Respuesta :
The equilibrium constant K is 17.78 × 10²⁷ at 255 K for the following reaction under acidic conditions.
4H⁺ (aq) + MnO₂ (s) + 2Fe⁺² (aq) → Mn⁺² (aq) + 2Fe⁺³ (aq) + 2H₂O (l)
What is the relation between standard electrode potential and equilibrium constant ?
The relation between standard electrode potential and equilibrium constant is as follows:
G = -2.303 RT log K
-nFE° = -2.303 RT logK
[tex]\log K = \frac{nFE^{\circ}_{\text{cell}}}{2.303RT}[/tex]
Here,
n = 2
F = 96500
[tex]E^{\circ}_{\text{Cell}} = E^{\circ}_{\text{reduced}} - E^{\circ}_{\text{oxidized}}[/tex]
= 0.77 V - 1.51 V
= -0.74 V
R = 8.314 J/K mole
T = 255 K
Now put the values in above formula we get
[tex]\log K = \frac{nFE^{\circ}_{\text{cell}}}{2.303RT}[/tex]
[tex]\log K = \frac{2 \times 96500 \times (-0.74)}{2.303 \times 8.314\ \text{J/K mole} \times 255\ K}[/tex]
[tex]\log K = \frac{-142820}{4882.521}[/tex]
log K = 29.25
K = 17.78 × 10²⁷
The equilibrium constant K is 17.78 × 10²⁷ at 255 K for the following reaction under acidic conditions.
4H⁺ (aq) + MnO₂ (s) + 2Fe⁺² (aq) → Mn⁺² (aq) + 2Fe⁺³ (aq) + 2H₂O (l)
Learn more about the Gibbs free energy here: https://brainly.com/question/17310317
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Question: Use the appropriate standard reduction potentials below to determine the equilibrium constant at 255 K for the following reaction under acidic conditions.
4H⁺ (aq) + MnO₂ (s) + 2Fe⁺² (aq) → Mn⁺² (aq) + 2Fe⁺³ (aq) + 2H₂O (l)
The equation that relates equilibrium constants with cell potentials is
[tex]\log K = \frac{nFE^{\circ}_{\text{cell}}}{2.303RT}[/tex]
Use this formula, along with the information in the appendix of your book to solve for K.
