Find the value of n in following ionic reaction:MnO4- + 8H+ + ne- →

1.	Find the value of n in following ionic reaction:MnO4- + 8H+ + ne- → Mn2+ + 4H2O
Answer» For determining the value of n, the oxidation number of Mn in left and right sides is calculated: In MnO₄^-, x + 4(-2) = -1 (Let oxidation no. of Mn = x) x - 8 = -1 x = -1 + 8 = +7 In Mn²⁺ x = +2 Since, there is decrease in oxidation number of 5. So, there is gain of 5 electrons in this reaction. Hence, the value of n = 5.

Find the value of n in following ionic reaction:MnO4- + 8H+ + ne- → Mn2+ + 4H2O

Answer»

For determining the value of n, the oxidation number of Mn in left and right sides is calculated:

In MnO₄^-,

x + 4(-2) = -1 (Let oxidation no. of Mn = x)

x - 8 = -1

x = -1 + 8 = +7

In Mn²⁺

x = +2

Since, there is decrease in oxidation number of 5. So, there is gain of 5 electrons in this reaction. Hence, the value of n = 5.