Write the net ionic equation for the cell reaction:

Question

A lead–acid automotive battery consists of voltaic cells in series. The cathode of each cell consists of lead(IV) oxide (PbO₂) packed in a metal grid. The anode of each cell is composed of lead. Both electrodes are immersed in sulfuric acid.

The half-reactions for this battery are:

PbO2​(s)+HSO−4​(aq)+3H+(aq)+2e−PbSO4​(s)+2H2​O(l)  E°red​ = 1.685 V

PbSO4​(s)+H+(aq)+2e−Pb(s)+HSO−4​(aq)  E°red​ = −0.356 V



Answer 1

PbO2(s) + 4H+ + 4e- right arrow Pb(s) + 2 H2O(liq)

the two oxygens from PbO2 lead to two water molecules, and the 4 protons and 4 electrons lead to the 4 hydrogens found in your 2 water molecules. one lead atom on the left leads to one Pb(s) on the right. everything else from the half reactions cancels out.

Answer 2

Anode: Pb(s) + SO₄^2-(aq) ==> PbSO₄(s)

Cathode: PbO₂(s) + SO₄^2-(aq) + 4H⁺(aq) ==> PbSO₄(s) + 2H₂O(l)

Overall: Pb(s) + PbO₂(s) + 2SO₄^2-(aq) ==> 2PbSO₄(s) + 2H₂O(l) net ionic equation