Chlorine dioxide (CIO_2), which is produced by the reaction below, has been test

Question

Chlorine dioxide (CIO_2), which is produced by the reaction below, has been tested as a disinfectant for municipal water treatment. 2 NaCIO_2(aq) + Cl_2(g) rightarrow 2 CIO2(g) + 2 NaCI(aq) Using data from the Standard Reduction Potentials table, calculate degree, Delta G degree, and K at 25 degree C for the production of CIO_2. One of the concerns in using CIO_2 as a disinfectant is that the carcinogenic chlorate ion (ClO_3^-) might be a by-product. It can be formed from the following reaction. CIO_2(g) rightarrow CIO_3^-(aq) + Cl^-(aq) Balance the equation for the decomposition of CIO_2. (Use the lowest possible coefficients. Include states-of-matter at 25 degree C and 1 atm in your answer.)

Explanation / Answer

Answer – We are given the reaction

2 NaClO2(aq) + Cl2(g) -----> 2 ClO2(g) + 2 NaCl(aq)

First we need to calculate the Eocell for this reaction

2 ClO2- ---> 2 ClO2 + 2e- Eooxi = -0.954V

Cl2 + 2e- ----> 2Cl-   Eored = 1.36 V

So, 2 NaClO2(aq) + Cl2(g) -----> 2 ClO2(g) + 2 NaCl(aq) Eocell = 0.406 V

?Go = -n*F*Eocell

       = - 2*96,485 C/mol *0.406 V

      = -78346 J

      = -78.3 kJ

Now we know,

?Go = -RT*lnK

-78346 J = 8.314 J/mol.K * 298 K * ln K

ln K = -31.62

so anitln from both side

K = 1.85*10-14

b) Now balancing the reaction –

ClO2(g) ----> ClO3- (aq) + Cl-(aq)

Step 1) Assign the oxidation state to each element and write the two half reaction

ClO2(g) ----> ClO3- (aq) + Cl-(aq)

Cl = +4             Cl = +5

                        Cl = -1

O = -2           O= -2

So,

         ClO2(g) ----> ClO3- (aq) ………oxidation half reaction

         ClO2(g) ----> Cl-(aq)……….reduction half reaction

Step 2) Balance the element other than O and H

                       ClO2(g) ----> ClO3- (aq)

                       ClO2(g) ----> Cl-(aq)

Step 3) Balance the O by adding 1 H2O for 1 O

                       ClO2(g) + H2O ----> ClO3- (aq)

                       ClO2(g) ----> Cl-(aq) + 2 H2O

Step 4) Balance the H by adding H+

                       ClO2(g) + H2O ----> ClO3- (aq) + 2H+

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

Step 5) Balance the charge by adding electron

                       ClO2(g) + H2O ----> ClO3- (aq) + 2H+ +1e-

                       ClO2(g) + 4H+ + 5e-----> Cl-(aq) + 2 H2O

Step 6) Balance the electron in both half reaction – Multiply by oxidation half reaction by 5

                       5 ClO2(g) +5 H2O ----> 5ClO3- (aq) + 10H+ +5e-

                       ClO2(g) + 4H+ + 5e-----> Cl-(aq) + 2 H2O

________________________________________________________________

   6 ClO2(g) + 3 H2O -------> 5ClO3- (aq) + Cl-(aq) + 6H+ (aq

Balanced redox reaction –

6 ClO2(g) + 3 H2O -------> 5ClO3- (aq) + Cl-(aq) + 6H+ (aq

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