Why is KI oxidized by H2SO4, whereas NaBr isn't oxidised? My textbook says that alcohols react with sodium bromide and sulphuric acid to form the corresponding alkyl bromides. At the same time they also mention that sulphuric acid mustn't be used with potassium iodide ($\ce{KI}$), because $\ce{HI}$ formed would be oxidized to $\ce{I2}$. Won't $\ce{HBr}$ formed from $\ce{NaBr}$/$\ce{H2SO4}$ oxidise to $\ce{Br2}$?
Iodide ions are much more powerful reducing agents than bromide ions:

$$\ce{2I- -&gt;I2 + 2e-} \hspace{1cm}E^{o}=-0.54\ \mathrm{V}$$

$$\ce{2Br- -&gt;Br2 + 2e-} \hspace{1cm}E^{o}=-1.0873\ \mathrm{V}$$

Iodide ions are so powerful, in fact, that they are capable of reducing sulfur from a +6 oxidation state in $\ce{H2SO4}$ all the way to -2 in $\ce{H2S}$ via the following half reaction:

$$\ce{HSO4- +9H+ +8e- -&gt; H2S + 4H2O} \hspace{1cm}E^{o}=+0.80\ \mathrm{V}$$

The full reaction being being:

$$\ce{HSO4- +9H+ +8I- -&gt; H2S + 4I2 +4H2O}\hspace{1cm}E^{o}=+0.26\ \mathrm{V}$$

A quick glance at the reduction potentials tells you that the analogous bromide reaction has a negative reduction potential and is therefore non-spontaneous.

[Source]