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!
Help with the redox reaction.
Moderators: Xen, expert, ChenBeier
- ChenBeier
- Distinguished Member

- Posts: 1754
- Joined: Wed Sep 27, 2017 7:25 am
- Location: Berlin, Germany
Re: Help with the redox reaction.
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
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
