Redox Reaction: Adding Excess Acidified KMnO4 to Ammonium Iron(II) Sulfate

AI Thread Summary
Excess acidified potassium permanganate is added to ammonium iron(II) sulfate, prompting a discussion on the overall redox equation for the reaction. Participants emphasize the importance of recognizing potassium permanganate as a strong oxidizing agent, which undergoes reduction from manganese in the +7 oxidation state to +2. The reduction half-equation is identified as MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O. The oxidation of iron(II) to iron(III) ions is also noted as a necessary component of the overall reaction. The conversation concludes with confirmation that combining these half-equations will yield the complete redox equation.
danago
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Homework Statement


Excess acidified potassium permanganate is added dropwise to an ammonium iron(II) sulfate solution. Write an overall redox equation for this process.

2. The attempt at a solution
Well I am not really sure what to do. I don't know what products are being formed. What should i do in a situation like this?
 
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Start with a balanced equation. Were you given the rules to follow for a redox reaction?
 
What is the one thing that you neeed to know about potassium permanganate?
 
How do i write a balanaced equation though? I don't know what the products are
 
Gokul43201 said:
What is the one thing that you neeed to know about potassium permanganate?

Its soluble in water?
 
Lots of things are soluble in water.

I quote from the first paragraph of the wiki on potassium permanganate:

Potassium permanganate is the chemical compound KMnO4. In this salt, manganese is in the +7 oxidation state. The salt is also known as "permanganate of potash" and "Condy's crystals". The permanganate ion is a strong oxidizing agent. It dissolves in water to give deep purple solutions, evaporation of which gives prismatic purple-black glistening crystals. It has a sweet taste and is odourless.
 
Ok, so its an oxidising agent, and therefore, itself reduced. I think this is the correct equation for its reduction:

<br /> MnO_4 ^ - _{(aq)} + 8H^ + _{(aq)} + 5e^ - \to Mn^{2 + } _{(aq)} + 4H_2 O_{(l)} <br />

Would i just combine that with the half equation of iron (II) being oxidised to iron (III) ions?
 
Yup, you got it! :approve:
 
Ahh ok. Thanks :)
 
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