Calculate the equilibrium constant?

Question

Using the standard reduction potentials listed in Appendix E in the textbook, calculate the equilibrium constant for each of the following reactions at 298 K .
a.
Cu(s)+2Ag^+(aq) ---->Cu^2+(aq)+2Ag(s)
b.
3Ce^4+(aq)+Bi(s)=H2o(l)---->3Ce^3+(aq)…
c.
N2H5^+(aq)+4Fe(CN)6^3-(aq)---->N2(g)+5…

Answer 1

Best Answer - Chosen by Voters

Cell Potential of the reaction and equilibrium constant are related as:
E° = R∙T/(n∙F)∙ln(K)
<=>
K = e^(n∙F∙E°/ (R∙T) )
(R universal gas constant, T absolute temperature, n number of electrons exchanged in reaction, F Faraday constant)

At T = 298K
F/(R∙T) = 96485C/mol/ (8.3145J/molK ∙ 298K) = 38.941V⁻¹
=>
K = e^(38.941V⁻¹ ∙ n ∙ E° )


a.
Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) ; E° = 0.340V
Ag⁺(aq) + e⁻ ⇌ Ag(s) ; E° = 0.7996V
=>
Cu(s) ⇌ Cu²⁺(aq) + 2 e⁻ ; E° = -0.34V
2Ag⁺(aq) + 2e⁻ ⇌ Ag(s) ; E° = 2∙7996V = 1.5992V
=>
Cu(s) + 2 Ag⁺(aq) ⇌ Cu²⁺(aq) + 2 Ag(s) ;
E° = -0.340V - 1.5992.4V = 1.2592V
=>
K = e^(38.941V⁻¹ ∙ 2 ∙ 1.2592V )
= e^( 98.069)
= 3.90×10⁴²


b.
Ce⁴⁺(aq) + e⁻ ⇌ Ce³⁺(aq) ; E° = 1.44V
Bi³⁺(aq) + 3e⁻ ⇌ Bi(s) ; E° = 0.32V
=>
3Ce⁴⁺(aq) + 3e⁻ ⇌ 3Ce³⁺(aq) ; E° = 3∙1.44V = 4.32V
Bi(s) ⇌ Bi³⁺(aq) + 3e⁻ ; E° = -0.32V
=>
3Ce⁴⁺(aq) + Bi(s) ⇌ 3Ce³⁺(aq) + Bi³⁺(aq)
E° = 4.32V - 0.32.4V = 4.0V
=>
K = e^(38.941V⁻¹ ∙ 3 ∙ 4.0V )
= e^( 467.3 )
too large for my pocket calculator


c.
i cant#t read the whole reaction equation.

Source(s):

Standard elektrode potentials
http://en.wikipedia.org/wiki/Standard_el…