Fe(OH)3 + OCl– → FeO42– + Cl–
When this reaction is balanced, how many electrons are transferred in each half-reaction and what is the coefficient of water (H2O)?
I assume the solution is basic
2Fe(OH)3 + 3OCl- + 4OH- → 2[FeO4]2- + 3Cl- + 5H2O ...
Well, to balance this reaction, we'll need to split it into half-reactions. Let's take a closer look:
Half-reaction 1: Fe(OH)3 → FeO42–
Half-reaction 2: OCl– → Cl–
In half-reaction 1, the Fe atom goes from a +3 oxidation state in Fe(OH)3 to a +4 oxidation state in FeO42–. This means that one electron is transferred.
In half-reaction 2, the OCl– ion goes from a -1 oxidation state to a -1 oxidation state in Cl–. Since the oxidation state doesn't change, no electrons are transferred.
So there you have it! Only one electron is transferred in each half-reaction. As for the coefficient of water, there isn't any water involved in this reaction. Looks like water's taking a break and letting the other compounds do the work!
To balance the given reaction, let's examine the changes in oxidation state for each element involved.
In the reaction:
Fe(OH)3 + OCl– → FeO42– + Cl–
For the left side of the equation:
The oxidation states are as follows:
Fe: +3
O: -2 (in OH–)
Cl: +1
For the right side of the equation:
The oxidation states are as follows:
Fe: +4 (in FeO42–)
O: -2
Cl: -1
To balance the oxidation states, we can see that Fe gains an electron (from +3 to +4) and Cl loses one electron (from +1 to -1).
So, in the half-reaction for Fe:
Fe(OH)3 → FeO42– + e–
1 electron is transferred.
In the half-reaction for Cl:
OCl– + e– → Cl–
1 electron is transferred.
Now, let's balance the number of atoms on each side of the equation:
Fe(OH)3 → FeO42– + e–
There is 1 Fe on both sides, 3 H on the left, and 4 O on the right, so we need to add 4 H2O on the left side to balance the oxygen atoms.
The balanced half-reaction for Fe is:
Fe(OH)3 + 4 H2O → FeO42– + e– + 4 OH–
And the balanced half-reaction for Cl is:
OCl– + e– + 2 H2O → Cl– + 4 OH–
Therefore, in each half-reaction, 1 electron is transferred, and the coefficient of water (H2O) is 4.
To balance the given redox reaction, follow these steps:
Step 1: Write down the unbalanced equation.
Fe(OH)3 + OCl– → FeO42– + Cl–
Step 2: Separate the overall reaction into two half-reactions—one for the oxidation process and one for the reduction process.
Oxidation half-reaction: Fe(OH)3 → FeO42–
Reduction half-reaction: OCl– → Cl–
Step 3: Balance the atoms except for oxygen and hydrogen in each half-reaction.
Oxidation half-reaction: Fe(OH)3 → FeO42–
To balance iron: Fe(OH)3 → FeO42– + 3e–
Reduction half-reaction: OCl– → Cl–
Since there is no unbalanced atom except for chlorine, we don't need to write any coefficients for this half-reaction.
Step 4: Balance the oxygen atoms by adding H2O molecules to the reactions.
Oxidation half-reaction: Fe(OH)3 → FeO42– + 3e–
To balance oxygen: Fe(OH)3 + 3H2O → FeO42– + 3e– + 6OH–
Reduction half-reaction: OCl– → Cl–
No oxygen atoms need to be balanced in this half-reaction.
Step 5: Balance the hydrogen atoms by adding H+ ions to the reactions.
Oxidation half-reaction: Fe(OH)3 + 3H2O → FeO42– + 3e– + 6OH–
To balance hydrogen: Fe(OH)3 + 3H2O → FeO42– + 3e– + 6OH– + 3H+
Reduction half-reaction: OCl– + 2H+ → Cl–
To balance hydrogen: OCl– + 2H+ → Cl– + H2O
Step 6: Balance the charge by adding electrons.
Oxidation half-reaction: Fe(OH)3 + 3H2O → FeO42– + 3e– + 6OH– + 3H+
No additional electrons are needed as the charge is already balanced.
Reduction half-reaction: OCl– + 2H+ → Cl– + H2O
To balance charge: OCl– + 2H+ + 2e– → Cl– + H2O
Finally, the balanced half-reactions and the balanced overall reaction are:
Oxidation half-reaction: Fe(OH)3 + 3H2O → FeO42– + 3e– + 6OH– + 3H+
Reduction half-reaction: OCl– + 2H+ + 2e– → Cl– + H2O
Overall balanced redox reaction:
Fe(OH)3 + OCl– + 3H+ → FeO42– + Cl– + 2H2O
In each half-reaction, 3 electrons are transferred. The coefficient of water (H2O) in the balanced reaction is 2.