View source for Carbon footprint to charge an iPod

== The Problem: ==

How much coal must you burn (and how much CO2 will you produce) to generate enough electrical energy to charge up your iPod/cellphone?

== The approximate half-cell reactions and overall reaction: ==
A typical lithium ion battery generates about 3.7 volts

xLi+ + xe- ⇄ Liºx(graphite) Eº = -3.05 volts

Li1-xMn2O4(s) + xLi++ xe- ⇄ LiMn2O4(s) Eº = +0.65 volts

Liºx(graphite) + Li1-xMn2O4(s) ⇄ LiMn2O4(s) Eºcell = +3.7 volts

== You can compute the maximum amount of work ==


Assumption #1: An average cell phone/iPod battery
contains about 7 grams (1 mole of Li)
Work that can be done by the battery
(during discharge/energy production):

wmax = -(coulombs) · Ecell ≈

(1 mole Li) · (96,500 coulombs/mole)·(3.7 volts)
≈ -357 kJoules (in fact you won’t get this much, you might get ca. 1/2 of this amount, since you won’t use all of the available Li° and you are limited by rates of electron transfer at both anode and cathode

Energy output ≈ -178.5 kJoules


== Now the work necessary to fully charge the battery: ==


LiMn2O4(s) ⇄ Liºx(graphite) + Li1-xMn2O4(s) Eºcell = -3.7 volts
Work that must be done to the battery (during charge/energy consumption):

wmax = -(coulombs) · Ecell ≈

(1 mole Li) · (96,500 coulombs/mole)·(3.7 volts)
≈ +357 kJoules (in fact you will have to do at least twice this much work to fully charge the battery
-- same problem: available Li+ and rates of
ET at both anode and cathode

Energy input ≈ +714 kJoules


== You go to standard thermodynamic tables to get the enthalpy for each of the reactants and products ==
{| class="wikitable"
|-
!  Formula
!  ΔHo,f
!  ΔGo,f
!  So
!  SoP
|-
|
|kJ/mol 
|kJ/mol 
|J/K·mol 
|J/K·mol
|-
|C(c,graphite) 
|0.0  
|0.0  
|5.740  
|8.527
|-
|C(c,diamond)  
|1.895  
|2.900  
|2.377  
|6.113
|-
|C(g)  
|716.682  
|671.257  
|158.096  
|20.838
|-
|CO2(g)  
|(-393.509)  
|(-394.359)  
|213.74 
|37.11
|-
|O2(g)  
|0.0  
|0.0  
|205.138  
|29.355
|-
|}

<math>\Delta G^{o,rxn} = \sum_m\Delta G^{o,f}(products) - \sum_n\Delta G^{o,f}(reactants)\,\!</math>

Assumption 

#2: coal = graphite
<math>C(graphite) + O2 \rightarrow CO2\,\!</math>

== Then to calculate the energy per unit mass of coal ==


ΔHo,rxn = ΣmΔHo,f(products) - ΣnΔHo,f(reactants)
= (-393.5) – ((0) + (0)) = -393.5 kJ/mole

ΔSo,rxn = ΣmSo,f(products) - ΣnΔSo,f(reactants)
= (213.7) – (5.74 + 205.1) = 2.9 J/mole = 0.0029 kJ/mole

ΔGrxn ≈ ΔHo,rxn - TΔSo,rxn (assume 1000K for the combustion
temperature) ≈ -393.5 kJ/mole - (1000)(.0029 kJ/mole)

ΔGrxn ≈ -396.4 kJ/mole


== Next to calculate transmission loss for power lines over a distance ==

<math>C(graphite) + O2 \rightarrow CO2\,\!</math>

ΔGrxn ≈ -396.4 kJ/mole

By the time it gets to Tucson, and to your wall outlet, you may have lost 2/3 of that available energy
(transmission losses):

Available energy to charge your battery ≈
-396.4 kJ/mole/3 =
-132 kJ/mole of carbon

Energy input to charge your battery ≈ +714 kJoules

Moles of carbon (coal) needed to charge your battery
≈ +714 kJoules/132 kJ/mole = 5.4 moles of carbon

Moles of carbon (coal) needed to charge your battery
≈ +714 kJoules/132 kJ/mole = 5.4 moles of carbon

5.4 moles of carbon >> 65 grams of carbon

'''454 grams/lb >>> requires 0.14 lb of coal(carbon,graphite)
to charge your iPod.'''

How much CO2 is generated? Mole CO2 = moles C
= 5.4 moles of CO2

Grams of CO2 = 44 grams/mole x 5.4 moles =
238 grams of CO2 :

'''You generate ca. ½ pound of CO2 per charge!'''