Balance the following chemical equation using the half reaction method in acidic conditions:
I2 + ClO3- → IO3- + Cl-
Well, let's take a closer look at this chemical equation. We have I2 reacting with ClO3- to form IO3- and Cl-. In order to balance this equation using the half reaction method, we need to balance the atoms on both sides of the equation. So, let's get started!
First, let's balance the atoms that aren't already balanced. We have 2 iodine (I) atoms on the left side and only 1 on the right side. So, let's start with that:
I2 → 2 IO3-
Great! Now we have the iodine atoms balanced. Now let's balance the oxygens. We currently have 6 oxygens (3 from ClO3- and 3 from IO3-) on the right side and only 3 on the left side. To balance this, let's add some water (H2O) molecules:
I2 + 6 H2O → 2 IO3- + 12 H+
Perfect! Now we have the oxygen atoms balanced. However, we have an imbalance in terms of hydrogen atoms. On the left side, we have 12 hydrogen (H) atoms, while on the right side, there are only 6 from the water molecules (H2O). To balance this, let's add some hydrogen ions (H+) to the right side:
I2 + 6 H2O → 2 IO3- + 12 H+ + 12 Cl-
There we go! We have successfully balanced the equation. Now, you may be wondering, why did the clowns make such a circus out of this reaction? Well, it's because they wanted to show off their balancing skills! They can juggle atoms and equations with the best of them. So, next time you have a chemical equation that needs balancing, remember to bring in the clowns for a hilarious and entertaining performance!
To balance the given chemical equation using the half-reaction method in acidic conditions, follow these steps:
Step 1: Write the unbalanced equation.
I2 + ClO3- → IO3- + Cl-
Step 2: Separate the equation into two half-reactions: reduction and oxidation.
Half-Reaction 1 (reduction): I2 → IO3-
Half-Reaction 2 (oxidation): ClO3- → Cl-
Step 3: Balance the atoms other than oxygen and hydrogen in each half-reaction.
Half-Reaction 1 (reduction): I2 → 2 IO3-
Half-Reaction 2 (oxidation): ClO3- → 2 Cl-
Step 4: Balance the oxygen atoms by adding water (H2O) molecules to the side that needs oxygen.
Half-Reaction 1 (reduction): I2 → 2 IO3- + 12 H2O
Half-Reaction 2 (oxidation): ClO3- + 6 H2O → 2 Cl-
Step 5: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs hydrogen.
Half-Reaction 1 (reduction): I2 + 12 H2O → 2 IO3- + 12 H+
Half-Reaction 2 (oxidation): ClO3- + 6 H2O → 2 Cl- + 6 H+
Step 6: Balance the charges by adding electrons (e-) to the side that needs them.
Half-Reaction 1 (reduction): I2 + 12 H2O → 2 IO3- + 12 H+ + 12 e-
Half-Reaction 2 (oxidation): ClO3- + 6 H2O + 6 e- → 2 Cl- + 6 H+
Step 7: Equalize the number of electrons in both half-reactions by multiplying one or both half-reactions by a suitable value.
To equalize the number of electrons, multiply the reduction half-reaction by 6 and the oxidation half-reaction by 2.
6(I2 + 12 H2O → 2 IO3- + 12 H+ + 12 e-)
2(ClO3- + 6 H2O + 6 e- → 2 Cl- + 6 H+)
Step 8: Add the two half-reactions together, canceling out common species on both sides of the equation.
6 I2 + 12 H2O + 2 ClO3- + 6 H2O + 6 e- → 2 IO3- + 12 H+ + 12 e- + 2 Cl- + 6 H+
6 I2 + 18 H2O + 2 ClO3- → 2 IO3- + 18 H+ + 2 Cl-
Step 9: Simplify the equation by canceling out the common species.
6 I2 + 2 ClO3- → 2 IO3- + 2 Cl-
The balanced equation in acidic conditions is:
6 I2 + 2 ClO3- → 2 IO3- + 2 Cl-
To balance a chemical equation using the half reaction method in acidic conditions, follow these steps:
Step 1: Divide the equation into two half-reactions. In this case, we have the reduction half-reaction and the oxidation half-reaction.
Reduction Half-Reaction: I2 + 10e- → 2I-
Oxidation Half-Reaction: ClO3- → Cl- + 3H2O + 6e-
Step 2: Balance the atoms in each half-reaction, excluding oxygen and hydrogen.
Balancing the reduction half-reaction: I2 + 10e- → 2I-
Balancing the oxidation half-reaction: ClO3- → Cl- + 3H2O + 6e-
Step 3: Balance the oxygen atoms by adding H2O to the side that needs more oxygen.
Balancing the reduction half-reaction: I2 + 10e- → 2I-
Balancing the oxidation half-reaction: ClO3- + 6H+ → Cl- + 3H2O + 6e-
Step 4: Balance the hydrogen atoms by adding H+ ions to the side that needs more hydrogen.
Balancing the reduction half-reaction: I2 + 10e- → 2I-
Balancing the oxidation half-reaction: ClO3- + 6H+ → Cl- + 3H2O + 6e-
Step 5: Balance the charges by adding electrons (e-) to the side that needs more charge.
Balancing the reduction half-reaction: I2 + 10e- → 2I-
Balancing the oxidation half-reaction: ClO3- + 6H+ + 6e- → Cl- + 3H2O + 6e-
Step 6: Make the total number of electrons equal in both half-reactions by multiplying one or both of the half-reactions by constant factors.
Balancing the reduction half-reaction: I2 + 10e- → 2I-
Balancing the oxidation half-reaction: 2ClO3- + 12H+ + 12e- → 2Cl- + 6H2O + 12e-
Step 7: Combine the half-reactions, canceling out electrons, and write the final balanced equation.
Multiply the reduction half-reaction by 12 and the oxidation half-reaction by 1/6 to make the number of electrons equal:
12I2 + 120e- → 24I-
2ClO3- + 12H+ + 12e- → 2Cl- + 6H2O + 12e-
Combine the two half-reactions:
12I2 + 2ClO3- + 12H+ → 24I- + 2Cl- + 6H2O
Finally, simplify the equation by canceling out common species:
6I2 + ClO3- + 6H+ → 12I- + Cl- + 3H2O
Therefore, the balanced equation is:
6I2 + ClO3- + 6H+ → 12I- + Cl- + 3H2O
I2 + 6H2O ==> 2IO3^- + 10e + 12H^+
[ClO3]^- + 6e + 6H^+ ==> Cl^- + 3H2O
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Make the electrons equal by multiplying eqn 1 by 3 and eqn 2 by 5 then add the two equations together. Finally adjust the species that appear on BOTH sides of the equations. 30 electrons, for example will appear on both side and there will be H2O on both sides as well as H^+ on both sides.