What is the ionic form of the following unbalanced equation #MnO_2 + HNO_2 -> Mn(NO_3)_2 + H_2O#?

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Answer 1 · 2017-10-23T08:14:10

You gots a redox reaction of the form....

#HONO(aq) +MnO_2(s) +H^(+) rarr Mn^(2+) +H_2O(l)+NO_3^(-)#

How do we get to this?

Well, #"manganese(IV) oxide"# is reduced to #Mn(II)#:

#MnO_2(s) +4H^(+) + 2e^(-) rarr Mn^(2+) +2H_2O(l)# #(i)#

And, #"nitrous acid"# is oxidized to #"nitrate ion"#:

#HONO(aq) +H_2O(l) rarr NO_3^(-)+ 3H^+ + 2e^(-)# #(ii)#

And we just add #(i) + (ii)# together......

#HONO(aq) +cancel(H_2O(l)) +MnO_2(s) +cancel4H^(+) + cancel(2e^(-))rarr Mn^(2+) +cancel2H_2O(l)+NO_3^(-)+ cancel(3H^+ + 2e^(-))#

To give finally.....

#HONO(aq) +MnO_2(s) +H^(+) rarr Mn^(2+) +H_2O(l)+NO_3^(-)#