KMnO_4 can be prepared from K_2MnO_4 as per the reaction 3MnO42-+2H_2O⇌2MnO4-+MnO_2+4OH^(-) The reaction can go to completion by removing OH^(-) ions…

Question

KMnO_4 can be prepared from K_2MnO_4 as per the reaction 3MnO42-+2H_2O⇌2MnO4-+MnO_2+4OH^(-) The reaction can go to completion by removing OH^(-) ions by adding

MARKS Content Team · Accepted Answer

Correct answer: (B) CO_2. As MnO4- can oxidise SO_2 , HCl but not CO_2 as in CO_2 C-atom has its highest oxidation state (+4) it means CO_2 can be used here.

$K M n O_{4}$ can be prepared from $K_{2} M n O_{4}$ as per the reaction
$3 M n O_{4}^{2 -} + 2 H_{2} O ⇌ 2 M n O_{4}^{-} + M n O_{2} + 4 O H^{-}$
The reaction can go to completion by removing $O H^{-}$ ions by adding

Solution

More d and f Block Elements Questions

KMnO4 can be prepared from K2MnO4 as per the reaction 3MnO42-+2H2O⇌2MnO4-+MnO2+4OH- The reaction can go to completion by removing OH- ions by adding

Solution

More d and f Block Elements Questions

$K M n O_{4}$ can be prepared from $K_{2} M n O_{4}$ as per the reaction
$3 M n O_{4}^{2 -} + 2 H_{2} O ⇌ 2 M n O_{4}^{-} + M n O_{2} + 4 O H^{-}$
The reaction can go to completion by removing $O H^{-}$ ions by adding