Respuesta :
Answer:
1.7 × 10 ^42
Explanation:
Using Nernst equation
E°cell = RT/nF Inq
at equilibrium
Q=K
E°cell = 0.0257 /n Ink= 0.0592/n log K
Fe2+(aq)+2e−→Fe(s) E∘= −0.45 V
Ag+aq)+e−→Ag(s) E∘= 0.80 V
Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)
balance the reaction
Fe → Fe²⁺ + 2e⁻ reversing for oxidation E° = 0.45 v
2 Ag⁺ +2e⁻ → 2Ag
n = 2 moles and K = equilibrium constant
E° cell = 0.80 + 0.45 = 1.25 V
E° cell = (0.0592 / n) log K
substitute the value into the equations and solve for K
(1.25 × 2) / 0.0592 = log K
42.23 = log K
k = 10^ 42.23
K = 1.7 × 10 ^42
The equilibrium constant for the given reaction at 25°C is [tex]\bold {1.7 \times 10 ^{42}}[/tex].
How to calculate the equilibrium constant?
It can be calculated by the Nernst equation, at equilibrium the
[tex]E_{cell} = \frac{R T}{n F} \ln Q[/tex]
Where,
[tex]E^0[/tex] = standard potential
R = universal gas constant
[tex]T[/tex] = temperature in kelvin
[tex]n[/tex] = ion charge (moles of electrons)
[tex]F[/tex] = Faraday constant
[tex]Q[/tex] = reaction quotient
At equilibrium, Q = K
[tex]E^o_{cell} = \frac{ 0.0592}{n } \ln k[/tex]
First, Calculate the standard cell potential,
[tex]\bold {2 Ag^+(aq) + 2 e^- \rightarrow 2 Ag(s)}[/tex] [tex]E^o_{reduction} = + 0.80 \rm \ V[/tex]
[tex]\bold {Fe(s) \rightarrow Fe2+ (aq) + 2 e^-}[/tex] [tex]E^o _{oxidation} = 0.45\rm \ V[/tex]
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[tex]\bold {Fe(s)+2Ag^+(aq)\rightarrow Fe_2^+(aq)+2Ag(s)}[/tex] [tex]E^o_{cell} = + 1.25 \rm \ V[/tex]
From the balanced chemical equation, the number of moles [tex]n = 2[/tex]
Put the values in the formula,
[tex]1.25 = \frac{ 0.0592}{2 } \ln k\\\\ k = 1.7 \times 10 ^{42}[/tex]
Therefore, the equilibrium constant for the given reaction at 25°C is [tex]\bold {1.7 \times 10 ^{42}}[/tex].
Learn more about the equilibrium constant:
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