Answer:
IO₃⁻ (aq) + 2H⁺ (aq) + e⁻ → IO₂ (s) + H₂O (l)
Explanation:
Iodate ion is IO₃⁻ (aq) (it's in aqueous state, since it's an ion)
Iodine dioxide is IO₂ (s) (it is given that it was in its solid state)
So, iodate ion becomes iodine dioxide in acidic medium
IO₃⁻ (aq) → IO₂ (s)
Introducing the acidic medium (H⁺)
IO₃⁻ (aq) + H⁺ (aq) → IO₂ (s) + H₂O (l)
To balance any half reaction, especially a redox reaction, it usually involves doing the stoichiometric balance of the reaction and the balance of charges for the reaction.
Balancing the reaction stoichiometrically,
IO₃⁻ (aq) + 2H⁺ (aq) → IO₂ (s) + H₂O (l)
We then balance the charges now,
(-1) + (2×(+1)) → 0 + 0
-1 + 2 → 0
+1 → 0
So, a charge of +1 becomes an overall charge of 0, hence, the addition of an electron to the the left hand side, balances the charges.
+1 + e⁻ → 0
+1 - 1 → 0
0 → 0, balanced!!
IO₃⁻ (aq) + 2H⁺ (aq) + e⁻ → IO₂ (s) + H₂O (l)
Hence, this is the balanced half reaction.
Hope this Helps!!!
