Ionic equations help needed Watch
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-
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