Balance #3b and 3c Fall 2017 Redox Problem Set Page 3 of 4 b) ci, (aq) CIO, (aq)

Question

Balance #3b and 3c Fall 2017 Redox Problem Set Page 3 of 4 b) ci, (aq) CIO, (aq) cr (aq) + c.) Cr(OH), (s) H o, (aq) CrO.. (aq) + 4.) In an aqueous solution, nitric oxide, NO, reacts with the triiodide ion, I, in a redox reaction to yield NO, and the iodide ion. A chemist titrates a 25.0 mL. sample of an acidic solution of NO with a 2.85 x 10 M aqueous solution of l. If she finds that requires 21.73 mL of the triodide solution to reach the equivalence point, what was the molar concentration of the NO in the original solution?

Explanation / Answer

3)

The given reactions are balanced through ion-electrode method:

b) Cl2(aq) ClO3-(aq) + Cl-(aq)

The reaction can be split into two half reactions as:

Oxidation half reaction                                                                        Reduction half reaction

Cl ClO3-                                                                                          Cl Cl-

Balancing the oxygen atoms: no oxygen atoms present

Cl + 3H2O ClO3-                                                                                             Cl Cl-

Balancing the hydrogen atoms:                                                          no hydrogen atoms present

Cl + 3H2O ClO3- + 6H+                                                                                Cl Cl-

Balancing the charges:                                                                       Balancing the charges:

Cl + 3H2O ClO3- + 6H+ + 5e-                                                         Cl +e- Cl-     

To equal the charges we have to multiply the reduction half reaction by 5, and then we have to add oxidation half reaction and reduction half reaction.

Cl + 3H2O ClO3- + 6H+ + 5e-

5Cl +5e- 5Cl-             

6Cl +3H2O ClO3- + 5Cl-+ 6H+

Balanced reaction is: 3Cl2(aq) +3H2O(aq) ClO3-(aq) + 5Cl-(aq)+ 6H+(aq)

c) The reaction given is, Cr(OH)3(s) + H2O2(aq) CrO4-(aq), I think something is missed in the equation, as per my assumption the reaction must be Cr(OH)3(s) + H2O2(aq) CrO4-(aq) + H2O.

The reaction can be split into two half reactions as:

Oxidation half reaction                                                            Reduction half reaction

Cr(OH)3 CrO4-                                                                   H2O2 H2O

Balancing the oxygen atoms: Balancing the oxygen atoms:

Cr(OH)3 + H2O CrO4- H2O2 H2O + H2O

Balancing the hydrogen atoms: Balancing the hydrogen atoms:

Cr(OH)3 + H2O CrO4-+ 5H+                                                       H2O2 + 2H+ 2H2O

Balancing the charges: Balancing the charges:

Cr(OH)3 + H2O CrO4-+ 5H++ 3e- H2O2 + 2H++2e- 2H2O

To equal the charges we have to multiply the reduction half reaction by 3, oxidation half reaction by 2; and then we have to add oxidation half reaction and reduction half reaction.

2Cr(OH)3 + 2H2O 2CrO4-+ 10H++ 6e-     

3H2O2 + 6H++6e- 6H2O

Balanced reaction is: 2Cr(OH)3 + 3H2O2 2CrO4-+4H++4H2O

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