soluition charpter 2

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EMA 4324 Problem Set 3a
2.3 Assuming standard states for all reactants and products,
determine the spontaneous direction of the following reactions
by calculating the cell potential (sic):
[a] Cu + 2HCl = CuCl2 + H2 [Cu + 2H+ = Cu+2 + H2]
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Jones is really asking for a calculation of cell voltage!!
[1] Cu+2 + 2e-= Cu, Eo(Cu+2/Cu) = 0.342 v;
[2] 2H+ + 2e-= H2, Eo(H+/H2) = 0.000 v
Reaction [a] proceeds spontaneously in reverse to what is written...
Vcell = 0.342v, reaction [1] proceeds as written [reduction]; reaction
[2] proceeds in reverse [oxidation] the reaction [written as a
reduction reaction] with the more positive potential will proceed as
the reduction reaction in a cell with the less positive [more
negative] reduction reaction proceeding as an oxidation [reduction
in reverse].
[b] Fe + 2HCl = FeCl2 + H2; Fe + 2H+= Fe+2 + H2
Fe+2 + 2e- = Fe, Eo(Fe+2/Fe) = -0.440v [by insp., oxidation] 
2H+ + 2e- = H, Eo(H+/H2) = 0.00 [by inspection, reduction]
Vcell = 0.000 - [-0.440] = 0.440 and Fe + 2H+ = Fe+2 + H2
[c] 2AgNO3 + Fe = Fe(NO3)2 + 2Ag
Ag+ + 1e- = Ag, Eo(Ag+/Ag) = 0.799v
Fe+2 + 2e- = Fe, Eo(Fe+2/Fe) = -0.440
and Fe + 2Ag+ = Fe+2 + 2Ag, Ecell = 0.799 - (-0.440) = 1.239v
[d] Ag + FeCl3 = FeCl2 + AgCl
Ag+ + 1e- = Ag, Eo(Ag+/Ag) = 0.799v
Fe+3+ 1e- = Fe+2, Eo(Fe+3/Fe+2) = 0.771v
Fe+2 + Ag+ = Fe+3 + Ag, Ecell = 0.799 - 0.771 = 0.028v
[e] 2Al + 3ZnSO4 = Al2(SO4)3 + 3Zn
Zn+2 + 2e- = Zn, Eo(Zn+2/Zn) = -0.763v
Al+3 + 3e- = Al, Eo(Al+3/Al) = -1.662v
2Al + 3Zn+2 = 2Al+3 + 3Zn, Ecell = -0.763 - (-1.662) = 0.899v
2.5 Write the electrochemical half-cell reactions for oxidation and
reduction during uniform corrosion in the following: 
[a] aluminum in air-free sulfuric acid
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Al+3 + 3e- = Al [-1.662 v, oxidation]; 
H+ + 1e- = 1/2H2[0.00 v, reduction]
[b] iron in air-free ferric sulfate solution
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Fe+2 + 2e- = Fe [-0.440, oxidation]
Fe+3 + 1e- = Fe+2 [+0.771, reduction]
SO4-2 + 6e- +8H+ = S + 4H20 [+0.357, depends.....]
[c] carbon steel in aerated seawater [careful here...]
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Fe+2 + 2e- = Fe [-0.440, oxidation]
O2 + 2H2O + 4e- = 4OH- [0.401, reduction]
H+ + 1e- = 1/2H2 [0.00 v, depends....]
Others???
[d] zinc-tin alloy in an oxygen saturated solution of cupric chloride,
stannic chloride and hydrochloric acid
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Zn+2 + 2e- = Zn [-0,763, oxidation]
O2 + 2H2O + 4e- = 4OH- [0.401, reduction]
Cu+2 + 2e- = Cu [+0.337, probable reduction]
Cu+2 + 1e- = Cu+1 [+0.153, depends...]
Sn+4 + 2e- = Sn+2 [+0.15, depends]
[e] copper in deaerated seawater [again, be careful...]
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Cu+2 + 2e- = Cu [+0.337, reduction!!!]
H+ + 1e- = 1/2H2 [0.00 v, oxidation!!!]
2.6 Copper immersed in concentrated pure hydrochloric acid is observed
to corrode rapidly with the evolution of gas bubbles. Is this contrary
to the results of problem 2.3a? Calculate a cell potential [sic] to
prove your answer, making any necessary assumptions.
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According to the chemical [CRC] handbook, the concentration of 
“ pure” HCl is 12.0 M. Assuming unit activity coefficients [poor
assumption], pH = -log10[12] = -1.08. Since E(H+/H2) = -0.0592pH,
E(H+/H2) = E(Cu+2/Cu) = 0.064 volts.
E(Cu+2/Cu) = 0.064 = 0.334 - [0.0592/2]log[1/[Cu+2]]
[Cu+2] = 7.56x10-10M!!!!!