Ferdiee
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Report Thread starter 4 years ago
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I'm trying to do this ionic equation but I can't quite understand what's happening, can somebody help me out?
The reaction is between potassium manganate (VII) with potassium iodide solution in the presence of dilute sulfuric acid.
At the moment what I think, but which may be completely or partially wrong is:
K^+ + 2MnO4^- + K^+ + I^- [arrow] K^+ + MnO2^4- + K^+ + I^- + 3O^2-
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knowsabit
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Report 4 years ago
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Firstly drop the spectator ion potassium as it's not needed (oxidation number doesn't change)

Build your answer up from two half equations

A) 2 I^- > I2 + 2e iodide oxidized to iodine
B) MnO4^- + 8H^+ + 5e > Mn^2+ + 4H2O permanganate reduced to manganese (II) under acidic conditions

You need 5 of A) to every 2 of B) to balance electrons (cancel out).

So overall 10 I^- + 2 MnO4^- + 16 H^+ > 5 I2 + 2 Mn^2+ + 8 H2O
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