Why is KI oxidized by H2SO4, whereas NaBr isn't oxidised?
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.

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