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Old 07-21-2019, 05:39 AM
bistander bistander is online now
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Originally Posted by Turion View Post
... So I take it this means you do not accept my theory that the energy is not consumed by the load? Or that the energy is still in the battery just equalized between the two sides?
Let's put it this way. I accept the science and theories which are well established and easily found in many reputable textbooks and college curriculums. Theories are like claims. You're expected to provide proofs. The authors and professors do this and have been for a very long time for the views I hold. Case in point; how the lead-acid cell works.


The present lead-acid cell consists, in a state of full charge, of a negative plate, or cathode, of spongy lead in a grid of hard lead, a positive plate, or anode, of PbO2*paste in a grid of hard lead, and an electrolyte of dilute sulphuric acid of specific gravity 1.28. This is a 37% solution, with 472.5 g/l of H2SO4. At full discharge, the electrolyte is of specific gravity 1.05, an 8% solution containing 84.18 g/l of acid. Both plates are coated with PbSO4. Approximately 4 moles of acid are used per litre, which corresponds to 213 A-h of charge (an ampere-hour is a current of one ampere flowing for one hour, or 3600 coulomb). Assuming that 4 moles of Pb are reacted at the cathode, and 4 moles of PbO2*at the anode, the total weight of active materials is about 3 kg. This gives a weight-to-capacity ratio of 14 g/A-h. Of course, this is much lower than is required for a practical battery, with case, electrode grids and other necessities. However, a limit of perhaps 25 g/A-h represents the maximum that can be expected of a lead-acid battery, and a limit of about 200 A-h per litre of electrolyte volume.

The cathode reaction is Pb + SO4++*→ PbSO4*+ 2e-. For each atom of lead, two electrons pass through the external circuit when the cell is delivering current. At the anode, the reaction is PbO2*+ 4H+*+ 2SO4++*+ 2e-*→ PbSO4*+ 2H2O. This reaction uses the two electrons sent by the cathode through the external circuit. For each two electrons, two molecules of acid are turned into two molecules of water and two molecules of lead sulphate. The electrode potential of the cathode reaction is -0.355V, and the electrode potential of the anode reaction is 1.685V, at standard concentrations. The net potential difference is 1.685 - (-0.355) = 2.040V. At the concentrations in a fully charged battery, the potential difference is closer to 2.2V, decreasing to 2.0V for a fully discharged battery.

It is easy to measure the specific gravity of the electrolyte with a hydrometer, and this gives an accurate estimate of the state of charge of the battery. This is one of the great advantages of the lead-acid cell. Note that the electrode reactions do not show any evolution of gases. With open cells, there is in fact some emission of H2*and O2, so the water lost in this way must be replenished regularly. This was once a regular duty in servicing a car, but modern batteries require very little care, and some are sealed, venting gas only when necessary. Also, ventilation was necessary to prevent the hydrogen from becoming an explosion hazard. When ordinary car batteries are charged rapidly, water is electrolyzed.