A) 1
B) 2
C) 5
D) 3
Correct Answer: A
Solution :
Oxidation of\[KMn{{O}_{4}}\]takes place in all the three medium acidic, basic and neutral. In all three medium the oxidation number is different. In acidic medium: \[2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\xrightarrow[{}]{{}}{{K}_{2}}S{{O}_{4}}\] \[+2MnS{{O}_{4}}+3{{H}_{2}}\text{O}+5O\] The net reaction is \[MnO_{4}^{-}\xrightarrow{{}}M{{n}^{2+}}\] change in oxidation number \[=7-2=5\] So electrons involved\[=5{{e}^{-}}\] In basic medium: \[2KMn{{O}_{4}}+2KOH\xrightarrow[{}]{{}}2{{K}_{2}}Mn{{O}_{4}}\] \[+{{H}_{2}}O+O\] Net reaction is \[\overset{+7}{\mathop{M}}\,nO_{4}^{-}\xrightarrow{{}}\overset{+6}{\mathop{M}}\,nO_{4}^{-2}\] Change in oxidation number \[=7-6=+1\] So electrons involved\[=1{{e}^{-}}\] In Neutral medium: \[2KMn{{O}_{4}}+{{H}_{2}}\text{O}\xrightarrow{{}}2KOH\] \[+2Mn{{O}_{2}}+3O\] Net reaction \[\overset{+7}{\mathop{M}}\,nO_{4}^{-}\xrightarrow{{}}\overset{+4}{\mathop{M}}\,n{{O}_{2}}\] Change in oxidation no.\[=7-4=+3\] So electrons involved\[=3{{e}^{-}}\]
You need to login to perform this action.
You will be redirected in
3 sec
