Question

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

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

which means it cannot oxidize any of the halides F2, Cl2, Br2, or I2. If the 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+ 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 V

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

The Nernst equation relates a nonstandard potential, E, to the standard potential, E∘, and the reaction quotient, Q, by

E=E∘−2.303RTnFlogQ=E∘−0.0592 VnlogQ

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

At 67.0 ∘C , what is the maximum value of the reaction quotient, Q, needed to produce a non-negative E value for the reaction

SO42−(aq)+4H+(aq)+2Br−(aq)⇌Br2(aq)+SO2(g)+2H2O(l)

In other words, what is Q when E=0 at this temperature?

Answer 1

given reaction is

S042- + 4H+ + 2Br- ---> Br2 + S02 + 2H20

anode reaction : oxidation

2 Br- ---> Br + 2e-

cathode reaction : reduction

S042- + 4H+ + 2e- ---> S02 + 2H20

we know that

Eo cell = Eo cathode - Eo anode

so

Eo cell = 0.2 - ( 1.07)

Eo cell = -0.87

now

we know that

E= Eo - ( 2.303RT/nF ) log Q

given

E = 0

so

Eo = ( 2.303RT/nF) log Q

here

n = 2 as two electrons are transferred

now

T = 67 C = 340 K

so

Eo = ( 2.303RT/nF) log Q

-0.87 = ( 2.303 x 8.314 x 340 / 2 x 96485 ) log Q

Q = 1.63 x 10-26

so

the max value of Q is 1.63 x 10-26