Consider the following half-reactions:
$\varepsilon$ ^o' = -0.315 V NAD⁺ + H⁺ + 2 e^- $\leftrightarrow$ NADH
$\varepsilon$ ^o' = 0.815 V 1/2 O₂ + 2 H⁺+ 2 e^- $\leftrightarrow$ H₂O
What is the $\varepsilon$ ^o'for the following redox equation?
O₂ + 2 NADH + 2 H⁺ $\to$ 2 H₂O + 2 NAD⁺

Question

Consider the following half-reactions:

\varepsilon

^o' = -0.315 V NAD⁺ + H⁺ + 2 e^-

\leftrightarrow

NADH

\varepsilon

^o' = 0.815 V 1/2 O₂ + 2 H⁺+ 2 e^-

\leftrightarrow

H₂O
What is the

\varepsilon

^o'for the following redox equation?
O₂ + 2 NADH + 2 H⁺

\to

2 H₂O + 2 NAD⁺

Quizplus · Accepted Answer

The standard electrode potential for a redox reaction can be calculated by subtracting the standard reduction potential of the reductant (which gets oxidized) from the standard reduction potential of the oxidant (which gets reduced). Here, the NADH is being oxidized (standard reduction potential = -0.315 V) and the 1/2 O2 is being reduced (standard reduction potential = 0.815 V). Therefore, the overall standard electrode potential is 0.815 V - (-0.315 V) = 1.130 V.

Consider the Following Half-Reactions ε\varepsilonε o' = -0 ↔\leftrightarrow↔ NADH ε\varepsilonε o' = 0

Consider the Following Half-Reactions $\varepsilon$ ^o' = -0 $\leftrightarrow$ NADH $\varepsilon$ ^o' = 0