Consider the two half-reactions below and their standard reduction potentials.NAD+ + H+ + 2e → NADHElo= -0.32 Va-Ketoglutarate + CO₂ + 2H+ + 2e → Isocitrate E' = -0.38 V(a) What is AE" for the spontaneous redox reaction that is, the reaction that actuallyoccurs under standard biochemical conditions (pH 7)?(b) Which of the following statements are correct under standard biochemical conditions?i. The concentration of H+ is 1.0 M.ii. The reaction NAD+ + Isocitrate → NADH + H+ + a-Ketoglutarate + CO₂ is favor-able.iii. NAD+ accepts electrons from isocitrate.iv. The NAD → NADH reaction actually occurs in reverse. The a-ketoglutarate →isocitrate reaction occurs as written.(c) Calculate AG" for the reaction in (a). (pH 7, 25°C, pressure, 1 atm.)(d) Suppose that the actual conditions are T = 25°C, pH = 7, CO₂ = 1 atm, [a-Ketoglutarate]10 mM, [NAD+] = 2.5 mM, and [NADH] = 0.5 mM.PAR= 2 mM, [Isocitrate]DEWhat is the value of AG under those conditions?(Hints: pH 7 is already included in AG". Similarly, 1 atm is the standard pressure ofCO2, which is also taken into account in AG.)

∆G'0 = -nF∆E'0 here; ∆G'0 is the change in standard Gibbs free energy. 'n' is the number of electrons transferred in the redox reaction. In our case, 2 electrons are transferred from Isocitrate to NAD+ . So, n=2 . 'F' is the Faradays constant. It has a value of 96,485 J V-1 mol-1 . The spontaneous redox reaction of concern, that occurs at standard biochemical condition is given below. Isocitrate + NAD+ → α-ketoglutarate + NADH + H+ + CO2Here we are given the 2 reduction half reactions involved. ∆E'0 = E'0(electron acceptor) - E'0(electron donor) The electron acceptor here is NAD+ and electron donor Isocitrate. E'0 is the standard reduction potential value. So, simply use the standard reduction potential values of the reduction half reactions involving Isocitrate and NAD+. These 2 half reactions and their E'0 values are given in question. So, we just need to use these values on the above equation. Do not reverse the sign! ∆E'0 = -0.32V-(-0.38V) = 0.06 VThe pH is 7 at standard biochemical conditions. pH = -log[H+]7=-log[H+][H+] = 1×10-7 M So, the concentration of H+ at standard biochemical condition is 1 x 10-7 M . We have found that the ∆E'0 of the reaction in option (ii) is 0.06 V i.e. a positive value. Hence, this reaction would have a negative ∆G'0 value. This means that this reaction would be favorable at standard biochemical condition. NAD+ gets reduced to NADH as it receives 2 electrons from isocitrate. NAD+ → NADH and Isocitrate →α-ketoglutarate are the conversions that occur here at standard biochemical condition. So, the correct options are (ii) and (iii) .∆G'0 = -nF∆E'0 =-2×96485 J V-1mol-1 × 0.06 V =-11578.2 J mol-1∆G = ∆G'0 + RT ln[products][reactants] here; 'R' is the universal gas constant and has a value of 8.315 J mol-1 K-1 'T' is temperature in kelvin. Given that temperature is 250C. Lets convert it to kelvin as shown below. 0K = 273.15 + 0C =273.15 + 250C =298.150K For the reaction of concern here, we can rewrite the equation for ∆G as shown below. ∆G = ∆G'0 + RT ln[α-ketoglutarate][NADH][Isocitrate][NAD+] Partial pressure of CO2 is 1 atm in standard biochemical condition. Also, the pH is 7 in standard biochemical condition. So, we do not need to consider H+ and CO2 concentration in the [product][reactant] part of the equation of ∆G, as it is already considered in the ∆G'0 component of the equation. Now, lets substitute the values and find the answer. ∆G = -11578.2+ 8.315×298.15 × ln2×0.510×2.5 = -19558.17 J mol-1