Half Reaction Problem

Question

These are the two half reactions:
Oxaloacetate+2H++2e-=malate (-0.166) =E0(V)
Fumarate+2H++2e-=succinate (+0.031)=E0(V)
1. Which of the oxidizing agents is stronger?
(Professor said fumarate)
2.which of the reducing agents is stronger?
(Malate)
3. What is the full reaction?
(malate+fumarate=oxaloacetate+succinate) Is this correct?
4. Which two agents will be present in highest concentration?
(Succinate and oxaloacetate?

Answer 1

What are given are the standard REDUCTION potentials. Thus...

Oxaloacetate + 2H⁺ + 2e- ==> malate  Eº = -0.166 V

Fumarate + 2H⁺ + 2e- ==> succinate  Eº = +0.031 V

 

The corresponding OXIDATION potentials are..

malate ==> oxaloacetate + 2H⁺ + 2e-  Eº = +0.166 V

succinate ==> fumarate + 2H⁺ + 2e-  Eº = -0.031 V

 

From these reactions and oxidation/reduction potentials, we can determine that..

(1)  The strongest oxidizing agent is FUMARATE because it is the one most easily reduced.

(2)  The strongest reducing agent is MALATE because it is the one most easily oxidized.

 

To obtain the full reaction, reverse the oxaloacetate => malate reduction reaction and make it an oxidation reaction. Then add the reduction and oxidation reactions together.

Fumarate + 2H⁺ + 2e- ==> succinate 

Malate ========> oxaloacetate + 2H⁺ +2e- 

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Fumarate + malate + 2H⁺ +2e- ==> succinate + oxaloacetate + 2H⁺ + 2e-  This is the FULL REACTION

You can cancel H⁺ and e- on both sides to obtain

Fumarate + malate ==> succinate + oxaloacetate  This is also the FULL REACTION

 

(4) Since the fumarate is being reduced to form succinate, and at the same time the malate is being oxidized to form oxaloacetate, the products, i.e. succinate and oxaloacetate will be in the highest concentrations at the end of the reaction.