Why is dilute sulphuric acid so nonoxidizing, and allows hydrogen to be reduced?
Standard reduction potentials seem to be:
HSO
4-+2e
-+3H
+ H
2SO
3+H
2O +0,16V
H
2SO
3+4e
-+4H
+ S
+3H
2O +0,50V
These 2 both being species applicable in acid conditions, pH 0 meaning H
+ concentration 1 M, meaning about 10 % H
2SO
4, it follows that SO
2 and H
2SO
3 should dismute in water. The total redox reaction of sulphuric acid ought to be:
HSO
4-+6e
-+7H
+ S
+4H
2O +0,39V
Strong enough oxidant to oxidize copper, and not allow hydrogen to be reduced.
Next step would be
S+2e
-+2H
+ H
2 +0,14V
Also, in presence of heavy metals, which form insoluble/stable sulphides, reduction of sulphate to sulphide should be favoured.
If cathode charge is applied to PbSO
4, why should the reaction be
PbSO
4+2e
-+H
+ Pb+HSO
4-rather than
PbSO
4+8e
-+8H
+ PbS+4H
2O
?