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Volumetric titrations involving KMnO(4),...

Volumetric titrations involving `KMnO_(4)`, are carried out only in presence of dilute `H_(2)SO_(4)` but not in the presence of HCI or `HNO_(3)` This is because oxygen produced from `KMnO_(4)` + dil: `H_(2)SO_(4)`, is used only for oxidizing the reducing agent. Moreover, H2SO4 does not give any oxygen of its own to oxidize the reducing agent. In case HCI is used, the oxygen produced from `KMnO_(4) + HCI` is partly used up to oxidize HCI and in case `HNO_(3)` is used, it itself acts as oxidizing agent and partly oxidizes the reducing agent. `KMnO_(4)`, in various mediums gives following products`Mn^(7+)overset(H^(+))toMn^(2+)`
`Mn^(7+)overset(H_(2)O)toMn^(4+)`
`Mn^(7+)underset(OH^(-))toMn^(6+)`
QNumber of electrons transferred in each case when `KMnO_(4)` acts as an oxidizing agent to give `MnO_(2)`, `Mn^(++)` `Mn(OH)_(3)`, and `MnO_(4)^(2-)`are respectively

A

3, 5, 4, 1

B

4, 3, 1,5

C

1, 3, 4, 5

D

5, 4, 3, 1

Text Solution

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The correct Answer is:
To solve the question regarding the number of electrons transferred when `KMnO4` acts as an oxidizing agent to give `MnO2`, `Mn^(2+)`, `Mn(OH)3`, and `MnO4^(2-)`, we will follow these steps: ### Step 1: Determine the oxidation states of manganese in the compounds involved. - In `KMnO4`, manganese (Mn) has an oxidation state of +7. - In `MnO2`, Mn has an oxidation state of +4. - In `Mn^(2+)`, Mn has an oxidation state of +2. - In `Mn(OH)3`, Mn has an oxidation state of +3. - In `MnO4^(2-)`, Mn has an oxidation state of +6. ### Step 2: Calculate the change in oxidation state for each reaction. 1. **From `KMnO4` to `MnO2`:** - Change in oxidation state: +7 to +4 - Number of electrons transferred = 7 - 4 = 3 electrons. 2. **From `KMnO4` to `Mn^(2+)`:** - Change in oxidation state: +7 to +2 - Number of electrons transferred = 7 - 2 = 5 electrons. 3. **From `KMnO4` to `Mn(OH)3`:** - Change in oxidation state: +7 to +3 - Number of electrons transferred = 7 - 3 = 4 electrons. 4. **From `KMnO4` to `MnO4^(2-)`:** - Change in oxidation state: +7 to +6 - Number of electrons transferred = 7 - 6 = 1 electron. ### Step 3: Summarize the results. - The number of electrons transferred for each case is: - From `KMnO4` to `MnO2`: **3 electrons** - From `KMnO4` to `Mn^(2+)`: **5 electrons** - From `KMnO4` to `Mn(OH)3`: **4 electrons** - From `KMnO4` to `MnO4^(2-)`: **1 electron** ### Final Answer: - The number of electrons transferred in each case is: - `MnO2`: 3 electrons - `Mn^(2+)`: 5 electrons - `Mn(OH)3`: 4 electrons - `MnO4^(2-)`: 1 electron ---
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