Question

Suppose you construct the following galvanic cell: Fe(s)|Fe²⁺(aq)||NAD⁺(aq)|NADH(aq)

Fe²⁺ + 2e^- → Fe E^o = -0.44V

NAD⁺ + 2e^- + 2H⁺ → NADH + H⁺ E^o' = -0.320V

(note the different standard states)

A) Will the cell generate current at biochemical standard state? At chemical standard state? At T = 4 ^oC, pH = 7? At T = 97 ^oC, pH = 7? Justify your answers.

B) How many protons can be moved across a membrane from pH 7.5 to pH 6.7 using your freshly charged galvanic cell?

If oxygen is present, the following reaction can occur: H⁺ + NADH + ½ O₂ → NAD⁺ + H₂O. Given the following standard reduction potential, O₂ + 4H⁺ + 4e^- → 2 H₂O E^o = +1.23V

C) If you start with 1M of NADH, what percent of it becomes oxidized after the reaction above reaches equilibrium (at biochemical standard state)?

D) You run your galvanic cell at concentrations of 1 mM NADH and 1 mM Fe²⁺. What is the overall cell potential? What are the concentrations of these species at equilibrium?

Answer 1

A: Fron the galvanic cell reaction it is clear that Fe(s) is oxidised to  Fe²⁺(aq), where as NAD+(aq) is reduced to NADH(aq).

For chemical standard state:

Oxidation reaction: Fe(s) ----> Fe²⁺(aq) + 2e-, E⁰(oxi) = + 0.44 V

Reduction reaction:  NAD+(aq) + 2e- + 2H⁺(aq) --- > NADH(aq) +H⁺(aq), E⁰(red) = - 0.320V

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Overall cell reaction: Fe(s) + NAD+(aq) + H⁺(aq) ----- > Fe²⁺(aq) + NADH(aq),  

E⁰(cell) =  E⁰(oxi) + E⁰(red) = + 0.44 V + (- 0.320V) = + 0.12 V

Since the value of E⁰(cell) is positive, the cell reaction is spontaneous at chemical standard state(concentration of reactants and products are 1 M)

Hence the cell will generate current at chemical standard state. (answer)

Also G⁰ = - nFE⁰(cell) = - 2x96500x( 0.12 V ) = - 23160 J

For biochemical Standard state:

Overall cell reaction: Fe(s) + NAD+(aq) + H⁺(aq) ----- > Fe²⁺(aq) + NADH(aq),  

Since H+ is consumed in the above reaction

G⁰¹ =  G⁰ + RTxln(1 / [H⁺(aq)]) =  G⁰ - RTxln[H⁺(aq)]

[H+(aq)] = 10^-7 M

=> G⁰¹ =  - 23160 J - 8.314 JK-1mol-1x298 Kxln[10^-7 M] = - 23160 J - ( - 39934) = +16774 J

Since G⁰¹ is positive, the reaction is not spontaneous at biochemical state.

Hence the cell will not generate current at biochemical standard state. (answer)

At T = 4 DegC and pH =7:

T = 4 degC = 273+4 = 277 K

G⁰¹ =  - 23160 J - 8.314 JK-1mol-1x277 Kxln[10^-7 M] = - 23160 J - ( - 37120J) = + 13960 J

Since G⁰¹ is positive, the reaction is not spontaneous at T = 4 DegC and pH =7

Hence the cell will not generate current at T = 4 DegC and pH =7 (answer)

At T = 97 DegC and pH =7:

T = 97 degC = 273+97 = 370 K

G⁰¹ =  - 23160 J - 8.314 JK-1mol-1x370 Kxln[10^-7 M] = - 23160 J - ( - 49582J) = + 26422 J

Since G⁰¹ is positive, the reaction is not spontaneous at T = 97 DegC and pH =7

Hence the cell will not generate current at  T = 97 DegC and pH =7 (answer)