How to Balance this Equation by Oxidation number method? #H_2SO_4+HI->H_2O+H_2S+I_2#

1 Answer
Jun 17, 2017

Answer:

Well, #"sulfur(VI+)"# is REDUCED to #S"(-II)"#......

Explanation:

Well, #"sulfur(VI+)"# is REDUCED to #S"(-II)"#......a formal 8 electron transfer.......

#"SO"_4^(2-)+8H^(+) +8e^(-) rarrS^(2-)+4H_2O(l)# #(i)#

And of course sulfide anion has the same oxidation state as the charge on the ion, i.e. #S(-II)#; a formal eight electron reduction.

And for every reduction there is a corresponding oxidation......

#I^(-) rarr 1/2stackrel(0)I_2(s)+e^(-)# #(ii)#

And we takes #(i)+8xx(ii)#:

#8I^(-) +SO_4^(2-)+8H^(+) rarr 4stackrel(0)I_2(s) +S^(2-) + 4H_2O(l)#

The which, I think, is balanced with respect to mass and charge; as indeed it must be if it is to represent physical reality.