Question

17.56: Chlorine Dioxide (ClO2) is produced by the following reaction of chlorate (ClO3-) with Cl- in acid solution:

2ClO3- (aq) + 2Cl-(aq) + 4H+(aq) --> 2ClO2(g) + Cl2(g) + 2H2O(l)

a) determine Eo for the reaction

b) The reaction produces a mixture of gases in the reaction vessel in which PClO2 = 2.0atm, PO2 = 1.0atm. Calculate [ClO3-] if at equilibrium T = 298K, [H+] = [Cl-] = 10.00M

Answer 1

ClO3^- (aq) + 2H^+ (aq) + e^- --------------> ClO2 (aq) + H2O      E0 = 1.175v

2Cl^- (aq) ----------------------> Cl2 + 2e^-                                         E0   = -1.358v

2ClO3^- (aq) + 4H^+ (aq) + 2e^- --------------> 2ClO2 (aq) +2 H2O      E0 = 1.175v

2Cl^- (aq) ----------------------> Cl2 + 2e^-                                              E0   = -1.358v

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2ClO3- (aq) + 2Cl-(aq) + 4H+(aq) --> 2ClO2(g) + Cl2(g) + 2H2O(l)    E0   = -0.183v

    n = 2

G⁰     = -n*E0cell*F

             = -2*-0.183*96500

             = 35319J

G⁰    = -RTlnK

35319    = -8.314*298*2.303logK

logQ       = 35319/-5705.8483

logQ      = -6.189

Q           = 10^-6.189 = 6.5*10^-7

Q          = PCl2 P²ClO2/[ClO3^-]^2[Cl^-]^2*[H^+]^4

6.5*10^-7   = 1*(2)^2/[ClO3^-]^2(10)^2(10)^4

[ClO3^-]^2   = 4/6.5*10^-7*10^6

                 = 4/6.5*10^-1

[ClO3^-]^2      = 6.153

[ClO3^-]       = 2.48M >>>>>answer

Ecell   = E0cell - 0.0592/n logQ

             = -0.183 -0.0592/2 logPcl2 P²ClO2/[ClO3^-]^2[Cl^-]^2

             = -0.183-0.0296log1*(2)^2/