Question

Given below is the Gibbs free energy change (\(\Delta G\)) in kilo Joules per electron equivalent (kJ e^-eq^-1) at pH 7.0, of organic and inorganic half reactions.
Acetate synthesis:
\[ \frac{1}{8}CO_2 + \frac{1}{8}HCO_3^{-} + H^{+} + e^{-} \rightarrow \frac{1}{8}CH_3COO^{-} + \frac{3}{8}H_2O \quad \Delta G = 27.4\ \text{kJ e}^{-}\text{eq}^{-1} \]
Reduction reaction:
\[ \frac{1}{4}O_2 + H^{+} + e^{-} \rightarrow \frac{1}{2}H_2O \quad \Delta G = -78.72\ \text{kJ e}^{-}\text{eq}^{-1} \]
The free energy change of acetate oxidation to \(CO_2\), \(H_2O\) and \(HCO_3^{-}\) is kJ e^-eq^-1. (rounded off to two decimal places)

Hint:

Flip the given acetate synthesis reaction to get oxidation (negate its delta G), then add it to the oxygen reduction delta G since both are on a per electron equivalent basis.

Answer 1

Correct Answer: -106.12

Step 1: Understand what each given half reaction represents.
The two half reactions are written per electron equivalent, which is the standard way half reactions are compared in redox energetics used for microbial growth and metabolism calculations. The acetate synthesis reaction, as written, is a REDUCTION half reaction: it consumes an electron (\(e^{-}\)) and \(H^{+}\) to reduce \(CO_2\) and \(HCO_3^{-}\) down to acetate. Its free energy is given as \(\Delta G_1 = +27.4\ \text{kJ e}^{-}\text{eq}^{-1}\).
The reduction reaction reduces oxygen to water, also consuming an electron: \(\Delta G_2 = -78.72\ \text{kJ e}^{-}\text{eq}^{-1}\). This plays the role of the electron ACCEPTOR half reaction when oxygen is the terminal electron acceptor.

Step 2: Reverse the acetate reaction to get the oxidation (donor) half reaction.
We are asked for acetate OXIDATION, meaning acetate is broken down to \(CO_2\), \(H_2O\) and \(HCO_3^{-}\), which is exactly the reverse of the given acetate synthesis reaction. Reversing a reaction flips the sign of \(\Delta G\):
\[ \frac{1}{8}CH_3COO^{-} + \frac{3}{8}H_2O \rightarrow \frac{1}{8}CO_2 + \frac{1}{8}HCO_3^{-} + H^{+} + e^{-} \quad \Delta G_{donor} = -27.4\ \text{kJ e}^{-}\text{eq}^{-1} \]
This is the electron DONOR half reaction: acetate gives up an electron as it is oxidized.

Step 3: Combine the donor and acceptor half reactions.
In half reaction energetics, the overall free energy of a coupled redox process (donor oxidized, acceptor reduced) is the sum of the donor half reaction \(\Delta G\) (written as oxidation) and the acceptor half reaction \(\Delta G\) (written as reduction), because the electron released by the donor is the same electron consumed by the acceptor:
\[ \Delta G_{overall} = \Delta G_{donor} + \Delta G_{acceptor} \]

Step 4: Substitute the values.
\[ \Delta G_{overall} = (-27.4) + (-78.72) \]
\[ \Delta G_{overall} = -106.12\ \text{kJ e}^{-}\text{eq}^{-1} \]

Final Answer:
The free energy change of the coupled oxidation of acetate, with oxygen as the electron acceptor, is \[ \boxed{-106.12\ \text{kJ e}^{-}\text{eq}^{-1}} \]