Equilibrium constant question!!!?
Standard reduction potentials
Reduction half-reaction E(V)
Ag+(aq) + e- ---> Ag (s) 0.80
Cu ^2+ (aq) + 2e- ---> Cu (s) 0.34
Sn^4+(aq) + 4e- ---> Sn (s) 0.15
2 H+(aq) + 2e- ---> H2 (g) 0
Ni^2+(aq) + 2e- ---> Ni (s) -0.26
Fe^2+ (aq) + 2e- --->Fe(s) -0.45
Zn^2+ (aq) +2e- ---> Zn (s) -0.76
Al^3+ (aq) + 3e- ---> Al (s) -1.66
Mg^2+ (aq) + 2e- ---> Mg (s) -2.37
Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 deg C) for the following reaction:
Fe (s) + Ni^2+ (aq) ---> Fe^2+ (aq) + Ni (s)
To calculate the equilibrium constant for the given reaction, we will use the Nernst equation, which relates the standard reduction potentials of species involved in a reaction to the equilibrium constant.
First, let's assign the oxidation states of the species in the reaction:
Fe(s) ---> Fe^2+(aq) + 2e-
Ni^2+(aq) + 2e- ---> Ni(s)
We can see that the reduction half-reaction for Fe is given, but for Ni, we need to flip the given reduction half-reaction.
So, the balanced overall reaction will be:
Fe(s) + Ni^2+(aq) ---> Fe^2+(aq) + Ni(s)
Now, let's determine the standard cell potential, E°, for the reaction:
E° = E°(cathode) - E°(anode)
E° = E°(Ni) - E°(Fe)
E° = E°(Ni) - (-0.45)
E° = E°(Ni) + 0.45
From the provided table, we know that E°(Ni) = -0.26 V.
Plug in the value and solve for E°:
E° = -0.26 + 0.45
E° = 0.19 V
Next, we will use the Nernst equation:
E = E° - (RT/nF) * ln(Q)
Where:
E = Cell potential under non-standard conditions
E° = Standard cell potential
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin (25°C = 298 K)
n = Number of electrons transferred in the balanced equation
F = Faraday constant (96485 C/mol)
Q = Reaction quotient
In this case, since we want to calculate the equilibrium constant, Q is the ratio of product concentrations to reactant concentrations at equilibrium:
Q = [Fe^2+(aq)] / [Ni^2+(aq)]
Now, let's plug in the values and calculate the equilibrium constant, K:
E = 0.19 - [(8.314 * 298) / (2 * 96485)] * ln(Q)
The natural logarithm of the reaction quotient can be calculated as follows:
ln(Q) = ln([Fe^2+(aq)] / [Ni^2+(aq)])
ln(Q) = ln[Fe^2+(aq)] - ln[Ni^2+(aq)]
Remember that ln(1) = 0, so if Q = K (at equilibrium), then E = 0.
Therefore, we can set E = 0 and solve for ln(K):
0 = 0.19 - [(8.314 * 298) / (2 * 96485)] * [ln[Fe^2+(aq)] - ln[Ni^2+(aq)]]
Simplify the equation and solve for ln(K):
0.19 = (8.314 * 298) / (2 * 96485) * [ln[Ni^2+(aq)] - ln[Fe^2+(aq)]]
ln(K) = [2 * 96485 * 0.19] / (8.314 * 298) * [ln[Ni^2+(aq)] - ln[Fe^2+(aq)]]
Finally, solve for K by taking the exponential of both sides of the equation:
K = e^(ln(K))
Using the calculated value of ln(K), we can find the equilibrium constant for the given reaction.