Given the following standard reduction potentials,
Fe^2+(aq) + 2e^- ==> Fe(s) E naught = -o.440V
FeS(s) + 2e^- ==> Fe(s) + s^2-(aq) E naught = -1.010V
determine the Ksp for FeS(s) at 25 degrees Celsius
Fe(s) ==> Fe^2+ + 2e^- (oxidation)
FeS(s) + 2E^- ==> Fe(s) + S^2-(aq) (reduction)
net reaction: FeS(s) + Fe(s) ==> Fe^2+(aq) + S^2-(aq) + Fe(s)
(-1.010V) - (-0.440V) = -.57
lnk = (1)(-0.57v)/(0.257v)
Ksp = e^-22.5 = 1.8x10^-10
To determine the Ksp for FeS(s) at 25 degrees Celsius using the given standard reduction potentials, follow these steps:
1. Identify the relevant redox reactions:
- The oxidation half-reaction: Fe(s) -> Fe^2+(aq) + 2e^-
- The reduction half-reaction: FeS(s) + 2e^- -> Fe(s) + S^2-(aq)
2. Write the overall balanced equation for the reaction:
FeS(s) + Fe(s) -> Fe^2+(aq) + S^2-(aq) + Fe(s)
3. Determine the overall standard cell potential (E°):
Subtract the reduction potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction.
E° = E°(reduction) - E°(oxidation)
E° = -1.010V - (-0.440V) = -0.57V
4. Use the Nernst equation to relate the standard cell potential (E°) to the equilibrium constant (K):
ln(K) = (n)(-E°) / (0.0257V)
In this case, n = 1 since there is a 1:1 ratio between FeS and Fe^2+ ions in the balanced equation.
5. Solve for ln(K):
ln(K) = (1)(-0.57V) / (0.0257V) = -22.5
6. Calculate K:
K = e^(-22.5) = 1.8x10^-10
Therefore, the Ksp for FeS(s) at 25 degrees Celsius is 1.8x10^-10.