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

Question

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

Avery Adams · Accepted Answer

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

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 there is an equivalent oxidation for each reduction.

#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, in my opinion, is balanced in terms of mass and charge, as it really needs to be in order to depict physical reality.

Caleb Adams · Answer

undefined 1. Assign oxidation numbers to each element in the equation.
2. Determine which elements are oxidized and which are reduced.
3. Write half-reactions for oxidation and reduction.
4. Balance the atoms in each half-reaction.
5. Balance the charges in each half-reaction by adding electrons.
6. Multiply the half-reactions by appropriate factors to make the number of electrons equal in both.
7. Add the balanced half-reactions together.
8. Cancel out any species that appear on both sides of the equation.
9. Check that the atoms and charges are balanced in the final equation.

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

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