Use the half-reaction method to balance the following redox reaction in acidic solution: As2O3(s) + NO3(aq)- ---> H3AsO4(aq) + N2O3(aq) How do you balance a redox reaction in acidic conditions?

enter image source here

1 Answer
Jan 14, 2018

Well you gots nitrate reduced to #N_2O_3#....

Explanation:

#2stackrel(+V)NO_3^(-) +6H^+ +4e^(-) rarr (O=)N-stackrel(+)N(=O)(-O)^(-)+3H_2O(l)#

...i.e. we gots #N(+III)# in #N_2O_3#....(I think that #N_2O_3# has a #N-N# bond not a bridging #N-O-N#...if anyone knows otherwise please correct me. The formal oxidation states are #N(+II)# and #N(+IV)#, i.e. #N(+III)# average.....

And meanwhile arsenic(III) oxide is oxidized to arsenate....

#As_2O_3(s) +5H_2O(l) rarr 2AsO_4^(3-)+10H^+ + 4e^(-)#

And so we adds the former to the letter to remove the electrons as virtual particles....

#As_2O_3(s) +cancel(5)2H_2O(l) +2NO_3^(-) +cancel(6H^+) +cancel(4e^(-))rarr 2AsO_4^(3-)+cancel(10)4H^+ + cancel(4e^(-))+(O=)N-stackrel(+)N(=O)(-O)^(-)+cancel(3H_2O(l))#

...to give...

#As_2O_3(s) +2H_2O(l) +2NO_3^(-) rarr 2AsO_4^(3-)+4H^+ +(O=)N-stackrel(+)N(=O)(-O)^(-)#

Please don't assume that I have done my sums right.....And note that when I do a redox equation I ALWAYS assume acidic conditions. If basic conditions are required I would simply add #4xxHO^-# to both sides.....