HW 14.117

Moderators: Chem_Mod, Chem_Admin

Jingyi Li 2C
Posts: 56
Joined: Fri Sep 29, 2017 7:06 am

HW 14.117

Postby Jingyi Li 2C » Sat Feb 17, 2018 9:07 pm

HW 14.117:
The body functions as a kind of fuel cell that uses oxygen from the air to oxidize glucose:
C6H12O6 (aq) + 6O2 (g) -->6CO2 (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.

Can someone explain this problem please? Thanks.

Clara Rehmann 1K
Posts: 53
Joined: Fri Sep 29, 2017 7:03 am

Re: HW 14.117

Postby Clara Rehmann 1K » Thu Feb 22, 2018 2:14 pm

Start with ΔG = -nFE and It = nF. Combining these equations, you get I=

Knowing that Eº = ~1 V, you can plug in values (ΔG = -10 x 106, E = 1, t = 86400) and and solve for a solution in amps

Juanyi Tan 2K
Posts: 33
Joined: Fri Sep 29, 2017 7:05 am

Re: HW 14.117

Postby Juanyi Tan 2K » Thu Feb 22, 2018 10:26 pm

There are examples in the textbook about how to use the equation n=It/F. The page number is 595. Also, the equation is related to delta G and E because delta G= -nFE, which can be written as It=nF=-delta G/E.

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

Who is online

Users browsing this forum: No registered users and 3 guests