Sulfuric acid is a very strong acid that can act as an oxidizing agent at high c

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, SO^2- _4 (aq) + 4H^+ (aq) + 2e^- rightwardsharpoonoverleftwardsharpoon SO_2(g) + 2H_2O(l), + 0.20 V Which means it cannot oxidize any of the halides F_2, Cl_2, Br_2 or I_2. If the H^+ ion concentration is increased, however, the driving force for the sulfuric acid reduction is also increased according to lessthanorequalto Chatelier'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: F_2 + 2e^- rightarrow 2 F^-, + 2.87 V Cl_2 + 2e^- rightarrow 2cl^-, + 1.36 V Br_2 + 2e^- rightarrow 2 Br^-, + 1.07 V I_2 + 2e^- rightarrow 2I_, + 0.54 V The Nernst equation allows us to determine what nonstandard conditions allow the reaction to occur (have a positive E vector value). The Nernst equation relates a nonstandard potential, E, to the standard potential, E degree, and the reaction quotient, Q, by E = E degree 2.303 RT/nF log Q = E degree - 0.0592 V/n log Q Where R = 8.314 J/(mol middot 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 65.0 degree C, what is the maximum value of the reaction quotient, Q needed to produce a non-negative E value for the reaction SO^2- _4 (aq) + 4H^+(aq) + 2Br^- (aq) rightwardsharpoonoverleftwardsharpoon Br_2 (aq) + SO_2 (g) + 2H_2O (l) In other words, what is Q when E = 0 at this temperature? Express your answer numerically to two significant figures.

Explanation / Answer

So42- + 4H+ +2e- --------------> So2 + 2H2O     E0 = 0.2v

2Br-    -----------------------> Br2 + 2e-    E0 = -1.07v

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SO42- + 4H+ + 2Br- ---------> So2 + 2H2o + Br2   E0cell = -0.87v

   E     = E0 -2.303RT/nFlogQ

    E = 0

R = 8.314J/mole-K

T   = 65C0   = 65+273 = 338K

F   = 96485c

n = 2

E     = E0 -2.303RT/nFlogQ

0   = E0 -2.303*8.314*338/2*96485 logQ

E0   = 2.303*8.314*338/2*96485 logQ

E0   = 0.03354logQ

-0.87 = 0.03354logQ

logQ   = -0.87/0.03354

logQ   = -25.94

Q   = 10^-25.94    = 1.15*10^-26

0     = -0.87-0.0592/2 logQ

-0.87    = 0.0296logQ

LogQ    = -0.87/0.0296

logQ    = -29.4

Q       = 10^-29.4   = 3.98*10^-30

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