which statement is false regarding the redox reactions shown? a. it requires energy to reduce nad⁺. b. nadh is a stronger electron donor than nitrite (no₂⁻) c. nitrate (no₃⁻) is a better terminal electron acceptor than nad⁺ d. none are false; all are true statements. electron acceptor → electron donor (mv) (a)

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# Brief Explanations:
- Option A: Reducing $ \text{NAD}^+ $ to $ \text{NADH} $ (e.g., in glycolysis or TCA cycle) requires energy input (in the form of electrons and protons), so this is true.
- Option B: $ \text{NADH} $ is an electron donor (it donates electrons to the electron transport chain), not an electron donor stronger than nitrite in a way that's misrepresented here—actually, $ \text{NADH} $ donates electrons, while nitrite ($ \text{NO}_2^- $) can act as an electron acceptor or donor, but the key is $ \text{NADH} $ is an electron donor, not a "stronger electron donor" in a false context. Wait, more accurately: $ \text{NADH} $ is an electron donor (loses electrons), while the statement says it's a "stronger electron donor than nitrite"—but nitrite can be an electron acceptor (e.g., in denitrification, nitrite is reduced, so it's an electron acceptor, not a donor in that case). So $ \text{NADH} $ is an electron donor, but the comparison to nitrite as a donor is incorrect because nitrite is typically an acceptor. Thus, this statement is false? Wait, no—let's re-express: Redox potential: $ \text{NADH} $ has a low (negative) redox potential, meaning it's a good electron donor. Nitrite ($ \text{NO}_2^- $) has a higher (less negative/more positive) redox potential, so as an electron acceptor, it's better, but as a donor, $ \text{NADH} $ is a better donor (since lower potential means more energy to donate electrons). Wait, maybe the error is in the term: $ \text{NADH} $ is an electron donor, but the statement says "stronger electron donor than nitrite"—but nitrite is not typically an electron donor (it's an acceptor, e.g., $ \text{NO}_2^- $ is reduced to $ \text{NO} $ or $ \text{N}_2\text{O} $ in denitrification, so it's accepting electrons). So $ \text{NADH} $ is a donor, nitrite is an acceptor, so comparing them as donors is incorrect—thus the statement is false? Wait, no, let's check each option again:

- Option C: Nitrate ($ \text{NO}_3^- $) has a higher (more positive) redox potential than $ \text{NAD}^+ $, so it's a better terminal electron acceptor (since it can accept electrons more readily, releasing more energy). So this is true.
- Option D: If A, C are true, and B is false, then D is wrong. Wait, maybe I messed up B. Let's recall: $ \text{NADH} $ is an electron donor (gives electrons to Complex I in ETC). Nitrite ($ \text{NO}_2^- $)—in some reactions, nitrite can be an electron donor (e.g., in nitrite-oxidizing bacteria, nitrite is oxidized to nitrate, so it's a donor). But the redox potential of $ \text{NADH} $ is around -0.32 V, and nitrite/nitrate couple ( $ \text{NO}_3^-/\text{NO}_2^- $) is around +0.42 V. So as a donor, $ \text{NADH} $ has a lower (more negative) potential, meaning it's a stronger donor (since it can donate electrons more easily to a higher potential acceptor). Wait, no—redox potential: the more negative the potential, the stronger the electron donor (because it has a higher tendency to lose electrons). So $ \text{NADH} $ (E°' ~ -0.32 V) is a stronger donor than nitrite (if nitrite is a donor, its E°' for $ \text{NO}_2^- \to \text{NO}_3^- $ is +0.42 V, which is a higher (more positive) potential, meaning it's a weaker donor (since it's more likely to accept electrons to be reduced, but if it's oxidized, it's a donor, but its potential is positive, so it's a weak donor). So $ \text{NADH} $ is a stronger donor than nitrite (as a donor). Wait, that would make B true? Then where's the false statement?

Wait, let's re-express each option:

A: Reducing $ \text{NAD}^+ $ to $ \text{NADH} $ (e.g., in catabolic reactions) requires energy (electrons + protons), so this is true (energy is needed to form $ \text{NADH} $ from $ \text{NAD}^+ $).

B: $ \text{NADH} $ is a stronger electron donor than nitrite ($ \text{NO}_2^- $)—as per redox potentials, $ \text{NADH} $ has a more negative E°', so it's a stronger donor (since it releases more energy when donating electrons). So B is true?

