balancing IO3^- (aq) + I^- (aq) > I2 (aq) , using the half reactions method
I will omit the aq and s. You add those to the finished equation.
IO3^- (aq) + I^- (aq) > I2 (aq)
2IO3^- + 10e + 12H^+ ==> I2 + 6H2O
2I^- ==> I2 + 2e
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Multiply equation 1 by 1 and equation 2 by 5. Add them. That give an equation that is TWICE what you wanted so divide all of the coefficients by 2 and you'll have it.
Ah, the magic trick of balancing chemical equations! Here's how we can balance the equation using the half-reaction method:
Step 1: Identify the oxidation and reduction half-reactions.
IO3^- (aq) is reduced to I2 (aq), so that's the reduction half-reaction.
I^- (aq) is oxidized to I2 (aq), so that's the oxidation half-reaction.
Step 2: Balance the atoms that are not oxygen or hydrogen.
Let's start with the reduction half-reaction:
IO3^- (aq) → I2 (aq)
There is only one iodine (I) atom on each side, so we're good there.
Step 3: Balance oxygen atoms by adding water (H2O) molecules.
On the reactant side, we have 3 oxygen atoms from IO3^- and 2 oxygen atoms from I2. So, we need to add 1 H2O on the product side to balance the oxygen atoms:
IO3^- (aq) → I2 (aq) + H2O (l)
Step 4: Balance hydrogen atoms by adding hydrogen ions (H+).
On the reactant side, we don't have any hydrogen atoms. So, we need to add 6 H+ ions on the reactant side to balance the hydrogen atoms:
6H+ + IO3^- (aq) → I2 (aq) + H2O (l)
Step 5: Balance the charge by adding electrons.
On the reactant side, the overall charge of IO3^- is -1, and on the product side, the overall charge of I2 is 0. So, we need to balance the charges by adding 6 electrons (e^-) on the reactant side:
6H+ + IO3^- (aq) + 6e^- → I2 (aq) + H2O (l)
Step 6: Balance the oxidation half-reaction.
Since we have 6 electrons on the reactant side, we need to multiply the oxidation half-reaction by 6 to balance the electrons:
6I^- (aq) → 6e^- + 3I2 (aq)
And there you have it! The balanced equation is:
6H+ + IO3^- (aq) + 6I^- (aq) → 3I2 (aq) + H2O (l)
To balance the equation IO3^- (aq) + I^- (aq) > I2 (aq) using the half-reaction method, follow these step-by-step instructions:
Step 1: Identify the oxidation and reduction half-reactions.
In this case, IO3^- (aq) is reduced to I2 (aq) and I^- (aq) is oxidized to I2 (aq).
Step 2: Balance the atoms other than O and H in each half-reaction.
The oxidation half-reaction has I^- on both sides, so it is already balanced.
The reduction half-reaction has I2 on the right side, so it is also balanced.
Step 3: Balance the oxygen atoms by adding H2O to the appropriate side of each half-reaction.
In the reduction half-reaction, there are 6 oxygen atoms on the left side, so add 6 H2O molecules to the right side.
IO3^- (aq) + 6H2O (l) -> I2 (aq)
Step 4: Balance the hydrogen atoms by adding H+ ions to the appropriate side of each half-reaction.
In the oxidation half-reaction, there are 8 hydrogen atoms on the right side, so add 8 H+ ions to the left side.
8H+ (aq) + I^- (aq) ->
Step 5: Balance the charges by adding electrons (e^-) to the appropriate side of each half-reaction.
In the oxidation half-reaction, there is a charge of -1 on the right side, so add one electron to the left side.
8H+ (aq) + I^- (aq) + e^- ->
Step 6: Multiply each half-reaction by a factor, if necessary, to make the number of electrons in each half-reaction the same.
In this case, the number of electrons is already the same (1 electron in each half-reaction).
Step 7: Combine the two half-reactions and cancel out any common terms.
Multiply the oxidation half-reaction by 6 and the reduction half-reaction by 1, to balance the number of electrons.
6(8H+ (aq) + I^- (aq) + e^-) + IO3^- (aq) + 6H2O (l) -> 6I2 (aq)
Step 8: Balance the final equation by adding coefficients as needed.
The final balanced equation is:
6IO3^- (aq) + 18H+ (aq) + 5I^- (aq) -> 6I2 (aq) + 9H2O (l)
And that's how you balance the equation IO3^- (aq) + I^- (aq) > I2 (aq) using the half-reaction method.
To balance the given chemical equation using the half-reaction method, follow these steps:
Step 1: Split the reaction into two half-reactions, one for the oxidation half-reaction and one for the reduction half-reaction.
IO3^- (aq) → I2 (aq) (Reduction half-reaction)
I^- (aq) → I2 (aq) (Oxidation half-reaction)
Step 2: Balance the atoms other than oxygen and hydrogen in each half-reaction. First, balance the atoms other than oxygen and hydrogen in the carbons, then the iodines.
Reduction half-reaction: IO3^- (aq) → I2 (aq)
Oxidation half-reaction: 2I^- (aq) → I2 (aq)
Step 3: Balance the oxygen atoms by adding water (H2O) molecules to the side of the half-reaction that lacks oxygen.
Reduction half-reaction: IO3^- (aq) + 3H2O (l) → I2 (aq)
Oxidation half-reaction: 2I^- (aq) → I2 (aq) + 2e^-
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side of the half-reaction that lacks hydrogen.
Reduction half-reaction: IO3^- (aq) + 6H+ (aq) + 3H2O (l) → I2 (aq)
Oxidation half-reaction: 2I^- (aq) → I2 (aq) + 2e^-
Step 5: Balance the charges by adding electrons (e^-) to the side of the half-reaction that needs them. The number of electrons added should be equal to the difference in charge between the reactant and the product.
Reduction half-reaction: IO3^- (aq) + 6H+ (aq) + 6e^- + 3H2O (l) → I2 (aq)
Oxidation half-reaction: 2I^- (aq) → I2 (aq) + 2e^-
Step 6: Multiply each of the half-reactions by appropriate coefficients to equalize the number of electrons transferred in the redox reaction. To make the number of electrons equal on both sides, multiply the oxidation half-reaction by a factor of 3.
Reduction half-reaction: 3IO3^- (aq) + 18H+ (aq) + 18e^- + 9H2O (l) → 3I2 (aq)
Oxidation half-reaction: 6I^- (aq) → 3I2 (aq) + 6e^-
Step 7: Add the two half-reactions to cancel out the electrons and get the overall balanced chemical equation.
3IO3^- (aq) + 18H+ (aq) + 6I^- (aq) + 9H2O (l) → 3I2 (aq) + 6I^- (aq) + 18H+ (aq)
Simplifying the equation:
IO3^- (aq) + 6H+ (aq) → 3I2 (aq) + 9H2O (l)
Finally, the balanced equation is:
IO3^- (aq) + 6H+ (aq) → 3I2 (aq) + 9H2O (l)
Note: The coefficients in front of the species in the balanced equation represent the relative number of moles of each.