Writing the Full Equation of Acidified KMnO4 & KI

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The discussion centers on the full equation for the reaction between acidified potassium permanganate (KMnO4) and potassium iodide (KI). The ionic equation provided is 2 MnO4- + 16 H+ + 10I- → 2Mn2+ + 8H2O + 5I2-, but there is a noted typo with a misplaced negative sign. Participants emphasize the importance of writing a complete balanced equation, suggesting that counterions should be included to maintain electrical neutrality. The consensus is that while the ionic equation is balanced, it requires correction for clarity. The conversation highlights the challenge of writing full equations and the necessity of balancing them correctly.
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I want to ask what is the full equation of acidified potassium permanganate and potassium iodide?

I have worked out the ionic equation...

2 MnO4- + 16 H+ + 10I- --> 2Mn2+ + 8H2O + 5I2-

How to write the full equation??

THanks
 
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There is a typo - "-" at teh end of teh reaction equation (but otherwise it is OK).

new hand said:
How to write the full equation??

I don't think it is always possible (and actually rarely matters). In general - add counterions on the left side to make the solution electrically neutral (so start with KMnO4, HCl and KI), copy counterions on the right, try to pair them with anions on the right.
 
Your equation isn't balanced.
Can you work out its half-reactions?
 
war485 said:
Your equation isn't balanced.

It is balanced, it just contains a typo.

If it is not balanced - please tell us what is wrong.
 
ah yes sorry, it is balanced when the - on I2 is removed. I have nothing more to add.
 

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