For the reaction
2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s)
the magnitude of the standard molar Gibbs free energy change, ΔᵣGₘᵒ = - ___________ kJ (Round off to the Nearest Integer).
[ E_(Fe²⁺/Fe(s))ᵒ = - 0.440V;, E_(Fe³⁺/Fe(s))ᵒ = - 0.036V; E_(I₂/2I -)ᵒ = 0.539V;, F = 96500C ]

Solution

Chemistry ma’am · 5 steps

Build the Fe³⁺/Fe²⁺ potential first by adding Δ G values, never E° values.

1. Missing couple. The reaction reduces Fe³⁺ to Fe²⁺, but that potential is not given. Build it from the two given half-reactions: Fe³⁺ + 3e⁻ → Fe minus Fe²⁺ + 2e⁻ → Fe gives Fe³⁺ + e⁻ → Fe²⁺.
2. Add free energies. Potentials do not add, but Δ G° = -nFE° does. So -1 × F × E°ₓ = -3F(-0.036) - [-2F(-0.440)], and F cancels.

   E°_(Fe³⁺/Fe²⁺) = 3(-0.036) - 2(-0.440) = -0.108 + 0.880 = 0.772 V

3. Cell potential. Fe³⁺ is reduced at the cathode and I⁻ is oxidised at the anode, so E°_(cell) = 0.772 - 0.539 = 0.233 V.
4. Gibbs energy. Two electrons move in the balanced reaction, so n = 2. Then Δᵣ G° = -2 × 96500 × 0.233 = -44 969 J.
5. So. Δᵣ G°ₘ ≈ -44.97 kJ, and its magnitude rounds to 45.

The common slip: 26. Subtracting the two given potentials directly gives 0.404 V for Fe³⁺/Fe²⁺. That makes E°_(cell) = -0.135 V and a magnitude of 26, not 45.

The idea to keep. To combine two half-reactions into a third half-reaction, add nE° values, which is the same as adding Δ G°. Only a full cell lets you subtract potentials directly.

Answer from the NTA final answer key. Solution by PrepWiser.