Question

The material covered in this homework applies to processes occurring at the Earth's surface. Show all your work. Balance the following oxidation reduction reactions (from Faure). i. MnO_4^- Cl^- leftrightarrow Mn^2+ + Cl_2 ii. As_2S_3 + NO_3^- leftrightarrow HASO_3 + S + NO iii. Cr_2O_7^2- + I^- leftrightarrow Cr^3+ + I_2 iv. CrO_2^- + ClO^- leftrightarrow CrO_4^2- + Cl^- v. CH_4^0 + SO_4^2- leftrightarrow HCO_3^- + HS^- Balance the following four overall redox reactions. Which reactions will generate acid and why is this a problem? i. Cu_5FeS_4 (bornite) + O_2 (g) leftrightarrow Cu^2+ + Fe(OH)_3 (s) + SO_4^2- ii. ZnS + O_2 (g) leftrightarrow Zn^2+ + SO_4^2- iii. Fe_3S_4 + O_2 (g) leftrightarrow Fe(OH)_3 (s) + SO_4^2- iv. FeS_2 Pyrite + NO_3^- leftrightarrow N_2 (g) + FeOOH_Goethite + SO_4^2-

Answer 1

(i)

MnO4- + Cl- ---------> Mn2+ + Cl2

Seperate oxidation and reduction half reactions

(a)

Cl^- ----------> Cl₂       oxidation half reaction

MnO₄^- ---------> Mn²⁺   Reduction Half Reaction

(b) Balance atoms other than O and H

2 Cl^- ----------> Cl₂

MnO₄^- ---------> Mn²⁺

(c)

To balance oxygen atoms in acidic medium add H2O molecules.

2 Cl^- ----------> Cl₂

MnO₄^- ---------> Mn²⁺ + 4 H₂O

(d) To balance hydrogen atoms in acidic medium add H⁺ ions.

2 Cl^- ----------> Cl₂

MnO₄^- + 8 H⁺---------> Mn²⁺ + 4 H₂O

(e) To balance charges add electrons

2 Cl^- ----------> Cl₂ + 2 e

MnO₄^- + 8 H⁺+ 5 e ---------> Mn²⁺ + 4 H₂O

(f) To equalise electrons in both half reactions multiply with suitabel numbers.

10 Cl^- ----------> 5 Cl₂ + 10 e

2 MnO₄^- + 16 H⁺+ 10 e ---------> 2 Mn²⁺ + 8 H₂O

(g) Add both half reactions to get balanced equation.

2 MnO₄^- + 16 H⁺+ 10 Cl^- ----------> 5 Cl₂ + 2 Mn²⁺ + 8 H₂O

this is the balanced equation.

follow the similar procedure for remaining