Question

In the reaction,$2KMnO_4 + 16 HCl \rightarrow 5Cl_2 + 2MnCl_2 + 2KCl+8H_2O$ reduction product is

• A. $Cl_2$
• B. $MnCl_2$
• C. $H_2O$
• D. $KCl$

Answer 1

Answer: (B)

$MnCl_2$

Answer 2

n this reaction O.N. of Mn changes from +7 in $KMn0_4$ to + 2 in $MnCl_2$. Thus. $MnCl_2$ is the reduction produce