Question

Balance the equation

AsO3 + NO3^- ------------- H3AsO4 + NO

Please show me how you solve it. I tried doing it, but im getting stuck on the last couple steps. Thank you

Answer 1

=> Break the reaction into two half-reactions, one for As and one for N.

As2O3 -----> H3AsO4
NO3- -----> NO

=>First, balance the main atoms, As and N.
As2O3 -----> 2H3AsO4
NO3- -----> NO

=> For an acidic solution, balance the O by adding H2O.
As2O3 + 5H2O -----> 2H3AsO4
NO3- -----> NO + 2H2O

=> Balance the H by adding H+
As2O3 + 5H2O -----> 2H3AsO4 + 4H+
NO3- + 4H+ -----> NO + 2H2O

=> Balance the charges by adding e- to the more positive side of each half-reaction.
As2O3 + 5H2O -----> 2H3AsO4 + 4H+ + 4e-
NO3- + 4H+ + 3e- -----> NO + 2H2O

=> Equalize the number of electrons gained and lost. Multiply the first equation by 3 and the second by 4.
3As2O3 + 15H2O -----> 6H3AsO4 + 12H+ + 12e-
4NO3- + 16H+ + 12e- -----> 4NO + 8H2O

=> Add the two half-reactions together.
3As2O3 + 15H2O -----> 6H3AsO4 + 12H+ + 12e- (oxidation occurs at the anode)
4NO3- + 16H+ + 12e- -----> 4NO + 8H2O (reduction occurs at the cathode)
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3As2O3 + 4NO3- + 4H+ + 7H2O -----> 6H3AsO4 + 4NO