a) I 2 (s) + Cu (s) ----> 2 I - (aq) + Cu 2+ (aq) Hint: Carry at least 5 signifi
ID: 635881 • Letter: A
Question
a)
I2(s) + Cu(s) ----> 2I-(aq) + Cu2+(aq)
Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. You may use the OWL references to find the values for physical constants.
Equilibrium constant: ?
delta G° for this reaction would be greater/less than zero
b)
2Ag+(aq) + Ni(s) -----> 2Ag(s) + Ni2+(aq)
Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. You may use the OWL references to find the values for physical constants.
Equilibrium constant: ?
delta G° for this reaction would be greater/less than zero
c)
What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the F2 pressure is 1.17 atm, the F- concentration is 3.99×10-3M, and the Hg2+ concentration is 4.67×10-4M ?
F2(g) + Hg(l)---->2F-(aq) + Hg2+(aq)
Answer: ? V
The cell reaction as written above is spontaneous for the concentrations given: TRUE/FALSE
Explanation / Answer
Part a
a voltaic cell
spontaneous cell reaction
larger voltage at cathode
smaller voltage at anode.
At anode oxidation reaction
Al (s) - ---> Al3+ (aq) + 3 e-
Eox = 1.660 V
At cathode reduction reaction
3 Ag+ (aq) + 3 e - ---> 3 Ag (s)
Ered = 0.799 V
Spontaneous cell reaction
3 Ag+ (aq) + Al (s) ----> 3 Ag (s) + Al3+ (aq)
Standard reduction potential
E° (cell) = Eox + Ered
= 1.66 + 0.799
= 2.459 V
Part b
I2(s) + Cu(s) ----> 2I-(aq) + Cu2+(aq)
Oxidation reaction at anode
Cu(s) ----> Cu2+(aq) + 2e-
Eox = - 0.34 V
Reduction reaction at cathode
I2(s) + 2e- ----> 2I-(aq)
Ered = 0.54 V
Standard reduction potential
E°cell = Eox + Ered
= - 0.34 + 0.54
= 0.20 V
E°cell > 0
G° = - nFE°cell
= - 2 mol x 96500 C/mol x 0.20 V
= - 38600 J/mol
G° < 0
G° = - RT ln K
K = exp (38600/8.314 x 298) = 5.837 x 10^6
Part c
2Ag+(aq) + Ni(s) -----> 2Ag(s) + Ni2+(aq)
Oxidation reaction at anode
Ni(s) ----> Ni2+(aq) + 2e-
Eox = 0.26 V
Reduction reaction at cathode
2Ag+(aq) + 2e- ----> 2Ag(s)
Ered = 0.80 V
Standard reduction potential
E°cell = Eox + Ered
= 0.26 + 0.80
= 1.06 V
E°cell > 0
G° = - nFE°cell
= - 2 mol x 96500 C/mol x 1.06 V
= - 204580 J/mol
G° < 0
G° = - RT ln K
K = exp (204580/8.314 x 298) = 7.259 x 10^35
Part d
F2(g) + Hg(l)---->2F-(aq) + Hg2+(aq)
Oxidation reaction
Hg(l)----> Hg2+(aq) + 2e-
Eox = - 0.855 V
Reduction reaction
F2(g) + 2e- ---->2F-(aq)
Ered = 2.87 V
Standard reduction potential
E°cell = Eox + Ered
= - 0.855 + 2.87
= 2.015 V
From the Nernst equation
E = E°cell - (0.0592/2) log[F-]2 [Hg2+] / [PF2]
= 2.015 - (0.0592/2) log[3.99*10^-3]2 (4.67*10^-4)/ [1.17]
= 2.258 V
E°cell > 0
G° < 0
The cell reaction is spontaneous - True
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