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Please answer all. 1)Enter the balanced chemical equation including states that

ID: 1003093 • Letter: P

Question

Please answer all.

1)Enter the balanced chemical equation including states that describes the electrochemical cell that is represented by the cell notation as shown.

Pt(s) l Fe2+(aq), Fe3+(aq) ll Hg2+(aq) l Hg(l)

St. Red. Pot. (V)

2)Enter the balanced chemical equation including states that describes the electrochemical cell that is represented by the cell notation as shown.

Pt(s) l Fe2+(aq), Fe3+(aq) ll Hg2+(aq) l Hg(l)

3)Enter the balanced chemical equation including states that describes the electrochemical cell that is represented by the cell notation as shown.

Pt(s) l SO42-(aq), S2O82-(aq) ll Br2(l) l Br-(aq) l Pt(s)

St. Red. Pot. (V)

4)Enter the balanced chemical equation including states that describes the electrochemical cell that is represented by the cell notation as shown.

Pt(s) l Pb2+(aq) l PbO2(s) l H+(aq) ll H+(aq) l MnO2(s) l Mn2+(aq) l Pt(s)

St. Red. Pot. (V)

Fe3+/Fe2+ +0.77 Hg2+/Hg +0.85

Explanation / Answer

1.Pt(s) l Fe2+(aq), Fe3+(aq) ll Hg2+(aq) l Hg(l)

Fe+2   ---------> Fe+3    +e-       E0 = -0.77v

Hg+2 + 2e- -------> Hg             E0 = 0.85V

2Fe+2   ---------> 2Fe+3    +2e-       E0 = -0.77v

Hg+2 + 2e- -------> Hg             E0 = 0.85V

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2Fe+2 + Hg+2 -------> 2Fe+3 + Hg    Ecell = 0.08V

2.

Fe---------> Fe+3 +3e-     E0 = 0.04v

I2 + 2e- -------> 2I-             E0 = 0.54V

2Fe---------> 2Fe+3 +6e-     E0 = 0.04v

3I2 + 6e- -------> 6I-             E0 = 0.54V

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2Fe + 3I2--------> 2Fe+3 + 6I-    Ecell = 0.58V

3. Pt(s) l SO42-(aq), S2O82-(aq) ll Br2(l) l Br-(aq) l Pt(s)

2SO42-    --------> S2O82- + 2e-   E0 = -2.01V

Br2 + 2e- -------> 2Br-                         E0 = 1.07V

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2So42- + Br2 ---------> S2O82- + 2Br- E0 = -0.94v

4.Pt(s) l Pb2+(aq) l PbO2(s) l H+(aq) ll H+(aq) l MnO2(s) l Mn2+(aq) l Pt(s)

Pb+2 + 2H2O---------> PbO2 +4H+ +2e-    E0 = -1.46v

MnO2 + 4H+ + 2e- ---------> Mn+2 + 2H2O   E0 = 1.21v

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Pb+2 + MnO2 --------> PbO2 + Mn+2          Ecell = -0.25v