I in HI has an oxidation state of -1. On the other side I2 is zero. So in going from 2I^- to I2, I lost 2electrons.
S in H2SO4 is +6. S in H2S is -2. In going from +6 to -2, there is a gain of 8 electrons. I do this intuitively but you can do it mathematically.
+6-(-2) = +8 so gain of 8.
-2-(0) = -2 so loss of 2.
Balance each of the following redox equations using the half-reaction method.
HI(s) + H2SO4 (aq) --> H2S(g) + I2 (s)
My only question about this equation is...how do I know how many electrons I need to put on both sides? Other than that I know how to do it.
2 answers
You carefully measure 20.00 of powder and add it to 81.00 of solution in a reaction flask of known mass. You notice bubbles as a reaction takes place. You then determine that the contents of the flask have a mass of 92.80 . The relevant equation is
Assuming no other reactions take place, what mass of was produced in this reaction?
Express the mass numerically in
Assuming no other reactions take place, what mass of was produced in this reaction?
Express the mass numerically in
1 answer