FeO₄²⁻→(+ 2.2V) Fe³⁺→(+ 0.70V) Fe²⁺→(- 0.45V) Fe⁰
E_(FeO₄²⁻/Fe²⁺)^θ is x × 10⁻³ V. The value of x is _________

Solution

Chemistry ma’am · 3 steps

Potentials do not add, but Δ G does: weight each E° by its number of electrons.

1. Count electrons. Iron is +6 in FeO₄²⁻, so going to Fe³⁺ takes n₁ = 3 electrons. Going from Fe³⁺ to Fe²⁺ takes n₂ = 1 electron, so the overall change to Fe²⁺ takes n = 4.
2. Add free energies. Each step has Δ G° = -nFE°, and these do add up. Cancelling -F from every term gives 4E° = 3E₁° + 1E₂°.

   E°_(FeO₄²⁻/Fe²⁺) = (3(2.2) + 1(0.70))/4 = 7.3/4 = 1.825 V

3. So. 1.825 V = 1825 × 10⁻³ V, so x = 1825. The -0.45 V step to Fe is not needed here.

The common slip: 1450. Taking a plain average of the two potentials, (2.2 + 0.70)/2 = 1.45 V, gives 1450. Each potential must be weighted by its electron count.

The idea to keep. To combine half-reactions into a new half-reaction, add nE° values, not E° values, then divide by the total electrons. E° is intensive; Δ G° is not.

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