I need help splitting this reaction into 1/2 reactions
3I2(s) + 2MnO2(s)+8 OH-(aq)<->
6I-(aq) + 2MnO4-(aq) + 4H2O(l)
Can you identify which components are oxidised, which are reduced and those that undergo no change?
Can you identify which components are oxidised, which are reduced and those that undergo no change?
I looked up the Ecell values in the appendix in the back of my text book and the MnO2(s)-> MnO4- reaction has a less positive value making it the one that got reversed and oxidized. So the I2 -> I- is the one that gets reduced
Thus having identified the substances being oxidised and reduced can you now write the half reactions
I2 + xe- -> 2I-
MnO2(s) -> MnO4- + xe-
for the Mn one it is easier if you include H+ and balance it as though it is in H+
MnO2(s) + H2O -> MnO4- + xe- + H+
then add OH- to each side to remove the H+ and form water.
does this help?
Yes it does. Thank you. The H2O and OH-'s were throwing me off, but I got it now
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