GATE 2020 BT Q.55: solution and answer

GATE 2020 · BTQ.55 · 2 marks · NAT

The mitochondrial electron transfer chain oxidizes NADH with oxygen being the terminal electron acceptor. The redox potentials for the two half-reactions are given below:

$NAD^{+} + H^{+} + 2 e^{-} \to NADH$ , $E^{0'} = - 0.32 V$

$\frac{1}{2} O_{2} + 2 H^{+} + 2 e^{-} \to H_{2} O$ , $E^{0'} = 0.816 V$

The free energy change associated with the transfer of electrons from NADH to $O_{2}$ is ______ kJ/mol (round off to 2 decimal places).

Given: $F = 96500 C/mol$ .

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Answer and solution

ANSWER

The free energy change for electron transfer from NADH to O2 is about -219.25 kJ/mol.

Official answer: -219.30 to -219.20Source: Official answer key, GATE 2020 BT (IIT Delhi)

Electrons flow spontaneously from the couple with the lower reduction potential ( $NAD^{+} / NADH$ , $- 0.32 V$ ) to the couple with the higher one ( $O_{2} / H_{2} O$ , $0.816 V$ ). So NADH is the electron donor and $O_{2}$ is the acceptor.
The free energy change is $Δ G^{0'} = - n F Δ E^{0'}$ , with $Δ E^{0'} = E_{acceptor}^{0'} - E_{donor}^{0'}$ .
$Δ E^{0'} = 0.816 - (- 0.32) = 1.136 V$ .
Both half-reactions carry 2 electrons per NADH oxidized, so $n = 2$ .
$Δ G^{0'} = - 2 \times 96500 C/mol \times 1.136 V = - 219248 J/mol$ .
Converting to kJ: $Δ G^{0'} = - 219.248 kJ/mol \approx - 219.25 kJ/mol$ . The negative sign shows the transfer is strongly exergonic.

KEY IDEA

Work out $Δ E^{0'}$ as acceptor minus donor, then use $Δ G^{0'} = - n F Δ E^{0'}$ with $n = 2$ electrons per NADH.

WATCH OUT FOR

Subtracting the potentials the wrong way round, which gives a positive $Δ G$ . Using $n = 1$ gives half the value. Forgetting to divide by 1000 to get kJ.

From the GATE 2020 BT paper. Every question of this paper · Download the full paper · All 13 Metabolism questions, with notes

GATE 2020 BT Q.55: solution and answer

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