Question

Sulfuric acid is a very strong acid that can act as an oxidizing agent at high concentrations (very low pHpH, or even negative pHpH values). Under standard conditions, sulfuric acid has a low reduction potential,

SO42−(aq)+4H+(aq)+2e−⇌SO2(g)+2H2O(l),   +0.20 VSO42−(aq)+4H+(aq)+2e−⇌SO2(g)+2H2O(l),   +0.20 V

which means it cannot oxidize any of the halides F2F2, Cl2Cl2, Br2Br2, or I2I2. If the H+H+ ion concentration is increased, however, the driving force for the sulfuric acid reduction is also increased according to Le Châtelier's principle. Sulfuric acid cannot oxidize the fluoride or chloride anions, but it can oxidize bromide and iodide anions when there are enough H+H+ ions present. The standard reduction potentials of the halogens are as follows:

F2+2e−Cl2+2e−Br2+2e−I2+2e−→→→→2F−,2Cl−,2Br−,2I−,+2.87 V+1.36 V+1.07 V+0.54 VF2+2e−→2F−,+2.87 VCl2+2e−→2Cl−,+1.36 VBr2+2e−→2Br−,+1.07 VI2+2e−→2I−,+0.54 V

The Nernst equation allows us to determine what nonstandard conditions allow the reaction to occur (have a positive EE value).

What pH is needed to produce this value of Q (Q=2.7×10−26) if the concentration and pressure values are [Br2]=2.50×10−4M, [Br−]=12.65M, [SO42−]=9.80M, and PSO2=3.50×10−5atm?

E=E∘−2.303RT/nF*logQ=E∘−0.0592 VnlogQ=E∘−0.0592V/n*logQ

where R=8.314 J/(mol⋅K)R=8.314 J/(mol⋅K), TT is the Kelvin temperature, n is the number of moles of electrons transferred in the reaction, and F=96,485 C/mol e−F=96,485 C/mol e−.