Generally, when working out the theoretical cell capacity, you work out the anode and cathode combined capacity using:
Q = (nF)/(3.6M)
Where Q is the gravimetric theoretical capacity (in mAh/g), n is the number of transferred electrons, F is faradays constant 96485 C/mol) and M is the molecular weight (in g/mol) of the portion of the electrode that holds the intercalate that released n electrons (e.g. in graphite it is 6 carbons). The 3.6 is a conversion factor to get the required mAh/g units.
The the non active component weights can then be added in, decreasing the theoretical mAh/g figure.
The issue I'm having a problem with is that in order to get a Wh/kg figure, you need to multiply by the PRACTICAL cell voltage. What if I want to do a calculation for a cell that is novel? It seems there is no way to figure out the wholly theoretical specific capacity.
Do you know what chemistry it is based on? Do you have very detailed information about the anode/cathode materials (to calculate the losses associated with mass transport,intercalation, etc.) What details do you have? I assume you know the two half-cell reactions and possibly the Gibbs' free energy?
A very lazy approach would be to make an estimate based on the delta of theoretical to practical voltages for that chemistry?
Or, if you know the type of processes occuring, you could check for typical values of the parasitic losses observed.
The cathode active materials are Na0.3WO3. The half cell reactions are as follows:
Anode:
7AlCl4- + Al Al2Cl7- + 3e-
Proposed Cathode:
AlCl4- + Na0.3WO3 AlCl4Na0.3WO3 + e-
Note sure what the Gibbs' free energy in this case. Is this able to be calculated as well? Not sure any tables would have this data...@Jules Lloyd Hammond
- Badges
- Science topic
- Similar topics
- Engineering
- Electrical Engineering
- Electrodes