Chemistry Forum

Hello everybody!
My teacher gave me an assignment for a redox reaction with the following elements:
KI+KMnO₄+KOH->I₂+K₂MnO₄+H₂O
The question arises: Am I the fool who can't equalize the coefficients of this reaction, or did the teacher make a mistake?
Thanks for earlier!

Find redox paar first

It's I-/I₂ and MnO₄-/ MnO₄ 2-

Oxidation 2 I- => I₂ + 2e-

Reduktion MnO₄- + e- => MnO₄ 2-

Second equation multiplied by 2 to get same electrons then do Addition

2 I- + 2 MnO₄- => I₂ + 2 MnO₄ 2-

2 KI + 2KMnO₄ => I₂ + 2 K₂MnO₄

No KOH and water are involved.

Additionally Permanganate can be reduced in alcaline solution.

4 MnO₄ - + 4 OH- => 4 MnO₄ 2- + O₂ + 2 H₂O

4 KMnO₄ + 4 KOH => 4 K₂MnO₄ + O₂ + 2 H₂O