Respuesta :
The reaction is:
6 Cr²⁺ + Cr₂O₇²⁻ + 14 H⁺ → 8 Cr³⁺ + 7 H₂O
E₀ = 1.33 - (-0.5) = 1.83 V
ΔG = - n f E₀
= - 6 * 96485 * 1.83
= - 1059405.3 J / mol
= - 1059.4 kJ / mol
6 Cr²⁺ + Cr₂O₇²⁻ + 14 H⁺ → 8 Cr³⁺ + 7 H₂O
E₀ = 1.33 - (-0.5) = 1.83 V
ΔG = - n f E₀
= - 6 * 96485 * 1.83
= - 1059405.3 J / mol
= - 1059.4 kJ / mol
Answer:
The correct answer is -1059.45 kJ.
Explanation:
The balanced equation is:
6Cr₂⁺ + Cr₂O₇²⁻ + 14H⁺ ⇒ 8Cr₃⁺ + 7H₂O
In the mentioned reaction 6 electrons are transferred
By calculating Ecell with the use of reduction potential of each cell:
Ecell = Eox + Ered
Ered = 0.50V
Eox = -Ered = -(-0.50V) = 0.50V
Ered = 1.33V
Ecell = 0.50V + 1.33V
Ecell = 1.83V
Now in order to calculate ΔG
n = 6, Faraday constant (f) = 9.68470 × 10⁴ = 96847 C mol⁻¹
ΔG = -nFE
ΔG = -6 × 96487 C mol-1 × 1.83V
ΔG = -1059.42 KJ
The relation between ΔG and ΔG°rxn
ΔG = ΔG° + RTlnQ
Under the standard condition Q = 1 and ΔG = ΔG°
Thus,
ΔG°rxn = ΔG = -1059.42 KJ