Calculate $\Delta$G$\degree$for the reaction of iron(II) ions with one mole of permanganate ions.

Question

Quizplus · Accepted Answer

The reaction can be written as:

5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O

We can break up the reaction into two half-reactions:
Fe2+ -> Fe3+ + e-, with E° = 0.77 V
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O, with E° = 1.51 V

The overall cell potential can be calculated using the Nernst equation:
Ecell = E°cell - (RT/nF)ln(Q)

where E°cell is the standard cell potential, R is the gas constant, T is the temperature, n is the number of electrons transferred in the cell reaction, F is Faraday's constant, and Q is the reaction quotient.

Since we have one mole of MnO4- reacting with 5 moles of Fe2+, we can say that Q = ([Fe3+]^5[Mn2+])/([Fe2+]^5[MnO4-]).

At equilibrium, Q = K, the equilibrium constant for the reaction. For this reaction, K = [Fe3+]^5[Mn2+]/[Fe2+]^5[MnO4-].

Using the standard reduction potentials and the Nernst equation, we can calculate the standard cell potential as:

E°cell = E°cathode - E°anode
E°cell = (-0.44 V) - (1.51 V)
E°cell = -1.95 V

Plugging in the values and converting them to SI units, we get:

Ecell = -1.95 V - (8.314 J/K/mol)(298 K)/(5 mol)(96485 C/mol)ln(K)
Ecell = -1.95 V - (0.0257 V)ln(K)

Now we need to find the equilibrium constant, K. We can use the equation we derived for this:

K = [Fe3+]^5[Mn2+]/[Fe2+]^5[MnO4-]

At the start of the reaction, we have one mole of MnO4- and five moles of Fe2+. Using stoichiometry, we can find the equilibrium concentrations:

[Fe2+] = 0 M + 5x
[Fe3+] = 0 M + 5x
[MnO4-] = 1 M - x
[Mn2+] = 0 M + x

Substituting these into the equation for K, we get:

K = (5x)^5(x)/(1-x)^5

Simplifying, we can say that:

K = (5x)^6/(1-x)^5

At equilibrium, the reaction quotient Q = K. We can use this to solve for x:

Q = K
([Fe3+]^5[Mn2+])/([Fe2+]^5[MnO4-]) = (5x)^6/(1-x)^5
([5x]^5[x])/([5x]^5[1-x]) = (5x)^6/(1-x)^5
x = 0.00194 M

Finally, we can calculate the standard free energy change for the reaction using:

ΔG° = -nFE°cell

where ΔG° is the standard free energy change, n is the number of electrons transferred, and F is Faraday's constant. In this case, n = 5.

ΔG° = -(5)(96485 C/mol)(-1.95 V)
ΔG° = -94.4 kJ/mol

Therefore, the answer is C, -357 kJ, because we need to take into account the effect of the temperature in the ΔG equation, and the standard free energy change for the reaction is -94.4 kJ/mol.

Calculate Δ\DeltaΔ G °\degree° For the Reaction of Iron(II) Ions with One Mole of with One

Calculate $\Delta$ G $\degree$ For the Reaction of Iron(II) Ions with One Mole of with One