How to calculate oxidation number of Pb in PbSO4?And Explain? +1 +3 0-1 +2 +1. ... Pb PbO2 PbSO4 H2O H2SO4. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). Lead, being a metal, has very set oxidation numbers. So, Pb is at +2 oxidation state. The total should equal when you add all oxidation numbers: Pb + (-2) = 0 ----> Pb = +2. H in H2 : 0. If you consider oxidation as the loss of electrons and gain of oxidation number, and reduction as the gain of electrons and decrease in oxidation number, the oxidation states are as follows: Pb : 0 (Elementary state) H in H2SO4 : +2. I went by the rule of the most electronegative atom getting their typical oxidation state (so O was assigned -2 (total -8)...then S was assigned -2...and finally Pb was given +10 to balance the molecule). It is suggested that a precipitation mechanism to form PbSO 4 occurs under some conditions prior to the solid state … Figuring out why these numbers are the way they are takes a fair amount of work. Could you tell me how you went about getting the oxidation number for Pb in PbSO4? PbSO4 --> Pb2+ + SO4 2- (A net dissociation reaction of the salt) SO4 has an oxidation number of -2. Therefore oxidation number of Pb is +2. Oxidation number of S=+6. Sulfide ion is at -2 oxidation state. Oxidation number of Pb is +2 The rates of PbSO4 formation on the Pb–0.08 mass% Ca–Sn alloys, which are the choice materials for grids in the valve-regulated lead–acid battery (VRLA), were inhibited by the presence of Sn. PbSO 4 salt does not have an overall charge. Identify the products and the reactants, and then their oxidation numbers. Sum of all oxidation number =0. Also, the answer would not be +10 if what you had said was correct (about the sulfur ion). But Sulfate ion has -2 charge. The anodic oxidation of solid Pb in H 2 SO 4 to form PbSO 4 has been investigated by rotating disc, potentiostatic pulse and ac impedance measurements. Pb in PbSO4 : +2. Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions: Pb(s) + PbO 2 (s) + 2H 2 SO 4 (aq) → 2PbSO 4 (s) + 2H 2 O(l) To neutral that -2 charge, +2 charge should be there and it is given by Pb 2+ ion. Asked by | 25th Mar, 2009, 06:56: PM. Oxidation = number goes down Reduction = number goes up. Pb. It can only be +2, or +4. Oxidation number of O=-2. Expert Answer: Let the oxidation number of Pb=x. pb +2 s-2 + h +1 2 o-1 2 → pb +2 s +6 o-2 4 + h +1 2 o-2 b) Identify and write out all redox couples in reaction. The system is compared with measurements on Pb(Hg) in H 2 SO 4 and with Pb(Hg), Hg, solid Pb in HClO 4 solution. In order to make the compound neutral, Pb must be +2. H2O2 + PbS --> PbSO4 + H2O. SO4 has charge of -2. Therefore x+6-8=0. 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