.. o electrons lost by the hydrogen atoms, thus becoming negatively charged (SO4–). These groups can combine with others of opposite charge to form other compounds. The lead-acid cell uses sulfuric acid as the electrolyte. The lead-acid storage battery is the most common secondary battery used today, and is typical of those used in automobiles.
The following will describe both the charging and discharging phase of the lead-storage battery and how sulfuric acid, as the electrolyte, is used in the process. The lead storage battery consists of two electrodes or plates, which are made of lead and lead peroxide and are immersed in an electrolytic solution of sulfuric acid. The lead is the anode and the lead peroxide is the cathode. When the battery is used, both electrodes are converted to lead sulfate by the following process. At the sulfate ion that is present in the solution from the sulfuric acid.
At the cathode, meanwhile, the lead peroxide accepts two electrons and releases the oxygen; lead oxide is formed first, and then lead joins the sulfate ion to form lead sulfate. At the same time, four hydrogen ions released from the acid join the oxygen released from the lead peroxide to form water. When all the sulfuric acid is used up, the battery is discharged produces no current. The battery can be recharged by passing the current through it in the opposite direction. This process reverses all the previous reactions and forms lead at the anode and lead peroxide at the cathode.Proposed Problem i) The concentration of sulfuric acid is 0.0443 mol/L.
The pH is: No. mol of hydrogen ions = 0.0443 mol/L x 2 = 0.0886 mol/L hydrogen ions pH = – log [H] = – log (0.0886) = – (-1.0525) = 1.05 Therefore, pH is 1.05. ii) The amount of base needed to neutralize the lake water is: volume of lake = 2000m x 800m x 50m = 800,000,000 m3 or 8×108 m3 since 1m3=1000L, therefore 8×1011 L 0.0443 mol/L x 8×1011 = 3.54 x 1010 mol of H2SO4 in water # mol NaOH = 3.54 x 1010 mol H2SO4 x 2 mol NaOH 1 mol H2SO4 = 7.08 x 1010 mol of NaOH needed Mass of NaOH = 7.08 x 1010 mol NaOH x 40 g NaOH 1 mol NaOH = 2.83 x 1012 g NaOH or 2.83 x 109 kg NaOH Therefore a total of 2.83 x 1012 g of NaOH is needed to neutralize the lake water.iii) The use of sodium hydroxide versus limestone to neutralize the lake water: Sodium hydroxide: Sodium hydroxide produces water when reacting with an acid, it also dissolves in water quite readily. When using sodium hydroxide to neutralize a lake, there may be several problems. One problem is that when sodium hydroxide dissolves in water, it gives off heat and this may harm aquatic living organisms. Besides this, vast amounts of sodium hydroxide is required to neutralize a lake therefore large amounts of this substance which is corrosive will have to be transported.
This is a great risk to the environment if a spill was to occur. The following equation shows that water is produced when using sodium hydroxide.2NaOH + H2SO4 –* Na2 SO4 + 2H2O Limestone: Another way to neutralize a lake is by liming. Liming of lakes must be done with considerable caution and with an awareness that the aquatic ecosystem will not be restored to its original pre-acidic state even though the pH of water may have returned to more normal levels. When limestone dissolves in water it produces carbon dioxide. This could be a problem since a higher content of carbon dioxide would mean a lowered oxygen content especially when much algae growth is present.
As a result, fish and other organisms may suffer. Limestone also does not dissolve as readily as sodium hydroxide thus taking a longer period of time to react with sulfuric acid to neutralize the lake. The equation for the neutralization using limestone is as follows: Ca CO3 + H2SO4 –* CaSO4 + H2O. iv) The effect of the Acid or excessive Base on the plant and animal life: You will probably find that there aren’t many aquatic living organisms in waters that are excessively basic or acidic. A high acidic or basic content in lakes kill fishes and other aquatic species. Prolonged exposure to acidic or excessively basic conditions can lead to reproductive failure and morphological aberration of fish. A lowered pH tends to neutralize toxic metals.
The accumulation of such metals in fish contaminates food chains of which we are a part as these metals can make fish unfit for human consumption. Acidification of a lake causes a reduction of the production of phytoplankton (which is a primary producer) as well as in the productivity of the growth of many other aquatic plants. In acidic conditions, zooplankton species will probably becompletely eliminated. In addition, bacterial decomposition of dead matter is seriously retarded in acidified lake waters. Other effects of acidic conditions arean overfertilization of algae and other microscopic plant lifecausing algae blooms. Overgrowth of these consumes quickly most of the oxygen in water thus causing other life forms to die from oxygen starvation.
When there are excessive base or acid in waters, not only do aquatic organisms get affected but animals who depend on aquatic plants to survive will starve too, since few aquatic plants survive in such conditions. Therefore each organism in the aquatic ecosystem is effected by excessive basic or acidic conditions because anything affecting one organism will affect the food chain, sending repercussions throughout the entire ecosystem. v) The factors that govern this plant’s location, if this plant employs 40% of the towns people: The major factors that would govern this plant’s location would be whether there is ready access to raw materials; whether the location is close to major transportation routes; whether energy resources are readily available and if there is an adequate water supply in the area. Since this plant would employ 40% of the towns people, the plant should be close to the town while still far enough so that in case of any leakage of the plant, the town will be within a safe distance of being severely affected. The factor of whether the general living conditions in the area are suitable for the workers should also be considered as well.