I know the answer is .947 V, but I have no clue how to get to that answer! Pleas

Question

I know the answer is .947 V, but I have no clue how to get to that answer! Please help! Nicotinamide Adenine Dimeleotide cation. NAD 1, can bereduced via haifreaction NAD 1+(aq)+ H 1+ (aq)+2 e 2- = NADH(aq) E degree = -0.320v Conider, however, the axidation of NADH via moleclar oxygen thet occurs in the mitochondria of acrobie cells where, for the following net balanced redox reaction, E degree (cell)=+1.135 V at 25 degree C: 2NADH(aq)= O 2(g)=2H 1+(aq) doubleheadarrow 2H 2 o(l)+@NAD 1+(aq) Calculate or Provide the following if: [NAD 1+]=4.19 times 10 -6M:[NADH]=8.72 times 10 -4M: and po 2=1.83 times 10 -3 atm The number of electrons transferred in the balanced redox reaction: The value of E degree (reduction potential) for the cathode The value of E degree if the reaction is buffered near blood pH = 7.30 (assume that the concentrations of other species are as given above):

Explanation / Answer

2NADH(aq) + O2(g) + 2H+(aq) ----------> 2H2O(l) + 2NAD+(aq) ; E0cell = 1.135 V

Now, in O2 , each oxygen atom has a oxidation number of 0

In H2O , oxygen atom is having a oxidation state of -2

Thus, number of electron transferred per O-atom = 0 - (-2) = 2

Hence, total electrons required to convert both O-atoms of O2 molecule to H2O = 2*2 = 4

Thus, number of electron transfer taking place per O2 molecule = 2*2 = 4................(a)

Now,

Oxidation half reaction (i.e. anode reaction):-

NADH(aq) --------> NAD+(aq) + 2e- + H+(aq) ; E0cell reduction = -0.32 V

Reduction half reaction (i.e cathode reaction):-

O2(g) + 4e- ----------> 2H2O(l) ; E0cell reduction = X V

Now, E0cell for overall reaction = E0cell reduction (cathode) - E0cell reduction (anode)

or, 1.135 = E0cell reduction (cathode) - (-0.32)

or, E0cell reduction (cathode) = 1.455 V.................(b)

Now, as per Nernst equation:-

Ecell = E0cell - (0.059/n)*log{[NAD+]2}/{NADH]2*[O2]*[H+]2} ; where n = total number of electron transfer taking place in the overall balanced reaction = 4

Now, pH = -log[H+]

Thus, [H+] = 10-pH = 10-7.3 = 5.012*10-8 M

Thus, Ecell = 1.135 - (0.059/4)*log{(4.19*10-6)2/{(8.72*10-4)2*(1.83*10-3)*(5.012*10-8)2}} = 0.948 V

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