[zephyrblows]

I saw a statement on a general chemistry textbook: "Neither HBr and HI can be prepared with H2SO4 because H2SO4 oxidizes Br^- and I^-."

According to the table of reduction potentials, however, the E⁰ of Br₂ and I₂ are both much higher than SO₄^2-:
I₂ + 2 e^- -----> 2 I^- ..E⁰=+0.53
Br₂ + 2 e^- -----> 2 Br^- ..E⁰=+1.07
SO₄^2- + 4 H⁺ + 2 e^- -----> SO₂ + 2 H₂O ..E⁰=+0.20

Why, then, is Br^- oxidized by concentrated H₂SO₄?

[sdekivit]

The table gives you only the standard electrodepotentials: all concentrations are 1M at T = 298 K and p = p₀

When the concentration deviates from the 1 M you have to use the Nernst equation. If you choose the concentration in such way, there can occur a redoxreaction between the two (when ln K > 0.92)

[sdekivit]

are you sure about the halfreaction of bromine ? I think it needs to be:

2Br(-) --> Br2(s) + 2e- E = +1.09 V

Nernst equation: E = E0 + (RT/nF) * ln K.

Now i don't understand the question:

How do I know Br- is oxidized by H2SO4 from the reduction potential table?

What do you want to know exactly?

[sdekivit]

hmm maybe somebody else has an opinion about this, but as far as i know the statement that H₂SO₄ oxidizes Br^- en I^- is strange because, as you noted, the standard electrode potential of sulfuric eacid is lower than that of Br2.

Or is this stament made in a particular context, i mean as an example of the theory ? his statement alone seems strange though.

[zephyrblows]

According to sdekivit's advice, I used the Nernst Equation, but got a strange logQ:

For Br^- in concentrated H₂SO₄, if Br^- is oxidized:
0 < (0.2-1.07) - (0.0592 / 2) logQ
0.87 < -0.0296 logQ
logQ <- 29.4

Is that wrong?

[zephyrblows]

Oh, I guess it's because the temperature isn't 298K.
The temperature at which they prepare HBr is higher...so...