Question

If the standard reduction potential for NAD+/NADH is -0.315 V and the standard reduction potential for oxaloacetate/malate is -0.166 V, what is the change in standard reduction potential (o’) for the oxidation of malate by NAD+?

Answer 1

The change in standard reduction potential for the oxidation of malate by NAD+ is -0.149 V, indicating the process occurs spontaneously under standard conditions.

Step-by-step explanation:

The change in standard reduction potential (Δεo’) for the oxidation of malate by NAD+ can be calculated by subtracting the standard reduction potential of the oxidized form from that of the reduced form. The given potentials are -0.166 V for oxaloacetate/malate and -0.315 V for NAD+/NADH.

Δεo’ = εo’(oxidized) - εo’(reduced) = εo’(NAD+/NADH) - εo’(oxaloacetate/malate) = -0.315 V - (-0.166 V) = -0.149 V.

This negative potential indicates that the overall process is spontaneous under standard conditions because the NAD+ is a stronger oxidizing agent than oxaloacetate.

Answer 2

The change in standard reduction potential for the oxidation of malate by NAD+ is calculated as -0.149 V, which is found by subtracting the standard reduction potentials of oxaloacetate/malate from NAD+/NADH.

Step-by-step explanation:

The question pertains to the calculation of the change in standard reduction potential (Δε´ˆ) for the oxidation of malate by NAD+. This is a redox reaction occurring in biochemical pathways, specifically during the citric acid cycle. The Δε´ˆ is calculated by subtracting the standard reduction potential of the agent getting reduced from that of the agent getting oxidized. In this case, the standard reduction potentials are -0.315 V for NAD+/NADH and -0.166 V for oxaloacetate/malate. The change in standard reduction potential for the oxidation of malate by NAD+ can be calculated as follows:

Δε´ˆ = E´ˆ (oxidizing agent) - E´ˆ (reducing agent)

Δε´ˆ = (-0.315 V) - (-0.166 V)

Δε´ˆ = -0.149 V

Therefore, the change in standard reduction potential for the oxidation of malate by NAD+ is -0.149 V.