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
| 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!