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Title: Reducing and oxidizing agent Post by: Annonn on Apr 1, 2022 For the following reaction, the reducing agent is ____(i)______ and the oxidizing agent is _____(ii)______.
Pb(s) + PbO2(s) + 2H2SO4(aq)→ 2PbSO4(s) + 2H2O(l) (i) Pb (ii) PbO2 (i) PbO2 (ii) Pb (i) PbO2 (ii) H2SO4 (i) H2SO4 (ii) PbO2 (i) Pb (ii) H2SO4 Title: Re: Reducing and oxidizing agent Post by: Yeah.Yeah. on Apr 1, 2022 Firstly, list out the oxidation number for all elements:
Pb (Pb =0) PbO2 (O = -2, Pb = +4) H2SO4 (H = +1, O = -2, S = +6) PbSO4 (Pb = +2, O = -2, S = +6) H2O (O = -2, H = +1) From Pb ----> PbSO4, the oxidation number changes from 0 to +2, thus Ob is oxidised. From PbO2 ---> PbSO4, the oxidation number changes from +4 to +2, thus Pb is reduced. PbO2 results in the oxidation of Pb, thus PbO2 is considered as an oxidizing agent. Pb results in the reduction of PbO2, thus, Pb is considered as an reducing agent. In summary: The reactant oxidized is Pb, and the oxidizing agent is PbO2 The reactant reduced is PbO2, and the reducing agent is Pb Title: Re: Reducing and oxidizing agent Post by: Annonn on Apr 1, 2022 Alright, so basically the first blank would be Pb and the second blank is PbO2?
(i) Pb (ii) PbO2 Title: Re: Reducing and oxidizing agent Post by: Yeah.Yeah. on Apr 1, 2022 👍
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