The body functions as a kind of fuel cell that uses oxygen from the air to oxidi

Question

The body functions as a kind of fuel cell that uses oxygen from the air to oxidize glucose:

C6H12O6(aq)+6O2(g)--> CO2 (g)+6H2O(l)

During normal activity, a person uses the equivalent of about 10 MJ of energy a day. Assume that this value represents delta G, and estimate the average current through your body in the course of a day, assuming that all the energy that we use arises from the reduction of O2 in the glucose oxidation reaction. See box 14.1

I'm not sure exactly what this question is asking... Am I supposed to find the reduction half- reaction that shows how energy we consume comes from the reduction of O2?

Explanation / Answer

average current through body in a day... is essentially flow of electrons per unit time, or "charge/time" which is tipically measured as

I [=] amp [=] C/s

time is already given, its per unit day, that is

1 day = 24 h = 24*3600 s = 86400 s

Now we need to calculate "Charge"

Relate dG as follows

dG = -n*F*Ecell

nF = Q, which is charge

so

dG = -Q*Ecell

Now, since we are required to assume this is solely the reduction of O2, this means:

O2(g) + 4 H+ + 4 e 2 H2O +1.229

that is

Ecell = 1.229 V

Find "nF" which is charge

dG = -Q*Ecell

10*10^6 J = -Q*(1.229)

Q = (10*10^6)/(1.229) = 8136696.50122 C

now

for an average day

I = Q/t

I = 8136696.50122/(86500) = 94.0658 C/s

I = 94.0658 amp

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