1. In the reaction, potassium permanganate (KMnO4) is dissolved in water, forming permanganate ions (MnO4-) and potassium ions (K+).
2. The permanganate ion (MnO4-) is then reduced by gaining electrons and reacting with hydrogen ions (H+), forming manganese (II) ions (Mn2+) and water (H2O).
3. The balanced half-reaction for the reduction of permanganate ion is:
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
4. This reduction half-reaction has a standard electrode potential (Eo) of +1.51V, indicating that the reduction of permanganate ion to manganese (II) ion is thermodynamically favorable.
5. As a result of the reduction reaction, the purple color of the permanganate ion is lost, and the solution becomes colorless due to the formation of manganese (II) ions.
6. Overall, the reaction involves the reduction of permanganate ions to form colorless manganese (II) ions.
Therefore, the reduction of permanganate ion involves the transfer of electrons and protons to form manganese (II) ions, resulting in the color change from purple to colorless.
Potassium permanganate, KMnO4(permanganate ion, MnO4-)
• MnO4-(aq) + 8H+(aq)+ 5e → Mn2+ (aq)+ 4H2O (l) Eo = +1.51V
• purple permanganate ion (MnO4-) is reduced to colorless manganese(II) ion (Mn2+)
explain step by step regarding the above reaction before coming to the final answer
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