Respuesta :
Answer:
Nitrate ions reacts with iodine and hydrogen ions in an acidic solution to produce nitrogen dioxide, iodate ions, and water. Balanced equation:
[tex]\rm 10\, {NO_3}^{-} + I_2 +8\, H^{+} \to 10\, NO_2 + 2\, {IO_3}^{-}[/tex].
Explanation:
The question states that reaction is taking place in an acidic solution. In other words, there are plenty of [tex]\rm H^{+}[/tex] ions around in the solution. When there appears to be more oxygen atoms on one side than the other, simply add
Find the oxidation states on the atoms
(To keep the answer under the character limit, this part is omitted. Please ask a new question if you need help finding the oxidation state in these chemicals.)
- [tex]\rm O[/tex]: [tex]-2[/tex] in all occurences.
- [tex]\rm N[/tex]: [tex]+5[/tex] in [tex]\rm {NO_3}^{-}[/tex] and [tex]+4[/tex] in [tex]\rm NO_2[/tex].
- [tex]\rm I[/tex]: [tex]0[/tex] in [tex]\rm I_2[/tex] and [tex]+5[/tex] in [tex]\rm {IO_3}^{-}[/tex].
Find the two half equations
Both [tex]\rm N[/tex] and [tex]\rm I[/tex] have changes in their oxidation states. Write a half equation for each one of the element.
The oxidation state of the iodine atoms here has become larger. As a result, iodine is oxidized; that would free up electrons. Since the change in oxidation state of each [tex]\rm I[/tex] atom is equal to [tex]5[/tex], each
Reactant side: [tex]\rm I_2[/tex]
- Reactant side: [tex]\rm 1\, I_2[/tex] molecule (two iodine atoms per molecule, [tex]2 \times 5 = 10[/tex] electrons transferred.)
- Product side: [tex]\rm 2\, {IO_3}^{-}[/tex] ions (since there are two [tex]\rm I[/tex] atoms in each [tex]\rm I_2[/tex] molecule) and [tex]\rm 10\, e^{-}[/tex].
At this point, the product side contains [tex]2 \times 3 = 6[/tex] more oxygen atoms than the reactant side. Since this reaction is taking place in an acidic solution, add [tex]2 \times 6 = 12[/tex] [tex]\rm H^{+}[/tex] ions to the product side (where there's one extra oxygen atom) and add [tex]1 \times 6 = 3[/tex] water molecules to the reaction side. Hence the oxidation half-reaction equation:
[tex]\rm I_2 + 6\, H_2O \to 2\, {IO_3}^{-} + 6\, H^{+} + 10\, e^{-}[/tex].
Similarly, nitrogen is reduced; one electrons would appear on the reactant side of its half-reaction equation.
- Reactant side: [tex]\rm 1\, {NO_3}^{-}[/tex] ion and [tex]1\, \rm e^{-}[/tex].
- Product side: [tex]\rm 1\, NO_2[/tex].
At this point, the reactant side contains [tex]\rm 1[/tex] more oxygen atom than the product side. Add [tex]\rm H^{+}[/tex] ions to the reactant side and a water molecule to the product side. Hence the reduction half-reaction equation:
[tex]\rm {NO_3}^{-} + 2\, H^{+} + 1\, e^{-}\to NO_2 + 1\, H_2O[/tex].
There shouldn't be any electrons in the net ionic equation.
In this case, there's one electron on the left-hand side (reactant side) of the reduction half reaction. However, there are ten electrons on the right-hand side (product side.) Hence, multiply all coefficients in the reduction half reaction by [tex]10[/tex] to obtain:
[tex]\rm 10\, {NO_3}^{-} + 20\, H^{+} + 10\, e^{-} \to 10\, NO_2 + 10\, H_2O[/tex].
Add the two half equations:
[tex]\begin{aligned} &\rm I_2 + 6\, H_2O + 10\, {NO_3}^{-} + 20\, H^{+} + 10\, e^{-} \\ &\rm \to 10\, NO_2 + 10\, H_2O + 2\, {IO_3}^{-} + 12\, H^{+} + 10\, e^{-} \end{aligned}[/tex].
Eliminate any duplicate:
[tex]\begin{aligned} &\rm I_2 + 10\, {NO_3}^{-} + 8\, H^{+} \to 10\, NO_2 + 4\, H_2O + 2\, {IO_3}^{-} \end{aligned}[/tex].
Make sure that there's no electron in the equation. Check if the number of atoms and the sum of the charges balance.
