Equivalent weight of KMnO_4 in its oxidising reactions in acidic medi…

Read the passage carefully and answer the Questions.

$KMnO_4$ is prepared by the fusion of $MnO_2$ with an alkali metal hydroxide and an oxidizing agent like $KNO_3$ to give a dark- green manganate ion which disproportionate to give permanganate as follows.

$2MnO_2+ 4KOH + O_2→2KMnO_4+2H_2O$

$3KMnO_4+ 4H^+→2KMnO_4 + MnO_2 + 2H_2O$

On heating $KMnO_4$ decomposes at 513 K to give $K_2MnO_4$. Permanganate ion is tetrahedral and diamagnetic. Acidified $KMnO_4$ acts a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.

Equivalent weight of $KMnO_4$ in its oxidising reactions in acidic medium is taken as

Answer & explanation

Correct answer: option 2

The correct answer is Option (2) → One-fifth of its molecular weight **

To find the equivalent weight of Potassium Permanganate ($KMnO_4$) in an acidic medium, we need to determine the n-factor (the number of electrons gained or lost per molecule).

The Calculation

In an acidic medium, the permanganate ion ($MnO_4^-$) is reduced to the manganese(II) ion ($Mn^{2+}$). Let's look at the change in the oxidation state of Manganese:

Oxidation state in $KMnO_4$: $Mn$ is in the +7 state.
Oxidation state in $Mn^{2+}$: $Mn$ is in the +2 state.
Difference (n-factor): $+7 \rightarrow +2 = 5$ electrons gained.

The half-reaction is:

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

Finding Equivalent Weight

The formula for equivalent weight is:

$\text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}}$

Since the n-factor is 5, the equivalent weight is one-fifth of its molecular weight.

Option-wise Explanation:

Option 1: One-third of its molecular weight

Would mean n-factor = 3

This happens only in neutral medium ($Mn^{7+} → MnO_2$)

Not acidic medium

Incorrect

Option 2: One-fifth of its molecular weight

Matches n-factor = 5 in acidic medium

Correct

Option 3: Half of its molecular weight

Would mean n-factor = 2

Not applicable for $KMnO_4$

Incorrect

Option 4: Equal to its molecular weight

Would mean n-factor = 1

Not possible for $KMnO_4$

Incorrect

Read the passage carefully and answer the Questions.

$2MnO_2+ 4KOH + O_2→2KMnO_4+2H_2O$

$3KMnO_4+ 4H^+→2KMnO_4 + MnO_2 + 2H_2O$

Equivalent weight of $KMnO_4$ in its oxidising reactions in acidic medium is taken as

Answer & explanation

Correct answer: option 2

The correct answer is Option (2) → One-fifth of its molecular weight **

To find the equivalent weight of Potassium Permanganate ($KMnO_4$) in an acidic medium, we need to determine the n-factor (the number of electrons gained or lost per molecule).

The Calculation

In an acidic medium, the permanganate ion ($MnO_4^-$) is reduced to the manganese(II) ion ($Mn^{2+}$). Let's look at the change in the oxidation state of Manganese:

Oxidation state in $KMnO_4$: $Mn$ is in the +7 state.
Oxidation state in $Mn^{2+}$: $Mn$ is in the +2 state.
Difference (n-factor): $+7 \rightarrow +2 = 5$ electrons gained.

The half-reaction is:

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

Finding Equivalent Weight

The formula for equivalent weight is:

$\text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}}$

Since the n-factor is 5, the equivalent weight is one-fifth of its molecular weight.

Option-wise Explanation:

Option 1: One-third of its molecular weight

Would mean n-factor = 3

This happens only in neutral medium ($Mn^{7+} → MnO_2$)

Not acidic medium

Incorrect

Option 2: One-fifth of its molecular weight

Matches n-factor = 5 in acidic medium

Correct

Option 3: Half of its molecular weight

Would mean n-factor = 2

Not applicable for $KMnO_4$

Incorrect

Option 4: Equal to its molecular weight

Would mean n-factor = 1

Not possible for $KMnO_4$

Incorrect