Question

1. Consider the following 2 half reactions

Fumarate +2H + 2e- --> succinate E= 0.031

Pyruvate +2H + 2e- --> lactate E= -0.185

a. In each reaction identify the oxidized form and the reduced form.

b. Write the overall reaction

c. Identify what is getting oxidized and what is getting reduced.

d. Calculate deltaE

e. Calculate deltaE under the following conditions; fumarate= 0.18mM, succinate= 0.32 mM, pyruvate= 0.25mM, lactate= 0.28nM at 37 degrees.

Answer 1

Fumarate + 2 H⁺ + 2 e^- ----------> Succinate         E = 0.031 V ……..(I)

Pyruvate + 2 H⁺ + 2 e^- -----------> Lactate            E = -0.185 V …….(II)

(a) Fumerate accepts two electrons and is converted to succinate. Since electrons are added to fumerate, hence, the process is reduction. Therefore, fumerate is the oxidized form and succinate is the reduced form.

Lactate is formed from pyruvate by the addition of two electrons to pyruvate. Again, since electrons are added to pyruvate, the process is reduction and thus, pyruvate is the oxidized form while lactate is the reduced form.

(b) Since pyruvate/lactate has a negative reduction potential, thus, it is extremely difficult to reduce pyruvate to lactate. Consequently, the reverse process, i.e, oxidation of lactate to pyruvate is easier. Thus, fumerate is reduced to succinate while lactate is oxidized to pyruvate in the overall reaction. The overall reaction is obtained by subtracting (II) from (I) to give

Fumerate + Lactate ----------> Succinate + Pyruvate

(c) Lactate is oxidized to Pyruvate while Fumerate is reduced to Succinate.

(d) The overall emf of the reaction is given as

ΔE = E(I) – E(II)

= (0.031 V) – (-0.185 V)

= 0.216 V

(d) The above, (d) denotes the emf of the overall reaction under standard conditions. Under non-standard conditions, the emf of the overall reaction is given as

ΔE = ΔE_std – RT/nF*ln [succinate][pyruvate]/[fumerate][lactate]

where T denotes the absolute temperature of the reaction; n is the number of electrons involved in the reaction and F = 96485 J/V.mol is the Faraday’s constant.

The square braces, [..] denote molar concentrations of the reactants and the products.

Since two electrons are involved in the reaction, hence, n = 2.

The reaction takes place at 37ºC; therefore, T = (37 + 273) K = 310 K.

Plug in values and get

ΔE =

(0.216 V) – (8.314 J/mol.K)*(310 K)/(2)(96485 J/V.mol)*ln [0.32 mM][0.25 mM]/[0.18 mM][0.28 mM]

= (0.216 V) – (0.0133 V)*ln (1.5873)

= (0.216 V) – (0.0133 V)*(0.4620)

= (0.216 V) – (0.0061446 V)

= 0.2098554 V

≈ 0.210 V

Fumarate + 2 H⁺ + 2 e^- ----------> Succinate         E = 0.031 V ……..(I)

Pyruvate + 2 H⁺ + 2 e^- -----------> Lactate            E = -0.185 V …….(II)

(a) Fumerate accepts two electrons and is converted to succinate. Since electrons are added to fumerate, hence, the process is reduction. Therefore, fumerate is the oxidized form and succinate is the reduced form.

Lactate is formed from pyruvate by the addition of two electrons to pyruvate. Again, since electrons are added to pyruvate, the process is reduction and thus, pyruvate is the oxidized form while lactate is the reduced form.

(b) Since pyruvate/lactate has a negative reduction potential, thus, it is extremely difficult to reduce pyruvate to lactate. Consequently, the reverse process, i.e, oxidation of lactate to pyruvate is easier. Thus, fumerate is reduced to succinate while lactate is oxidized to pyruvate in the overall reaction. The overall reaction is obtained by subtracting (II) from (I) to give

Fumerate + Lactate ----------> Succinate + Pyruvate

(c) Lactate is oxidized to Pyruvate while Fumerate is reduced to Succinate.

(d) The overall emf of the reaction is given as

ΔE = E(I) – E(II)

= (0.031 V) – (-0.185 V)

= 0.216 V

(d) The above, (d) denotes the emf of the overall reaction under standard conditions. Under non-standard conditions, the emf of the overall reaction is given as

ΔE = ΔE_std – RT/nF*ln [succinate][pyruvate]/[fumerate][lactate]

where T denotes the absolute temperature of the reaction; n is the number of electrons involved in the reaction and F = 96485 J/V.mol is the Faraday’s constant.

The square braces, [..] denote molar concentrations of the reactants and the products.

Since two electrons are involved in the reaction, hence, n = 2.

The reaction takes place at 37ºC; therefore, T = (37 + 273) K = 310 K.

Plug in values and get

ΔE =

(0.216 V) – (8.314 J/mol.K)*(310 K)/(2)(96485 J/V.mol)*ln [0.32 mM][0.25 mM]/[0.18 mM][0.28 mM]

= (0.216 V) – (0.0133 V)*ln (1.5873)

= (0.216 V) – (0.0133 V)*(0.4620)

= (0.216 V) – (0.0061446 V)

= 0.2098554 V

≈ 0.210 V