E °cell combing half reactions

Question

Use the following half-reactions to write three spontaneous reactions.

Calculate E °cell for each reaction, and rank the strengths of the oxidizing and reducing agents.

(1) Al³⁺(aq) + 3e^- → Al (s) E° cell = -1.66 V

(2) N₂O (g) + 2e^- → 2NO₂‾ (aq) E °cell = 0.867 V

(3) SO₄^2- (aq) + H₂O (l) + 2e- → SO₃^2- (aq) + 2OH (aq) E °cell = 0.93 V

Answer 1

For the reaction to be spontaneous, there has to be a positive (+) Eº. That is, the difference between the reduction and oxidation reaction must be positive. In equation (2) of the question, it is written as a reduction reaction (electrons on the left), however, N is being oxidized going from N₂O to NO₂^-. So there is something wrong with the equation. I will assume the standard reduction reaction is

2NO₂^- + 6H⁺ + 4e- => N₂O(g) + 3H₂O

(1). SO₄^2- (aq) + H₂O (l) + 2e- → SO₃^2- (aq) + 2OH^- (aq) E °cell = 0.93 V (cathode, reduction)

Al³⁺(aq) + 3e^- → Al (s) E° cell = -1.66 V (anode, oxidation)

Balanced redox rxn: 3SO₄^2- (aq) + 3H₂O (l) + 2Al(s) → 3SO₃^2- (aq) + 6OH^- (aq) + 2Al³⁺(aq)

Eº = 0.93 V + 1.66 V = +2.59 V

(2). SO₄^2- (aq) + H₂O (l) + 2e- → SO₃^2- (aq) + 2OH (aq) E °cell = 0.93 V (cathode, reduction)

2NO₂^- + 6H⁺ + 4e- => N₂O(g) + 3H₂O E °cell = 0.867 V (anode, oxidation)

Balanced redox rxn: 2SO₄^2- (aq) + N₂O(g) + H₂O → 2SO₃^2- (aq) + 2NO₂^- + 2H⁺

Eº = 0.93 V - 0.867 V = +0.063 V

(3). Al³⁺(aq) + 3e^- → Al (s) E° cell = -1.66 V

2NO₂^- + 6H⁺ + 4e- => N₂O(g) + 3H₂O E °cell = 0.867 V

Follow the same procedure as above using 2NO₂^- + 6H⁺ + 4e- => N₂O(g) + 3H₂O as the cathode (reduction) . Eº should be +2.53V

Oxidizing strength: Al³⁺ > NO₂^- > SO₄^2-

Reduction strength: SO₄^2- > NO₂^- > Al₃⁺