A) fuse with KOH/air, electrolytic oxidation
B) fuse with KOH/air, electrolytic reduction
C) fuse with cone HN03/air, electrolytic reduction
D) All of the above
Correct Answer: A
Solution :
\[Mn{{O}_{2}}+4KOH+{{O}_{2}}\xrightarrow[{}]{{}}2{{K}_{2}}Mn{{O}_{4}}+{{H}_{2}}O\]Step I \[MnO_{4}^{2-}\xrightarrow[{}]{Electrolytic\text{ }oxidation}MnO_{4}^{-}\] Step IIYou need to login to perform this action.
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