Reduction of the metal centre in aqueous perm | vedclass

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Question

(A) In neutral medium:$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$(Involves $3$ electrons)In alkaline medium:$MnO_4^- + e^- \rightarrow MnO_4^{2

Vedclass · Accepted Answer

$A, C, D$

Vedclass · Answer

(A) In neutral medium:$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$(Involves $3$ electrons)In alkaline medium:$MnO_4^- + e^- \rightarrow MnO_4^{2-}$$MnO_4^{2-} + 2H_2O + 2e^- \rightarrow MnO_2 + 4OH^-$Overall: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$(Involves $3$ electrons)In acidic medium:$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$(Involves $5$ electrons)Comparing with the options:$A$: $3$ electrons in neutral medium (Correct)$B$: $5$ electrons in neutral medium (Incorrect)$C$: $3$ electrons in alkaline medium (Correct)$D$: $5$ electrons in acidic medium (Correct)Thus,the correct combination is $A, C, D$.

Reduction of the metal centre in aqueous permanganate ion involves

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