Using the standard reduction potentials, calculate the equilibrium constant for each of the following reactions at 298 K.
10Br- + 2MnO4- + 16H+ -> 2Mn2+ + 8H20 + 5Br2
Work:
2Br- -> Br2 + 2e- (This would be multiplied by 5 to balance out with the upper equation.)
MnO4- + 8H+ + 5e- -> Mn2+ +4H20 (This would be multiplied by 2 to balance out with the upper equation.)
The standard reduction potential for Br- is +1.065.
The standard reduction potential for MnO4- is +1.51.
Look up that equation, nEF = -lnK (I'm not at home and can't confirm that equation so check it out) but that is the one you want to use.
I looked back through my notes and found the following equations:
K = e^(-delta G/RT)
Delta G = -nFE
I know that E = standard reduction of the reduced - standard reduction of the oxidized
I've tried plugging everything I have in, but I'm not getting the right answer. I can't figure out what it is that I am doing wrong.
I have E = 1.51 - 1.065 and n = 10
So I have that
1.51 V - 1.065 V = .445 V = E
and that
ΔG = - 10 mol * 96485 J / mol*V * .445 V
ΔG = -429358 J
so then plugging in:
K = e^(-(-429358)/(8.314 J/mol*K * 298 K))
K = 1.82 * 10^75
The 'correct' answer appears to be 1.5 * 10^75 , but that's not far off, and my homework program accepted my answer as correct.
To calculate the equilibrium constant (K) for the given reaction, we need to use the Nernst equation, which relates the equilibrium constant to the standard reduction potentials (E°) of the half-reactions involved.
The Nernst equation is given by:
E = E° - (RT / nF) * ln(Q)
Where:
E is the cell potential or electromotive force (emf),
E° is the standard cell potential,
R is the gas constant (8.314 J/(mol*K)),
T is the temperature in Kelvin,
n is the number of moles of electrons transferred,
F is the Faraday constant (96485 C/mol),
ln is the natural logarithm,
and Q is the reaction quotient.
Let's calculate the equilibrium constant step-by-step:
1. Calculate the cell potential (E) for the overall reaction using the standard reduction potentials (E°).
E° (overall) = E° (cathode) - E° (anode)
E° (overall) = (1.51 V) - (-1.065 V) [Note: The standard reduction potential for Br2=1.065 V]
E° (overall) = 2.575 V
2. Calculate the reaction quotient (Q) for the given reaction.
This involves multiplying the individual half-reactions by their respective stoichiometric coefficients:
Q = [Mn2+]^2 / [Br-]^10
3. Plug the values into the Nernst equation and solve for ln(Q):
E = E° - (RT / nF) * ln(Q)
2.575 V = 0 - [(8.314 J/(mol*K)) * (298 K) / (10 mol * 96485 C/mol)] * ln(Q)
ln(Q) = -2.40855
4. Exponentiate both sides of the equation to solve for Q:
Q = e^(-2.40855)
Q ≈ 0.0896
Therefore, the equilibrium constant (K) for the given reaction at 298 K is approximately 0.0896.