balance the following equation in acidic conditions. phases are optional. mno4 + hno2
MnO^-4+HNO2-->NO^-3+Mn^2+
WHICH ELEMENT GOT OXIDIZED?
REDUCE?
WHICH SPECIES WAS THE OXIDIZING AGENT?
REDUCING AGENT?
Homework Answers
MnO4- --->Mn2+
HNO2 + H2O ---> NO3- +3H+ + 2e-
MnO4- + 8H+ + 4e- ---> Mn2+ + 4H2O
LHS = 1 x MnO4^- + 8 H+ = 7+
RHS = 1 x Mn^2+ = 2+
add 5e- to products side to balance the number of electrons
MnO4- + 8H+ + 5e- ? Mn2+ + 4H2O
5 x (HNO2 + H2O --->NO3- +3H+ + 2e-) ---------1
2 x (MnO4- + 8H+ + 5e- ---> Mn2+ + 4H2O) -------2
5HNO2 + 5H2O + 2MnO4^- +16H+ +10e ---> 5NO3- +15H+ +10e- + 2Mn^2+ + 8H2O
cancelling the similar terms of the equation we get a balanced reaction as
5HNO2 + 2MnO4^- + H+ --->5NO3- + 2Mn^2+ + 3H2O
elements getting oxidised =
elements getting reduced
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