Question

Which of the following reaction(s) does the permanagate ion shows in neutral medium?

• A. $Mn{O}_4^{-}+4H^{+}+3e^{–}\to Mn{O}_2+2H_{2}O$
• B. $Mn{O}_4^{-}+1e^{–}\to Mn{O}_4^{2-}$
• C. $MnO{4}^{-}+8H^{+}+5e^{–}\to M{n}^{2+}+4H_{2}O$
  
• D. (b) and (c) above.

Answer 1

The correct option is A $Mn{O}_4^{-}+4H^{+}+3e^{–}\to Mn{O}_2+2H_{2}O$

Permanganate undergoes different reactions at acidic, neutral and basic medium.

In acidic medium:
$MnO{4}^{-}+8H^{+}+5e^{–}\to M{n}^{2+}+4H_{2}O$

In basic medium:
$Mn{O}_4^{-}+1e^{–}\to Mn{O}_4^{2-}$

In neutral medium:
$Mn{O}_4^{-}+4H^{+}+3e^{–}\to Mn{O}_2+2H_{2}O$

Example:
Oxidation of iodide to iodate:
$2Mn{O}_4^{-}+I^{–}+H_{2}O\to 2Mn{O}_2+2O{H}^{–}+I{O}_3^{-}$

Hence, option (b) is correct.