a) ΔG° for (1) = -7.4 × 10⁴ J
ΔG° for (2) = 8.5 × 10⁴ J
The relation between ΔG° and E° is
ΔG°= - nFE° Farad(F ) = 96500 Coulomb
1C × Volt = 1 Joule
For (1) reaction, Fe³⁺ (aq) + e⁻ ⇄ Fe²⁺(aq) , n = 1
(ΔG°)₁ = - nFE°
= -1 × 96500×0.77 J
= - 74.305 J = -7.4×10⁴ J
For (2) reaction, Fe²⁺ (aq) + 2e⁻ ⇄ Fe (s), n = 2
(ΔG)₂ = -2× 96500 × (-0.44V)
= 84,920 J = 8.5 × 10⁴ J
b) For (3) reaction, Fe³⁺ (aq) + 3e⁻ ⇄ Fe(s)
(ΔG°)₃ = (ΔG°)₁ + (ΔG°)₂
= -7.4 × 10⁴J + 8.5 × 10⁴ J
= 1.1 × 10⁴ J
c) Fe³⁺ (aq) + 3e⁻ ⇄ Fe(s) , n = 3
(ΔG°)₃ = (ΔG°)₁ + (ΔG°)₂
- 3FE° = - F × 0.77 V + 2 F ( 0.44 V )
3 E° = 0.77 V - 0.88 V
E° = -0.037 V
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Question: You are given the following three half- reactions:
1) Fe³⁺ (aq) + e⁻ ⇄ Fe²⁺(aq) E° = + 0.77 V
2) Fe²⁺ (aq) + 2e⁻ ⇄ Fe (s) E° = - 0.44 V
3) Fe³⁺ (aq) + 3e⁻ ⇄ Fe(s)
a) Calculate ΔG° for (1) and (2) from their E° values
b) Calculate ΔG° for (3) from (1) and (2)
c) Calculate E° for (3) from its ΔG°
