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 +0.85Explanation / 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