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In the redox reactin xKMnO(4)+NH(3)rar...

In the redox reactin
`xKMnO_(4)+NH_(3)rarryKNO_(3)+MnO_(2)+KOH+H_(2)O`

A

x = 4, y = 6

B

x = 3, y = 8

C

x = 8, y = 6

D

x = 8, y = 3

Text Solution

Verified by Experts

The correct Answer is:
D
`overset(+7)(KMnO_4)rarroverset(+4)(MnO_2)" "...(i)` (Reduction)
`overset(-3)(NH_3)rarroverset(+5)(KNO_3)" "....(ii)` (Oxidation)
There is change in oxidation no. of 3 units in eq. (i) and 8 units in equation (ii). So multiplying eq. (i) by 8 and eq. (ii) by 3 and adding:
`8KMnO_(4) +3NH_3 rarr 8MnO_(2) + 3KNO_3`
Adding KOH molecule to balance K – atoms and `H_2O` molecule to balance H and O atoms:
`8KMnO_(4) +3NH_3 rarr 8MnO_(2) + 3KNO_3 + 5KOH + 2H_2O`
Hence x 8 , y = 3
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