Question

choose one: oxidation reduction part 1 (2 points) + i₂(s) + 6h₂o(l) → 2io₃⁻ + 12h⁺(aq) + choose one: oxidation reduction

Explanation:

Oxidation is the loss of electrons or an increase in oxidation state. Reduction is the gain of electrons or a decrease in oxidation state. In the reaction $I_2(s)+6H_2O(l)\longrightarrow 2IO_3^{-}+12H^{+}(aq)$, the oxidation state of iodine in $I_2$ is 0 and in $IO_3^{-}$ it is +5. Since the oxidation state of iodine increases, this is an oxidation process.

Answer:

Oxidation