How do I solve an oxidation with multiple reactants?
MnO4+(Cl-)+H3O->Mn+Cl2+H20
And why is elemental chlorine formed?
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Because Permanganate is a strong OM and the chloride guarantees an electron. Look into the electrochemical voltage series where what’s up.
How to solve this? In which partial equations are set up and these are solved in series
Can you do that?
Yeah, I can, but then you wouldn’t learn anything. Look at your records and work with the oxidation figures.
HCl dissociated in aqueous solution to give H3O+ and Cl-
But how do I split this 16hcl into individual ions? I mean, if I replace this 16 HCl by 16Cl and 16H3O+, the equation is not properly balanced. But thank you!
Redox Reaction Permanganate + Hydrochloric Acid Learning Confluence Podcast Chemistry High Level – YouTube
I still don’t understand. Please help me. This oxonium and water confused me..
@MeisterRuelps, UserMod Light I still don’t understand. Please help me.
What does the water do? So the oxidizing numbers change?
1e- +H3O->H20?
I’ve found some videos on Yt that fit your situation. Search again 😁
MnO4+5e->Mn2+
Cl=Cl2+1e
With oxoniun and water, nothing happens?!
On Youtube I find no videos describing MY Fall. Besides, I don’t have any proper documents.
See your documents 🙂
Thank you. How do I do these partial reactions now?
Oh, no!
Cl- has -1, Mn2+ has +2
@MasterRuelps, but how do I do it with three educts?
MnO4:Mn=7. O=-2
Cl-=0
H3O H:1,O=-2
Mn2+=0
CL2 =
H2O:H=1; O=-2?!