Finding ΔG and K for a given redox reaction

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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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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?