ENCE 433 Dr. Alba Torrents

ENVIRONMENTAL ENGINEERING ANALYSIS Fall 1996

Homework Set # 8 SOLUTION

REDOX

1.

(a) According to rule 6: Oxidation state Mn + 4 x (-2) = -1

Oxidation state of Mn = +7

(b) According to rule 5: Oxidation state of C + 4 x (-1) = 0

Oxidation state of C = +4

(c) According to rules 4 and 6: 2 x (Ox. st. Cr.) + 7x(-2) = -2

Oxidation state of Cr = +6

(d) KClO3 is a salt form with K+ and ClO3-

Oxidation state of Cl + 3 x (-2) = -1

Oxidation state of Cl = +5

(e) In methanol:

Oxidation state of C + (-2) + 4x(1) = 0

Oxidation state of C = -2

(f) In the cyanide ion, we utilize rule 5. N is the more electronegative of these two elements, and the common negative oxidation state of N is -3 ( as in NH3). This makes the oxidation state of carbon +2, since +2 + (-3) = -1, the charge on the ion.

2.

Sulfur dioxide gets oxidize by the oxidant (residual chlorine is the oxidant). The residual chlorine compounds get reduce.

The half reactions are:

H₂SO₃ + H₂O SO₄^-2 + 4 H⁺ + 2 e^-

HOCl + H⁺ + 2e^- Cl^- + H₂O

NH₂Cl +2e^- + 2H⁺ Cl^- + NH₄⁺

The overall balanced reactions are:

HOCl + H₂SO₃ SO₄^-2 + Cl^- + 3 H⁺

NH₂Cl + H₂SO₃ SO₄^-2 + Cl^- + NH⁺₄ + 2 H⁺

3.

(a)

(b)

c)

The most reducing environment is (a)

4.

Reduction: MnO₄^-2 + 8 H⁺ + 5 e^- Mn⁺² + 4 H₂O E^o = 1.49 V

Oxidaton: 2 Br^- Br₂ + 2 e^- E^o = -1.09 V

Overall: 2MnO₄^-2 + 16 H⁺ + 10 Br ^- 5Br₂ + 2Mn⁺² + 8 H₂O DE^o = 0.40V

Since the calculated standard cell potential is positive, the reaction proceeds spontaneous as written when all the substances are in their standard state. Permanganate ions is a very strong oxidazing agent and oxidizes Br^- to Br₂ .