in balancing I3- --- I- + IO3- wat is reduced to what
a)I3 goes from 0 to -1 to form some I-
b) I3 goes from 0 to +5 to form IO3-
so you need five (a) for each (b)
2I3>>5I- + IO3-
yes you need to work on oxygen.
I think I caught the first question early today. Here is what I do. The I3^- just complicates things, at least for me.
IO3^- + I^- ==> I + I
I goes from +5 on the left to zero on the right (only one of the I atoms used on the right).
I^- goes from -1 on the left to zero on the right (the other I atom on the right). Balance those two half equations, then add I^- to both sides to show I3^-.
To determine what is being reduced in the reaction, we need to identify the species that gains electrons. In this case, I3-, I-, and IO3- are involved in the reaction. We can assign oxidation states to each element to track the electron transfer.
The oxidation state of I in I3- is +5, while the oxidation state of I in I- is -1. The oxidation state of O in IO3- is -2.
Considering these oxidation states, we can observe that I in I3- is being reduced from +5 to -1. This means that I3- is the species being reduced in the reaction.
To balance this reaction, we need to add electrons to the side containing I3-. The number of electrons required is determined by the difference in oxidation states, which is 6 (from +5 to -1). Therefore, we add 6 electrons to the side with I3-:
I3- + 6e- → I-
Now, we can proceed to balance the remaining atoms and charges in the overall reaction using the method of trial and error or the half-reaction method.