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.

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.

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

To determine the number of electrons needed to balance a redox equation, you can follow these steps:

1. Assign oxidation numbers to each element in the equation. In this case:
HI(s) → H+1 I-1
H2SO4(aq) → H+1 S+6 O-2 4H+1

2. Identify the elements that change their oxidation numbers. In this case, the iodine goes from -1 on the left side (as part of HI) to 0 on the right side (as part of I2), while the sulfur goes from +6 on the right side (as part of H2SO4) to -2 on the left side (as part of H2S).

3. Write the two separate half-reactions for oxidation and reduction:
a) Oxidation half-reaction: I-1 → I2
b) Reduction half-reaction: S+6 → S-2

4. Balance the atoms in each half-reaction, excluding the oxygen and hydrogen elements. For the oxidation half-reaction, you have one iodine on the left side and two iodines on the right side. Thus, you need to multiply the left side by 2 to balance it: 2I-1 → I2. The reduction half-reaction is already balanced in terms of atoms.

5. Now, balance the oxygen atoms by adding water (H2O) molecules to the side that lacks oxygen. In this case, only the reduction half-reaction requires oxygen balancing. On the left side of the equation, there are 4 oxygen atoms due to the presence of H2SO4. To balance it, add 4 water molecules to the right side: S+6 → S-2 + 4H2O.

6. Next, balance the hydrogen atoms by adding hydrogen ions (H+) to the side that lacks hydrogen. Here, only the reduction half-reaction requires hydrogen balancing. On the left side, there are 4 hydrogen atoms due to the presence of H2SO4. To balance it, add 8 hydrogen ions to the right side: S+6 → S-2 + 4H2O + 8H+.

7. Now, balance the charges by adding the appropriate number of electrons (e-) to each side of the half-reactions. Start by balancing the charges in the reduction half-reaction. Since the sulfur element goes from +6 to -2, a total of 8 electrons needs to be added on the left side: S+6 + 8e- → S-2 + 4H2O + 8H+.

8. Finally, balance the charges in the oxidation half-reaction. The number of electrons should match the reduction half-reaction. In this case, you need to add 8 electrons on the right side: 2I-1 → I2 + 8e-.

Now, the equation is balanced, and you can proceed with the remaining steps of balancing the full redox equation.

I hope this explanation helps you understand how to determine the number of electrons needed to balance a redox equation using the half-reaction method.