SME
Biochemistry · Problem type

Free energy from reduction potentials

Use this for “Two electrons pass from succinate (E°′ = +0.031 V) to O₂ (E°′ = +0.815 V). Find ΔG°′ and how many ATP it could pay for.”

Diagram

−0.4−0.20+0.2+0.4+0.6+0.8+1E°′ (V)DonorE°′(donor) = −0.32 VAcceptorE°′(acceptor) = +0.815 Ve⁻ΔE°′ = +1.135 V
ΔE°′
=
E°′(acceptor)
−
E°′(donor)
=
+0.815
V
−
(−0.32
V)
=
+1.135
V.
ΔG°′
=
−nFΔE°′
=
−2
×
96,485
×
1.135
1000
=
−219
kJ/mol.
The electrons fall down the tower, from the donor to the acceptor, on their own.
This is an example. Type your own values to start a new problem.
ΔG°′ (kJ/mol)
=
−
n
F(
E°′(acceptor) (V)
−
(
E°′(donor) (V)
)
)
ΔE°′
Potential difference
calculated · V
NATP
Most ATP it could make
calculated

Formulas

ΔE°′ = E°′(acceptor) − E°′(donor)
+1.135 = 0.815 − (−0.32)
ΔG°′
=
−
n
×
96.485
×
ΔE°′
−219 = −2 × 96.485 × 1.135
NATP = −ΔG°′ ÷ 30.5
7.181
=
−
−219
30.5

Assumptions

•
Electrons flow from the donor (lower E°′) to the acceptor (higher E°′), so ΔE°′ = E°′(acceptor) − E°′(donor).
•
ΔG°′ = −nFΔE°′ with F = 96.485 kJ/(mol·V) gives kJ/mol.
•
n is the electrons each molecule passes: 2 for NADH, FADH₂ and ½O₂.
•
Each ATP made from ADP + Pᵢ costs 30.5 kJ/mol at standard conditions, so −ΔG°′ ÷ 30.5 is the most ATP.

Step-by-step

WE KNOW
E°′(acceptor) = +0.815 V, E°′(donor) = −0.32 V, n = 2
FIND
potential difference (ΔE°′), standard free energy change (ΔG°′), most ATP it could make (NATP)
Step 1 · Find potential difference (ΔE°′)
Use ΔE°′ = E°′(acceptor) − E°′(donor)
Electrons fall from the donor’s potential to the acceptor’s; the gap is acceptor minus donor.
ΔE°′ = 0.815 − (−0.32)
ΔE°′ = +1.135 V
Step 2 · Find standard free energy change (ΔG°′)
Use ΔG°′ = −n × 96.485 × ΔE°′
Multiply
−
n
by
F,
96.485
kJ/(mol·V),
and
by
ΔE°′.
ΔG°′ = −2 × 96.485 × 1.135
ΔG°′ = −192.97 × 1.135
ΔG°′ = −219.02 kJ/mol
Step 3 · Find most ATP it could make (NATP)
Use NATP = −ΔG°′ ÷ 30.5
Share the free energy released among ATPs at 30.5 kJ/mol each.
NATP
=
−
−219.02
30.5
NATP = 7.181 → An upper limit: cells capture less (about 2.5 ATP for each NADH pair sent to O₂).
Check
Put the numbers back in. Both sides should match.
+1.135 = 0.815 − (−0.32) ✓
−219.02 = −2 × 96.485 × 1.135 ✓
7.181
=
max(0,
−
−219.021
30.5
)
✓
Topic check
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