C: Nitrate ($ \text{NO}_3^- $) has a higher E°' than $ \text{NAD}^+ $ ( $ \text{NO}_3^-/\text{NO}_2^- $ is +0.42 V, $ \text{NAD}^+/\text{NADH} $ is -0.32 V), so nitrate is a better terminal electron acceptor (since it can accept electrons and release more energy). So C is true.

D: None are false—all are true. But that can't be. Wait, maybe I made a mistake with B. Wait, $ \text{NADH} $ is an electron donor, but nitrite is an electron acceptor (in many cases). So the statement says "NADH is a stronger electron donor than nitrite"—but if nitrite is an acceptor, not a donor, then comparing them as donors is incorrect. So maybe the statement is false? Wait, no—nitrite can be a donor (e.g., in nitrite oxidation to nitrate, nitrite is the donor). So in that case, $ \text{NADH} $ (donor, E°' -0.32 V) vs. nitrite (donor, E°' +0.42 V for $ \text{NO}_2^- \to \text{NO}_3^- $). A more negative E°' means stronger donor (since it has a higher tendency to lose electrons). So $ \text{NADH} $ is a stronger donor than nitrite (as a donor). So B is true.

Then A: reducing $ \text{NAD}^+ $ to $ \text{NADH} $ requires energy—yes, because $ \text{NAD}^+ $ is a higher energy state? No, wait: $ \text{NADH} $ has more energy than $ \text{NAD}^+ $, so to form $ \text{NADH} $ from $ \text{NAD}^+ $, energy (in the form of electrons and protons) is required. So A is true.

C: Nitrate is a better terminal electron acceptor than $ \text{NAD}^+ $—yes, because its redox potential is higher, so using it as an acceptor releases more energy. So C is true.

Thus, all statements are true, so D is correct? Wait, but the question is which is false. Wait, maybe I misread B: "NADH is a stronger electron donor than nitrite"—but $ \text{NADH} $ is a donor, nitrite is an acceptor, so the comparison is invalid. So the statement is false? Wait, no—if nitrite is a donor (in nitrite oxidation), then the comparison is valid. This is confusing. Alternatively, maybe the false statement is B. Wait, let's check standard redox concepts:

- Electron donors are oxidized (lose electrons), electron acceptors are reduced (gain electrons).

- $ \text{NADH} $ is oxidized to $ \text{NAD}^+ $ (donates electrons), so it's a donor.

- Nitrite ($ \text{NO}_2^- $): in denitrification, it's reduced (accepts electrons), so it's an acceptor. In nitrite oxidation (by nitrite-oxidizing bacteria), it's oxidized (donates electrons) to nitrate.

- Redox potential: the more negative the E°', the stronger the donor (since it has a higher tendency to lose electrons). $ \text{NADH}/\text{NAD}^+ $ E°' is -0.32 V, $ \text{NO}_2^-/\text{NO}_3^- $ E°' is +0.42 V. So as a donor, $ \text{NADH} $ is stronger (more negative E°') than nitrite (which has a positive E°' as a donor, meaning it's a weak donor). So B is true.

Then A: reducing $ \text{NAD}^+ $ to $ \text{NADH} $ requires energy—yes, because you're forming a higher-energy molecule, so energy (from food, etc.) is needed. So A is true.

C: Nitrate ($ \text{NO}_3^- $) has a higher E°' than $ \text{NAD}^+ $ ( $ \text{NO}_3^-/\text{NO}_2^- $ is +0.42 V, $ \text{NAD}^+/\text{NADH} $ is -0.32 V), so nitrate is a better terminal electron acceptor (since it can accept electrons and release more energy). So C is true.

Thus, all statements are true, so D is the answer (None are false; all are true). But that seems odd. Wait, maybe the error is in A: "It requires energy to reduce $ \text{NAD}^+ $"—actually, in biological systems, reducing $ \text{NAD}^+ $ to $ \text{NADH} $ is an exergonic process when coupled with other reactions (like glycolysis or TCA cycle), but the reduction itself (the half-reaction) has a positive free energy change? Wait, the half-reaction $ \text{NAD}^+ + \text{H}^+ + 2e^- \to \text{NADH} $ has a standard reduction potential (E°') of -0.32 V, so the free energy change $ \Delta G°' = -nFE°' $, so $ \Delta G°' = -2 \times 96485 \times (-0.32) \approx +61.7 \, \text{kJ/mol} $, which is endergonic (requires energy). So to reduce $ \text{NAD}^+ $, energy is required, so A is true.

Therefore, all statements A, B, C are true, so D is correct. But the question is which is false—so if D says "None are false; all are true", then D is the correct answer (since the other options are true).
# Answer:
D. None are false; all are true statements.

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