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Annonn Annonn
wrote...
Posts: 213
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A year ago
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
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Replies
wrote...
Valued Member
A year ago
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
Annonn Author
wrote...
A year ago
Alright, so basically the first blank would be Pb and the second blank is PbO2?

(i) Pb                      (ii) PbO2
wrote...
Valued Member
A year ago
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