 kJ/mol.

Calculate Δ\DeltaΔ G ∘\circ∘ for the Electrochemical Cell Fe(s) | Fe2+(aq) || Sn4+(aq)

Calculate $\Delta$ G^$\circ$ for the Electrochemical Cell Fe(s) | Fe²⁺(aq) || Sn⁴⁺(aq)