How do I balance a redox reaction for NaI + H2SO4 = H2S + I2 + Na2SO4 + H2O? I don't understand how to balance the electron loss for Iodine.

Answer 1

First we note that we have 1 I on the left and 2 on the right. In order to compare apples with apples we need to have the same number; therefore, we stick a 2 on the left so we start with the same number of I atoms.
2I^- ==>I2
So we have changed from -2 (for two I atoms) to zero (for two I atoms). That is a loss of two electrons.
2I^- ==> I2 + 2e
That first step is almost never addressed in redox instructions and it is the difference between success and failure. It is common to do this with I^- ==> I2, Cr2O7^2- ==> Cr^3+, Cl2 ==> Cl^-, S2O4^2- ==> S4O6^2-, C2O4^2- ==> CO2, etc.