Balance the folloing redox reaction in basic solution.
I-(aq)+ ClO-(aq) → I2 (aq) + Cl-(aq)
Balance the folloing redox reaction in basic solution.
I
-
(aq) + ClO-
(aq) → I2 (aq) + Cl-
(aq)
ClO^- + 2e + H2O==> Cl^- + 2OH^-
2I^- ==> I2 + 2e
------------------------------------------
2I^- + ClO^- + H2O ==> I2 + Cl^- + 2OH^-
You can add the phases.
To balance a redox reaction in basic solution, follow these steps:
Step 1: Separate the reaction into half-reactions: one for the oxidation process and one for the reduction process. In this case, the oxidation half-reaction is the oxidation of iodide ion (I-) to iodine (I2), and the reduction half-reaction is the reduction of chlorate ion (ClO-) to chloride ion (Cl-).
Oxidation half-reaction: I-(aq) → I2(aq)
Reduction half-reaction: ClO-(aq) → Cl-(aq)
Step 2: Balance the atoms involved in the half-reactions, other than hydrogen (H) and oxygen (O).
In the oxidation half-reaction, there is only one iodine (I) atom on each side, so it is already balanced.
In the reduction half-reaction, there is one chlorine (Cl) atom on each side, so it is also balanced.
Step 3: Balance the oxygen (O) atoms by adding water (H2O) molecules to the side that needs it.
In the reduction half-reaction, there is one oxygen atom on the reactant side but no oxygen atoms on the product side. To balance it, add one water molecule to the product side:
ClO-(aq) + H2O(l) → Cl-(aq)
Step 4: Balance the hydrogen (H) atoms by adding hydrogen ions (H+) to the side that needs it.
In the oxidation half-reaction, there are no hydrogen atoms involved.
In the reduction half-reaction, there are two hydrogen atoms on the reactant side but none on the product side. To balance it, add two hydrogen ions to the product side:
ClO-(aq) + H2O(l) → Cl-(aq) + 2H+(aq)
Step 5: Balance the charge by adding electrons (e-) to the side that needs it.
In the oxidation half-reaction, iodide ion (I-) loses one electron. Therefore, we need to add one electron to the product side:
I-(aq) → I2(aq) + e-
In the reduction half-reaction, there is no charge imbalance.
Step 6: Balance the electrons in the half-reactions. To do this, multiply each half-reaction by a factor that makes the number of electrons equal in both half-reactions. In this case, the number of electrons is already equal, so no further multiplication is needed.
Step 7: Combine the half-reactions. Make sure the number of electrons is the same in both half-reactions. In this case, both half-reactions have one electron, so they can be directly combined.
3I-(aq) + 5ClO-(aq) + 6H2O(l) → 3I2(aq) + 5Cl-(aq) + 6OH-(aq)
Step 8: Balance the charges by adding hydroxide ions (OH-) to the side that needs it. Since the reaction is in a basic solution, you can use hydroxide ions. Add the same number of hydroxide ions as the number of hydrogen ions (H+) that are present, but on the opposite side of the equation.
3I-(aq) + 5ClO-(aq) + 6H2O(l) → 3I2(aq) + 5Cl-(aq) + 6OH-(aq)
Step 9: Check the balance of atoms and charges to ensure the equation is balanced.
There are 3 iodine (I) atoms, 5 chlorine (Cl) atoms, 12 hydrogen (H) atoms, 6 oxygen (O) atoms, and the charges are balanced on both sides of the equation.
Therefore, the balanced equation for the given redox reaction in basic solution is:
3I-(aq) + 5ClO-(aq) + 6H2O(l) → 3I2(aq) + 5Cl-(aq) + 6OH-(aq)