Question

What will be the equivalent weight of KMnO4 in acidic, basic and neutral medium?

Answer 1

The Mn in KMnO4 exists in +7 state.

In Acidic medium, This Mn+7 goes to Mn+2 state and hence there is a net gain of 5 electrons. 
Now, equivalent weight = [molar mass] /[number of electrons gained or lost]
so, eq wt = 158/5 = 31.6 g.

Now, for alkaline medium, there are two possibilities
1) faintly alkaline/neutral:
 Mn+7 changes into Mn+4 therefore, gain of 3 electrons
 Hence, eq wt = 158/3 = 52.67g

Answer 2

In Acidic medium, This Mn+7 goes to Mn+2 state and hence there is a net gain of 5 electrons.
Now, equivalent weight = [molar mass] /[number of electrons gained or lost]
so, eq wt = 158/5 = 31.6 g.

Now, for alkaline medium, there are two possibilities
1) faintly alkaline/neutral:
Mn+7 changes into Mn+4 therefore, gain of 3 electrons
Hence, eq wt = 158/3 = 52.67g.
2) highly alkaline:
Mn+7 changes into Mn+6 therefore, gain of 1 electron
Hence, eq wt = 158/1 = 158g

In many cases though if it's written alkaline, it mostly means the neutral one; for the highly alkaline thing, it'll especially mention it.

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