Determine E o for the following reaction, using the given standard reduction pot

Question

Determine Eo for the following reaction, using the given standard reduction potentials:
Cu2+(aq) + Ti(s) ? Cu(s) + Ti2+(aq)
Eo for Ti2+(aq) = -1.63 V
Eo for Cu2+(aq) = 0.34 V

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Determine Eo for the following reaction, using the given standard reduction potentials:
Pb2+(aq) + Hg(l) ? Pb(s) + Hg2+(aq)
Eo for Hg2+(aq) = 0.85 V
Eo for Pb2+(aq) = -0.13 V

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Determine Eo for the following reaction, using the given standard reduction potentials:
Ti2+(aq) + Mn(s) ? Ti(s) + Mn2+(aq)
Eo for Mn2+(aq) = -1.18 V
Eo for Ti2+(aq) = -1.63 V

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Consider the following half-reactions. What is the maximum Eo of a cell constructed using two of these half-equations.

I2(s) + 2 e- ? 2 I-(aq) Eo = 0.53 V
S2O82-(aq) + 2 e- ? 2 SO42-(aq) Eo = 2.01 V
Cr2O72-(aq) + 14 H+ + 6 e- ?Cr3+(aq) + 7 H2O(l) Eo = 1.33 V

Explanation / Answer

E cell = Eo cathode - Eo anode

Ecell = 0.34+1.63 = 1.97 V

2.

Ecell = -0.13- 0.85 = -0.98 V

3.
Ecell= =1.63 +1.18 = -0.45 V

4. max E cell will be for the cell which has max Eo for cathode and min Eo for anode,
so , max Eo cell will be = 2.01 - 0.53 = 1.48 V

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