Answer
$H^{+}$(aq) , $Cu^{2+}$(aq), $Pb^{2+}$(aq) can oxidize $O_{2}$ from $H_{2}O$.
Work Step by Step
We have a standard reduction potential of reactions as given below.
$ O_{2}(g) → H_{2}O$ $ E^{\circ} = +1.23V $
$ H^{+} (aq) → H_{2}O$ $ E^{\circ} = 0.00V $
$ ClO_{3}^{-}(aq) → Cl^{-}(aq) $ $ E^{\circ} = +1.45V$
$ Cl_{2(g)} → Cl^{-}(aq)$ $ E^{\circ} = +1.36V $
$ Cu^{2+}(aq) → Cu_{(s)} $ $ E^{\circ} = +0.15V $
$ Pb^{2+}(aq) → Pb_{(s)} $ $ E^{\circ} = -0.13V $
For the oxidation of $ H_{2}O → O_{2}(g)$, the oxidizing agent should have a standard reduction potential value less than +1.23 V.
Hence $H^{+}$(aq) , $Cu^{2+}$(aq), $Pb^{2+}$(aq) which have lower standard reduction potential than +1.23 V oxidize $O_{2}$ from $H_{2}O$.
