This is what I have. Could you please correct my work? Thank you.
Compute the equilibrium constant for Ni^2+ (aq) and Cd (s).
reduction
Ni²⁺(aq) + 2 e⁻ ⇌ Ni(s) ; E°_red = -0.25V
-oxidation
Cd(s) ⇌ Cd²⁺(aq) + 2 e⁻ ; E°_ox = 0.40V
(negative value of the listed values, cause we consider reaction in opposite direction)
Overall
Ni²⁺(aq) + Cd(s) ⇌ Ni(s) + Cd²⁺(aq) + 2 e⁻ ; E = 0.40V
E° = E_ox + E°_red = 0.15V
∆G° = -n∙F∙E°
(n number of electrons exchanged, F faraday's constant)
∆G° = -R∙T∙ln(K)
n∙F∙E° = R∙T∙ln(K)
=>
K = e^{ n∙F∙E°/ (R∙T) }
standard temperature T = 298K
n = 2
K = e^{ 2 ∙ 96485J/Vmol ∙ 0.15V / (8.3145J/mol ∙ 298K) }
= 1.18×10⁵
1.18X10^10
I ran through the work and obtained 1.18 x 10^5, your first answer before you re-posted with 1.18 x 10^10
You don't list any concns in the reaction. I have assumed they must be 1 M and E for the half cell is Eo for the half cell.
Overall, your work is quite good and you have correctly applied the principles of redox reactions to calculate the equilibrium constant. However, I noticed a minor error in your calculation of the equilibrium constant. Let me correct that for you:
You correctly determined that the standard electromotive force (E°) of the overall reaction is 0.15V. To calculate the equilibrium constant (K), you need to use the equation:
K = e^(n*F*E° / (R*T))
where:
- n is the number of electrons exchanged (which is 2 in this case)
- F is Faraday's constant (96485 J/Vmol)
- R is the ideal gas constant (8.3145 J/mol K)
- T is the temperature in Kelvin (which is 298K in this case)
Now, let's substitute the values into the equation:
K = e^(2 * 96485 J/Vmol * 0.15V / (8.3145 J/mol K * 298K))
Using these values, we can calculate K:
K = e^(28947 J / (24.755 J/K))
K = e^(1169.53)
K ≈ 3.22 × 10^507
So, the corrected value for the equilibrium constant is approximately 3.22 × 10^507.
Remember to check your significant figures and round the final answer appropriately based on the level of precision required.