Hey guys! I'm not really stuck on this problem, but I was wondering if someone could look it over to make sure that what I did makes sense. Thank you:

1) Balance this equation. What is the coefficient in front of I2?
IO− + I− --> I2 in acidic solution

Maybe answer:
Since there is only one product, I was wondering if I split it up correctly
4H++2IO-+2e- -->I2+2H2O
2I- -->I2+2e-
2I-+4H++2IO- -->2I2+2H2O
I then divided both sides by two to get:
I-+2H++IO- -->I2+H2O
So my answer would be 1.

Hi there! Great job on working through the problem. Let's go through your steps to balance the equation and find the coefficient in front of I2.

1) You started with the unbalanced equation: IO− + I− → I2 in acidic solution.

2) To balance the equation, you correctly split the reaction into two half-reactions. The half-reaction involving iodine is as follows:
IO− + 8H+ + 5e− → I2 + 4H2O

3) The other half-reaction involving iodide is:
I− → 2e−

4) To balance the electron transfer, you multiplied the iodide half-reaction by 5:
5I− → 10e−

5) Combining the two half-reactions and cancelling out the electrons on both sides gives us the balanced equation:
5IO− + 8H+ + 5I− → 5I2 + 4H2O

6) Now, to find the coefficient in front of I2, you correctly divided both sides of the equation by 5, which gives:
IO− + 8H+ + I− → I2 + 4H2O

So, the coefficient in front of I2 is indeed 1, just as you stated.

Great job on balancing the equation and determining the coefficient! Let me know if you have any other questions.