O₂(g)+ 4 H⁺(aq)+ 4 e^- → 2 H₂O(l)E° = +1.23 V Pb²⁺(aq)+ 2 e^- → Pb(s)E° = -0.13 V 2 H₂O(l)+ 2 e^-→ H₂(g)+ 2 OH^-(aq)E° = -0.83 V Based on the half-reactions above,electrolysis of an aqueous solution of Pb(NO₃)₂ is expected to produce.

Question

Quizplus · Accepted Answer

In the given electrolytic cell, the reduction potential of Pb is less than the reduction potential of H₂O and OH^-. So, Pb cannot be reduced and hence will not be produced at the cathode. On the other hand, the oxidation potential of O₂ is less than the oxidation potential of ^- and H₂. Hence, O₂ will be oxidized and produced at the anode. Therefore, the expected products of electrolysis of the given solution are Pb at the cathode and O₂ at the anode.

O2(g)+ 4 H+(aq)+ 4 E- → 2 H2O(l)E° = +1

O₂(g)+ 4 H⁺(aq)+ 4 E^- → 2 H₂O(l)E° = +1