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There is a reaction in which succinate reacts with FAD to give fumarate and FADH. In this reaction, which substance is oxidized, and which is reduced? Which substance is the oxidizing agent, and which is the reducing agent?

Short Answer

Expert verified
Succinate is oxidized and is the reducing agent. FAD is reduced and is the oxidizing agent.

Step by step solution

01

Identify the Reactants and Products

The reactants in the given reaction are succinate and FAD, while the products are fumarate and FADH.
02

Understand Oxidation and Reduction

Remember that oxidation is the loss of electrons, and reduction is the gain of electrons. The substance that loses electrons is oxidized, and the substance that gains electrons is reduced.
03

Determine Oxidation States

In this context, the oxidation state of succinate increases when it becomes fumarate, indicating it is oxidized. FAD accepts electrons and becomes FADH, indicating it is reduced.
04

Identify Oxidizing and Reducing Agents

The substance being oxidized (succinate) is the reducing agent because it donates electrons. The substance being reduced (FAD) is the oxidizing agent because it accepts electrons.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation
Oxidation is a chemical process where a substance loses electrons. It's often associated with gaining oxygen or losing hydrogen in a reaction. In biochemical contexts, such as cellular respiration, oxidation plays a vital role in energy production. For example, in the exercise reaction where succinate turns into fumarate, succinate is oxidized. This means that succinate loses electrons, allowing it to transform into fumarate. Remember, the key indicator of oxidation is the loss of electrons or an increase in the oxidation state of the substance.
Reduction
Reduction is the opposite of oxidation— it involves the gain of electrons. In a biochemical reaction, when a molecule is reduced, it typically gains hydrogen or loses oxygen. In our given exercise, FAD is reduced to FADH. This occurs because FAD gains electrons, becoming FADH in the process. To easily remember, oxidation is losing (OIL) and reduction is gaining (RIG) — OIL RIG. Understanding this concept provides clarity on how energy transformations and electron transfers occur within cells.
Oxidizing Agent
An oxidizing agent is a substance that promotes oxidation by accepting electrons from another substance. It itself gets reduced in the process. This means that an oxidizing agent often gains electrons during a reaction. In the exercise mentioned, FAD is the oxidizing agent because it takes electrons from succinate, allowing succinate to be oxidized to fumarate. By gaining these electrons, FAD is reduced to FADH. Think of the oxidizing agent as the 'electron acceptor.' It helps other substances lose electrons and undergo oxidation.
Reducing Agent
A reducing agent is the chemical that promotes reduction by donating electrons to another substance. It itself gets oxidized in the process. In the biochemical reaction described, succinate acts as the reducing agent. This is because succinate donates electrons to FAD, leading to the formation of fumarate. By giving away its electrons, succinate is oxidized while FAD is reduced to FADH. The reducing agent is essential in driving the reduction process, acting as the 'electron donor.'

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Most popular questions from this chapter

The following half reactions play important roles in metabolism. \\[ \begin{array}{c} ^{1} /_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 e^{-} \rightarrow \mathrm{H}_{2} \mathrm{O} \\ \mathrm{NADH}+\mathrm{H}^{+} \rightarrow \mathrm{NAD}^{+}+2 \mathrm{H}^{+}+2 e^{-} \end{array} \\] Which of these two is a half reaction of oxidation? Which one is a half reaction of reduction? Write the equation for the overall reaction. Which reagent is the oxidizing agent (electron acceptor)? Which reagent is the reducing agent (electron donor)?

Identify the molecules oxidized and reduced in the following reactions and write the half reactions. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}+\mathrm{NADH} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{NAD}^{+}\) (b) \(\mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{+}(\mathrm{aq})+\mathrm{Fe}^{3+}(\mathrm{aq})\)

The \(\Delta G^{\text {os }}\) for the reaction Citrate \(\rightarrow\) Isocitrate is \(+6.64 \mathrm{kJ} \mathrm{mol}^{-1}=+1.59 \mathrm{kcal} \mathrm{mol}^{-1} .\) The \(\Delta G^{\circ}\) for the reaction Isocitrate \(\rightarrow \alpha\) -Ketoglutarate is \(-267 \mathrm{kJ} \mathrm{mol}^{-1}=-63.9 \mathrm{kcal} \mathrm{mol}^{-1}\) What is the \(\Delta G^{\circ}\) for the conversion of citrate to \(\alpha\) -ketoglutarate? Is that reaction exergonic or endergonic, and why?

Show that the hydrolysis of ATP to AMP and \(2 P_{i}\) releases the same amount of energy by either of the two following pathways. Pathway 1 \\[ \begin{array}{l} \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{ADP}+\mathrm{P}_{\mathrm{i}} \\ \mathrm{ADP}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{AMP}+\mathrm{P}_{\mathrm{i}} \end{array} \\] Pathway 2 \\[ \begin{array}{c} \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{AMP}+\mathrm{PP}_{\mathrm{i}}(\mathrm{Pyrophosphate}) \\ \mathrm{PP}_{\mathrm{i}}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{P}_{\mathrm{i}} \end{array} \\]

There is a reaction in carbohydrate metabolism in which glucose- 6 -phosphate reacts with \(\mathrm{NADP}^{+}\) to give 6 -phosphoglucono- 8 -lactone and NADPH. In this reaction, which substance is oxidized, and which is reduced? Which substance is the oxidizing agent, and which is the reducing agent?

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