✔ 最佳答案
The two chemicals used would be potassium permanganate solution (KMnO4, because it is the only oxidizing agent among the 4 chemicals) and potassium iodide solution (KI, because it is the stronger reducing agent among the 4 chemicals).
The two solutions are put separately into two beakers. Dip a graphite rod to each beaker, and the two graphite rods are connected, by using crocodile clips and conducting wire, to the two ends of a voltmeter. Use a salt bridge to connect the two beakers.
On the side of potassium permanganate solution, potassium permanganate ions gain electrons from the extenal circuit, and are reduced to manganese(II) ions.
MnO4-(aq) +8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
The purple KMnO4 solution changes to colourless.
On the side of potassium iodide solution, iodide ions and are oxidized to iodine solution, and give electrons to the external circuit.
2I-(aq) → I2(aq) + 2e-
The brown iodine solution changes to colourless.
Overall equation :
2MnO4-(aq) +16H+(aq) + 10I-(aq) → Mn2+(aq) + 4H2O(l) + 5I2(aq)