Redox reactions

We looked at a few redox reactions today in class, some of you wanted the solutions posted. Enjoy.

For each pair of half cells, write the balanced spontaneous (standard) reaction and calculate the spontaneous (standard) cell voltage. {Actually, you'd do the second part of that process before you could do the first part}

Cu⁺¹|Cu (E⁰_red = +0.521V) and Fe³⁺|Fe²⁺ (E⁰_red = +0.771V)

Cu half must be reversed to be the oxidation half rxn

E⁰_cell = E⁰_red + E⁰_ox = 0.771V + (-0.521V) = +0.250V

Cu(s) ⇄ Cu⁺¹(aq) + 1e^⊖(aq)

1e^⊖ + Fe³⁺(aq) ⇄ Fe²⁺(aq)

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Cu(s) + Fe³⁺(aq) ⇄ Fe²⁺(aq) + Cu⁺¹(aq)

BrO₃^-1|Br₂ (E⁰_red = +1.478V) and Pt²⁺|Pt (E⁰_red = +1.188V)

Pt half must be reverse to be the oxidation half rxn

E⁰_cell = E⁰_red + E⁰_ox = 1.478V + (-1.188V) = +0.290V

5 { Pt(s) ⇄ Pt⁺²(aq) + 2e^⊖(aq) }

12 H^⊕(aq) + 10e^⊖(aq) + 2 BrO₃^-1(aq) ⇄ Br₂(l) + 6 H₂O(l)

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12 H^⊕(aq) + 5 Pt(s) + 2 BrO₃^-1(aq) ⇄ Br₂(l) + 5 Pt⁺²(aq) + 6 H₂O(l)

Cr⁺²|Cr (E⁰_red = -0.403V) and Pb⁺⁴|Pb⁺² (E⁰_red = +1.69V)

Cr half must be reverse to be the oxidation half rxn

E⁰_cell = E⁰_red + E⁰_ox = 1.69V + (0.403V) = +2.09V

Cr(s) ⇄ Cr⁺²(aq) + 2e^⊖(aq)

2e^⊖ + Pb⁺⁴(aq) ⇄ Pb⁺²(aq)

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Cr(s) + Pb⁺⁴(aq) ⇄ Pb²⁺(aq) + Cr⁺²(aq)

TeO₂|Te (E⁰_red = +0.604V) and Se|H₂Se (E⁰_red = -0.115V)

Se half must be reverse to be the oxidation half rxn

E⁰_cell = E⁰_red + E⁰_ox = 0.604V + (0.115V) = +0.719V

2 { H₂Se(aq) ⇄ Se(s) + 2e^⊖(aq) + 2 H^⊕(aq) }

4 H^⊕(aq) + 4e^⊖ + TeO₂(s) ⇄ Te(s) + 2 H₂O(l)

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4 H^⊕(aq) + 2 H₂Se(aq) + TeO₂(s) ⇄ Te(s) + 2 Se(s) + 4 H^⊕(aq) + 2 H₂O(l)

2 H₂Se(aq) + TeO₂(s) ⇄ Te(s) + 2 Se(s) + 2 H₂O(l)

MnO₄^-1|Mn⁺² (E⁰_red = +1.23V) and ClO₄^-1|ClO₃^-1 (E⁰_red = +1.201V)

Cl half must be reverse to be the oxidation half rxn

E⁰_cell = E⁰_red + E⁰_ox = 1.23V + (-1.201V) = +0.03V

5 { H₂O(l) + ClO₃^-1(aq) ⇄ ClO₄^-1(aq) + 2e^⊖(aq) + 2 H^⊕(aq) }

2 { 8 H^⊕(aq) + 5e^⊖ + MnO₄^-1(aq) ⇄ Mn²⁺(aq) + 4 H₂O(l) }

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616 H^⊕(aq) + 5 H₂O(l) + 2 MnO₄^-1(aq) + 5 ClO₃^-1(aq) ⇄ 5 ClO₄^-1(aq) + 2 Mn²⁺(aq) + 38 H₂O(l) + 10 H^⊕(aq)

6 H^⊕(aq) + 5 H₂O(l) + 2 MnO₄^-1(aq) + 5 ClO₃^-1(aq) ⇄ 5 ClO₄^-1(aq) + 2 Mn²⁺(aq) + 3 H₂O(l)