MnO4–+ Cl– → MnO2 + Cl2
Well, it looks like manganese is trying to break up bromance between chlorate and chloride. MnO4– says "Adios, amigo" and grabs Cl–, leaving behind MnO2 and Cl2. Guess even chemical reactions can have a messy breakup!
The given chemical equation is:
MnO4– + Cl– → MnO2 + Cl2
This reaction represents the reduction of permanganate ion (MnO4–) by chloride ion (Cl–) to form manganese dioxide (MnO2) and chlorine gas (Cl2).
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.
Here is how to balance the equation step-by-step:
Step 1: Balance the atoms other than oxygen and hydrogen.
MnO4– + 5Cl– → MnO2 + Cl2
By adding 5 chloride ions (Cl–) on the left side, we ensure that the number of chlorine atoms is balanced.
Step 2: Balance oxygen atoms using water molecules (H2O).
MnO4– + 5Cl– → MnO2 + Cl2 + 4H2O
By adding 4 water molecules (H2O) on the right side, we ensure that the number of oxygen atoms is balanced.
Step 3: Balance hydrogen atoms using hydrogen ions (H+).
MnO4– + 5Cl– + 8H+ → MnO2 + Cl2 + 4H2O
By adding 8 hydrogen ions (H+) on the left side, we ensure that the number of hydrogen atoms is balanced.
Now the equation is balanced:
MnO4– + 5Cl– + 8H+ → MnO2 + Cl2 + 4H2O
The provided chemical equation represents a redox reaction between permanganate ion (MnO4-) and chloride ion (Cl-) in an acidic solution. The reactants are MnO4- and Cl-, and the products are MnO2 and Cl2.
To determine the oxidation states of the atoms involved in the reaction, we can start by assigning the oxidation state of oxygen. In most compounds, oxygen has an oxidation state of -2. Since there are four oxygen atoms in the permanganate ion (MnO4-), the total oxidation state contributed by oxygen is -8.
Now, let's assume the oxidation state of manganese (Mn) in MnO4- is x. Since the overall charge of the ion is -1, the sum of the oxidation states of all the atoms should equal the charge of the ion. Therefore, we have:
x + (-8) = -1
x - 8 = -1
x = +7
So, in the permanganate ion (MnO4-), the oxidation state of manganese is +7.
Next, let's assign the oxidation state of chlorine in Cl-. Since it is a halogen, chlorine usually has an oxidation state of -1. Thus, the oxidation state of chlorine in Cl- is -1.
In MnO2, the compound is neutral, so the sum of the oxidation states of manganese and oxygen atoms should be zero. Assuming the oxidation state of manganese in MnO2 is y, we have:
y + 2*(-2) = 0
y - 4 = 0
y = +4
Therefore, the oxidation state of manganese in MnO2 is +4.
Finally, in Cl2, it is a diatomic molecule, and each chlorine atom has the same oxidation state. So, each chlorine atom in Cl2 has an oxidation state of 0.
To balance the redox reaction, we need to equalize the number of electrons gained and lost. The permanganate ion (MnO4-) is reduced to MnO2, which means it gains electrons. By comparing the oxidation states of manganese in both species, we can deduce that MnO2 gains 3 electrons.
On the other hand, the chloride ion (Cl-) is oxidized to chlorine gas (Cl2), implying it loses electrons. By comparing the oxidation states of chlorine in both species, we can conclude that Cl- loses 1 electron.
Therefore, to balance the equation, we need to multiply the half-reactions by their respective coefficients. The balanced equation becomes:
8H+ + MnO4- + 5e- -> MnO2 + 4H2O
2Cl- -> Cl2 + 2e-
MnO4^- + 3e + 2H2O ==> MnO2 + 4OH^-
2Cl^- ==> Cl2 + 2e
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Multiply equation 1 by 2 and multiply equation 2 by 3, then add them.