Finding ΔG and K for a given redox reaction

AI Thread Summary
To calculate ΔG and the equilibrium constant K for the given redox reaction, the standard reduction potentials of the half-reactions are combined to find ΔErxn, resulting in 1.162 V. The equation ΔG = -nFΔE is used, where n represents the moles of electrons transferred. The discussion confirms that n equals 2, as the electrons cancel out when balancing the half-reactions. The calculations proceed with n = 2 to determine ΔG and subsequently K using the equation ln K = nFE/RT. The importance of balancing the reaction and correctly identifying n is emphasized for accurate results.
anisotropic
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Homework Statement



Redox reaction:
B-hydroxybutarate + 1/2O2 -> acetoacetate + H2O

Half reaction 1:
O2 + 4H+ + 4e- -> 2H2O (E = +0.816 V)

Half reaction 2:
acetoacetate + 2H+ + 2e- -> B-hydroxybutarate (E = -0.346 V)

Using standard reduction potentials given,

  • calculate ΔG
  • calculate the equilibrium constant, K

Conditions are the biochemist's standard state, pH 7, 298.15 K.

Homework Equations



ΔG = -nFΔE

ln K = nFE/RT

The Attempt at a Solution



ΔErxn = (0.816 V) + (0.346 V)
ΔErxn = 1.162 V

ΔG = -nFΔE
ΔG = -n(96485 C/mol)(1.162 V)

n = ? (moles of electrons transferred)

By writing out both half reactions, the electrons cancel out. Should there not be electrons left over on one side of the equation in order to determine the value of n?
 
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No, if electrons are left, reaction is not balanced.
 
So 2 mol of e- per mol of product (n = 2)?
 
anisotropic said:
So 2 mol of e- per mol of product (n = 2)?
Can someone please confirm this?
 
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