Finding ΔG and K for a given redox reaction

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Discussion Overview

The discussion revolves around calculating the Gibbs free energy change (ΔG) and the equilibrium constant (K) for a specific redox reaction involving B-hydroxybutarate and acetoacetate. The context includes the use of standard reduction potentials and conditions typical in biochemistry, specifically at pH 7 and 298.15 K.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • Post 1 presents the redox reaction and provides the half-reactions along with their standard reduction potentials, seeking to calculate ΔG and K.
  • Post 1 raises a question regarding the number of electrons transferred (n) and whether electrons should remain after balancing the half-reactions.
  • Post 2 asserts that if electrons are left over, the reaction is not balanced, implying that the balancing process should result in no excess electrons.
  • Post 3 and Post 4 both propose that n equals 2, suggesting that 2 moles of electrons are transferred per mole of product, and seek confirmation of this value.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the value of n, with some proposing it to be 2 while others emphasize the importance of balancing the reaction correctly. The discussion remains unresolved as participants seek confirmation on the number of electrons transferred.

Contextual Notes

The discussion does not clarify the assumptions behind the balancing of half-reactions or the implications of the chosen standard state conditions on the calculations.

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?
 
Last edited:
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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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