Plz EXPLAIN
Q4. In the reaction : PbS(s) + 4H2O(aq) PbSO4 (s) + 4H2O(). Which is oxidizing agent and which is reducing agent?
Dear student
The reaction should be :
PbS(S) + 4H2O2(aq) → PbSO4(s) + 4H2O(l)
Reactants:
Pb in PbS is +2
S in PbS is -2
H in H2O2 is +2
O in H2O2 is -1
Products:
Pb in PbSO4 is +2
S in SO42- is +6
H in H2O is +1
O in H2O is -2
So the oxidation state of Pb and H have remained constant.
The S went from -2 to +6, was oxidised and is the reducing agent
The O went from -1 to -2, was reduced and is the oxidising agent.
Regards
The reaction should be :
PbS(S) + 4H2O2(aq) → PbSO4(s) + 4H2O(l)
Reactants:
Pb in PbS is +2
S in PbS is -2
H in H2O2 is +2
O in H2O2 is -1
Products:
Pb in PbSO4 is +2
S in SO42- is +6
H in H2O is +1
O in H2O is -2
So the oxidation state of Pb and H have remained constant.
The S went from -2 to +6, was oxidised and is the reducing agent
The O went from -1 to -2, was reduced and is the oxidising agent.
Regards
