Solveeit Logo
Login

Question

Question: The logarithm of the equilibrium constant, \(\log K_{eq,}\)of the net cell reaction of the cell, \(...

The logarithm of the equilibrium constant, log⁡Keq,\log K_{eq,}of the net cell reaction of the cell,

X(s)∣X2+∣⥂∣Y+∣Y(s)(givenEcell0=1.20V),X(s)|X^{2 +}| ⥂ |Y^{+}|Y(s)(givenE_{cell}^{0} = 1.20V), is

A

47.2

B

40.5

C

21.4

D

12.5

Answer

40.5

Explanation

Solution

ΔG0=−nE0F=−RTln⁡Keq⇒ln⁡Keq=nE0FRT=nE0RT/F\Delta G^{0} = - nE^{0}F = - RT\ln K_{eq}\mathbf{\Rightarrow}\mathbf{\ln}\mathbf{K}_{\mathbf{eq}}\mathbf{=}\frac{\mathbf{n}\mathbf{E}^{\mathbf{0}}\mathbf{F}}{\mathbf{RT}}\mathbf{=}\frac{\mathbf{n}\mathbf{E}^{\mathbf{0}}}{\mathbf{RT/F}}

log⁡Keq=nE02.303RTF=nE00.0592(at250C)\mathbf{\log}\mathbf{K}_{\mathbf{eq}}\mathbf{=}\frac{\mathbf{n}\mathbf{E}^{\mathbf{0}}}{\mathbf{2.303}\frac{\mathbf{RT}}{\mathbf{F}}}\mathbf{=}\frac{\mathbf{n}\mathbf{E}^{\mathbf{0}}}{\mathbf{0.0592}}\mathbf{(}\text{at}\mathbf{2}\mathbf{5}^{\mathbf{0}}\mathbf{C)}; log⁡Keq=2×1.200.0592=40.5\mathbf{\log}\mathbf{K}_{\mathbf{eq}}\mathbf{=}\frac{\mathbf{2}\mathbf{\times}\mathbf{1.20}}{\mathbf{0.0592}}\mathbf{= 40.5}