The lead IV oxide formula or the lead 4 oxide formula - PbO2 is the formula of an oxide. The oxidation state of lead is +4. Moreover, it is also called lead dioxide, or Plumbic oxide or anhydrous Plumbic acid. The plumbic oxide is given as a dark-brown colour crystalline powder, which is insoluble inside the alcohol and water. It dissolves in hydrochloric acid, dilute nitric acid, oxalic acid, and other various acids also. However, it is a widely useful oxide while making matches, explosives, and electrodes. It is a strong oxidizing agent.
The accurate and exact mass and the monoisotopic mass of the Lead IV oxide are given as 239.966 g/mol. The hydrogen bond acceptors amount is equal to 2 and the hydrogen bond donors count is equal to 0. In addition, this compound has a single covalently bonded unit and is canonicalized.
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We obtain the lead IV oxide or lead dioxide commercially by a number of methods, such as oxidation of red lead means, Pb3O4 in the alkaline slurry in the chlorine atmosphere, reaction of the lead (II) acetate with the “chloride of lime” that is, calcium hypochlorite, which is the reaction of Pb3O4 and with the nitric acid affords the dioxide:
Pb3O4 + 4 HNO3 → PbO2 + 2 Pb (NO3)2 + 2 H2O
PbO2 compound reacts with the sodium hydroxide to form the hexahydroxy plumbate (IV) ion [Pb(OH)6]2−, which is soluble with water.
An alternate synthesis technique can be electrochemical. Here, the lead dioxide will produce on the pure lead in the dilute sulfuric acid when polarized anodically at the electrode potential near up to +1.5 V at the level of room temperature. We use this specific procedure for large-scale industrial production of the PbO2 anodes.
Then, the copper and lead electrodes immerse in the sulfuric acid, which is flowing at a rate of 5–10 L/min. The electrodeposition can be carried out galvanostatically by the current application of up to 100 A/m2 for nearly 30 minutes.
The chemical formula for lead IV oxide is given as plumbic oxide.
As an electrolyte, a 38 percent solution of sulphuric acid is used. Here, the density is given as 1.294 g mL-1) and the battery holds up to 3.5 L of the acid. During battery discharge, the density of the H2SO4 falls to 1.139 g mL-1.
Explain what amount of electricity we require in terms of Faraday to carry out a reduction of 1 mole of the PbO2?
Pb holds the oxidation state of +4 in the PbO2 and reduces to PbSO4, where the oxidation state is given as +2. Therefore, 2 electrons (+4 – +2) electrons are given as the amount that we require here for this reduction reaction to take place:
Pb4+ + 2e- → Pb2+
Since the 2 electrons are welcome, they can be used up in this reduction process. Thus, by Faraday’s law of electrolysis:
1 mole of the Pb requires 2F electricity to carry out the reduction process.