Question

In the reaction: I2 + ClO3− → IO3− + Cl−, indicate:a. the element that is oxidized: b. the element that is reduced: c. the reducing agent: d. the oxidizing agent: e. Balance the reaction

Answer 1

1) Which element is oxidized

`I_2+ClO^-_3\rightarrow IO^-_3+Cl^-`

Oxidation numbers

I2: 0

ClO3-: for Cl +5 and for O is -2

IO3-: for I is +5 and for O -2

Cl-: -1.

According to the oxidation number

a. Iodine (I) has been oxidized. It changed from 0 to +5.

b. Chlorine has been reduced. It changed from +5 to -1

2) reducing agent and oxidizing agent

c. A reducing agent donates electrons. In the reaction, Iodine is the reducing agent.

d. An oxidizing agent accepts electrons. In the reaction, Chlñorine is the oxidizing agent.

3) Balancing the chemical equation

`I_2+ClO^-_3\rightarrow IO^-_3+Cl^-`

Step 1: break the reaction into two half-reactions.

Oxidation half-reaction

`I_2\rightarrow IO^-_3`

Reduction half-reaction

`ClO^-_3\rightarrow Cl^-`

Step 2: balance all elements EXCEPT for hydrogen and oxygen

Oxidation half-reaction

`I_2\rightarrow2IO^-_3`

Reduction half-reaction

`ClO^-_3\rightarrow Cl^-`

Step 3: Balance OXYGEN. We do so by adding water molecules to the half-reactions as needed.

Oxidation half-reaction

`I_2+6H_2O\rightarrow2IO^-_3`

Reduction half-reaction

`ClO^-_3\rightarrow Cl^-+3H_2O`

Step 4: Balance HYDROGEN. We do so by adding protons (H+) to the half-reactions as needed.

Oxidation half-reaction

`I_2+6H_2O\rightarrow2IO^-_3+12H^+`

Reduction half-reaction

`ClO^-_3+6H^+\rightarrow Cl^-+3H_2O`

Step 5: Balance CHARGES. We do so by adding electrons

Oxidation half-reaction

`I_2+6H_2O\rightarrow2IO^-_3+12H^++10e^-`

Reduction half-reaction

`ClO^-_3+6H^++6e^-\rightarrow Cl^-+3H_2O`

Step 6: Multiply each half-reaction in such a way we can cancel the electrons.

Oxidation half-reaction

`3\cdot(I_2+6H_2O\rightarrow2IO^-_3+12H^++10e^-)`

New oxidation half-reaction

`3I_2+18H_2O\rightarrow6IO^-_3+36H^++30e^-`

Reduction half-reaction

`5\cdot(ClO^-_3+6H^++6e^-\rightarrow Cl^-+3H_2O)`

New reduction half-reaction

`5ClO^-_3+30H^++30e^-\rightarrow5Cl^-+15H_2O`

Step 7: combine the half-reactions. We have 30e in the reactants and 30e in the products. We can cancel them and combine the remaining species.

New oxidation half-reaction

`3I_2+18H_2O\rightarrow6IO^-_3+36H^++30e^-`

New reduction half-reaction

`5ClO^-_3+30H^++30e^-\rightarrow5Cl^-+15H_2O`

Overall reaction

`3I_2+18H_2O+5ClO^-_3+30H^+\rightarrow6IO^-_3+36H^++5Cl^-+15H_2O`

Step 8: Balance the reaction by reducing the number of water molecules and protons.

Overall reaction

`3I_2+3H_2O+5ClO^-_3^{}\rightarrow6IO^-_3+6H^++5Cl^-`

e. Balance the reaction

`3I_2+3H_2O+5ClO^-_3^{}\rightarrow6IO^-_3+6H^++5Cl^-`

.