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Me17 Electrolysis of sulfuric acid with lead electrodes by a photo voltaic current

(Only for experienced teachers of chemistry!
Sulfuric acid is very corrosive. Work on a tray.
Sulfuric acid will be electrolysed powering two lead electrodes with a photo voltaic cell.
The result will be tested by connecting the electrodes by a LED.
1. Connect three wide 5-ml vials with sticky tape, place them on a tray.
2. Fill one of these vials with sulfuric acid 20%.
3. Prepare two sandpapered sheets of lead (1 x 6 cm), put them into the acid, place a plastic strip between them for insulation.
4. Photo 2: Connect the electrodes with a solar cell (3V. 80 mA). 5. Illuminate it with as much light as possible.
6. Photos 1 and 3: Observe changes of the electrodes. 7. Fix the short feeder of a LED with the negative electrode, the nlong one with the positive.
Observations:      1. Gas bubbles at both electrodes.
                            2. A brown cover on the positive electrode (photo 3).
                            3. Shining of the LED (right photo).
Explanations:
1. Hydrogen gas forms at the negative lead electrode: 4 e^- + 4 H+(aq) --reduction--> 2 H₂(g),
2. Lead of the anode reacts leaving lead dioxide 2 H₂O+Pb--oxidation--> PbO₂+4 H⁺(aq)+4e^-
3. These reactions resulted in the formation of a Galvanic cell powering a LED.
    During fueling the LED another indirect redox reaction takes place:
    Positive electrode:          2 e^- + Pb⁴⁺ --reduction-->Pb²⁺
     Negative electrode:                  Pb ---oxidation--> 2e^- + Pb²⁺.
      total reaction:       PbO₂+Pb+2H₂SO₄--redox reacation-->2PbSO₄+ 2H₂O+energy