The different oxidation state which show spontaneous disproportionation in acidic medium are:
A
Mn6+ and Mn3+
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B
Mn6+ and Mn2+
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C
Mn7+ and Mn3+
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D
Mn4+ and Mn2+
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Solution
The correct option is AMn6+ and Mn3+ A particular oxidation state will show disproportionation reaction if cell formed by this will have positive emf value. So Mn+6 will show as its emf is positive. 3Mn6+→2Mn7++Mn4+;
Eo=EoOPMn6+/Mn7++EoRPMn6+/Mn4+
=−0.56+2.26
=1.70V Also, Mn+3 will show such reaction as its emf is also positive as 2Mn3+→Mn4++Mn2+;