Question

Consider the following unbalanced redox reactions. In each case, separate the whole reactions into half-reactions, balance the half reactions, and combine to yield a balanced whole reaction. Compute the net reduction potential (E°) and pε° values for the net reactions and indicate whether the reaction is spontaneous as written. The reduction potential for Fe(OH)_3(s)/Fe^2+ is -0.181 V and pε is -3.08.
a. Fe^2 + + NO_3 = Fe(OH)_3(s) + N_2
b. Mn^2 + + Fe(OH)_3(s) = MnO_2(s) + Fe^2+

Answer 1

Answer & Explanation:

(a)

$Fe^{2+} +NO_{3}^{-} => Fe{(OH)}_3 + N_2$

reducing agent = Fe²⁺

Oxidizing agent = NO₃⁻

oxidation

Fe²⁺  ⇒ Fe(OH)₃

reduction

NO₃⁻  ⇒ N₂

Oxidation Half Reaction

(redox reactions are balanced by adding appropriate H⁺ and H₂O atoms)

Fe²⁺ ⇒ Fe(OH)₃

Balance O atoms

Fe²⁺ + 3H₂O ⇒ Fe(OH)₃

Balance H atoms

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺

balance Charge

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺ + e⁻..............(1)

reduction Half Reaction

NO₃⁻ ⇒ N₂

Balance N atoms

2NO₃⁻ ⇒N₂

Balance O atoms by adding appropriate H₂O

2NO₃⁻ ⇒ N₂ + 6H₂O

Balance H atoms

2NO₃⁻ + 12H⁺ ⇒ N₂ + 6H₂O

Balance Charge

2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O.................(2)

Combine Equation (1) and (2)

(1) × 10: 10Fe² + 30H₂0 ⇒ 10Fe(OH)₃ + 30H⁺ + 10e⁻

(2) × 1:   2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O

(1) + (2): 10Fe² + 30H₂0 + 2NO₃⁻ + 12H⁺ + 10e⁻ ⇒10Fe(OH)₃ + 30H⁺ + 10e⁻ +              

            N₂ + 6H₂O

            10Fe² + 24H₂0 + 2NO₃⁻  ⇒ 10Fe(OH)₃ + 18H⁺ + N₂

this is the balanced reaction

REDUCTION POTENTIAL

10Fe²⁺(aq) + 10e⁻ ⇒ 10Fe(OH)₃(aq)             E°ox = 10(-0.44) = -4.4V

2NO₃⁻(aq) -  2e⁻ ⇌ N₂(g) + 18H⁺       E°red = 2(+0.80) = +1.6

10Fe² + 24H₂0 + 2NO₃⁻  ⇒ 10Fe(OH)₃ + 18H⁺ + N₂    E°cell = -2.8V

E°cell = E°red + E°ox