Complete and balance the equation for this reaction in acidic solution.

MnO^-4+HNO2-->NO^-3+Mn^2+

WHICH ELEMENT GOT OXIDIZED?
REDUCE?
WHICH SPECIES WAS THE OXIDIZING AGENT?
REDUCING AGENT?

HNO2 ---> NO3-
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