O=-2
Pb=x
x+(-2)*2=0
x+4-=0
-4+4=0
Pb4-
No, Pb is not a transition metal and it has 2 oxidation states
It'll decompose by heating, releasing carbondioxide: Pb(II) or plumbous carbonate: Pb(CO3) --> PbO + CO2 or Pb(IV) or plumbic carbonate: Pb(CO3)2 --> PbO2 + 2CO2
86.62(8)%The molar masses of the constituent elements must be known.M(Pb) = 2.072(1) x 102 gmol-1M(O) = 1.59994(3) x 10 gmol-1From these and their respective molar ratios in the compound, the total molar mass of the compound must be calculated.M(PbO2) = M(Pb) + 2M(O)M(PbO2) = (2.072 x 10 gmol-1 + 2(1.59994 gmol-1)) x 10M(PbO2) = 2.392(1) x 102 gmol-1From the molar mass of the element in question and its ratio, and the compound, their total masses present in one mole of compound must be calculated.m = nMm(Pb) = 1 mol x 2.072 x 102 gmol-1m(Pb) = 2.072(1) x 102 gm(PbO2) = 1 mol x 2.392(1) x 102 g (always 1 mole)m(PbO2) = 2.392(1) x 102 gFrom these, the percentage of the element in question present in the compound can be calculated.m(Pb) x 100%/m(PbO2) = 2.072 g x 100%/2.392 gm(Pb) x 100%/m(PbO2) = 86.62(8)%
Dr M KanagasabapathyAsst. Professor in ChemistryRajus college, Affiliated to M K UniversityRajapalayam (TN) India------------------------------------------------------------------------------------------------------Lead acid battery contains the following1. Lead PB2. Lead oxide PbO3. Sulfuric acid H2SO4During discharging Pb Oxidizes to Pb 2+ and PbO2 reduces to Pb2+ from Pb 4+PbO2 + Pb + H2SO4 = 2 PbSO4 + H2O (~ 2.0 V per cell)During charging the reverse reaction takes placePb2+ reduces to lead at cathode and another molecule of Pb2+ oxidizes to PB4+ or PbO2 by reacting with distilled water.2 Pb(2+) + H2O2 = Pb + PbO2 (> 2.0 V per cell)The important thing is the voltage for recharging should be slightly higher than the discharging voltage.During discharging the battery functions as electrochemical cell.And during charging it becomes an electrolytic cell.By,Dr M KanagasabapathyAsst. Professor in ChemistryRajus college, Affiliated to M K UniversityRajapalayam (TN) India
K = +1 oxidation state Cl = +3 oxidation state O = -2 oxidation state
-2 for each O +4 for Pb
O is 2- and there are two of them so Pb would have to be 4+
Pb + PbO2 + H2SO4 --> PbSO4 + H2O
No, Pb is not a transition metal and it has 2 oxidation states
It has to be Pb(NO3)2 with NaCl as Pb has a +II oxidation state and NO3 has -I oxidation state. The reaction is the following: Pb(NO3)2 +2NaCl ----> PbCl2 + 2NaNO3
Tin and Lead are both metals with oxidation numbers of 2+ and 4+, so the two equations for Lead (Pb) are 2Pb + O2 ---> 2PbO and Pb + O2 ---> PbO2 The equations for Tin (Sn) look the same ... just substitute Sn where you see Pb.
Assuming the 2 oxidation state of lead. Pb + 2HNO3 --> Pb(NO3)2 + H2
PbO2....Pb has a 4+ charge and therefore can be reduced to Pb2+ in PbO. Since it can be reduced it is an oxidizing agent.
the positive plate is Pb the negative is PbO2 and the acid is H2SO4.
PbO2 > Pb+O2
The formula for plumbous oxide is PbO.
The ionic chemical formula of lead(IV) oxide is: (Pb)4+ + 2 O-.