Write a balanced half-reaction for the reduction of iodate ion to solid iodine dioxide in acidic aqueous solution. Be sure to add physical state symbols where appropriat

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!!!

the answer is c i believe

did anyone actually do this? like if anyone has submitted this i rly need .