In: Other
a)
Consider the following reaction at 298K.
2
Cu2+(aq) +
Hg (l)
------>2
Cu+(aq) +
Hg2+(aq)
Which of the following statements are correct?
Choose all that apply.
n = 1 mol electron
The reaction is product-favored.
delta Go < 0
Eocell < 0
K < 1
b)
Consider the following reaction at 298K.
3
Hg2+(aq) +
2 Cr (s)
------> 3 Hg
(l) + 2
Cr3+(aq)
Which of the following statements are correct?
Choose all that apply.
K > 1
n = 2 mol electrons
The reaction is reactant-favored.
Eocell > 0
delta Go > 0
c)
What is the calculated value of the cell potential at 298K for
an electrochemical cell with the following reaction, when the
H2 pressure is
6.47×10-3 atm, the
H+ concentration is
1.43M, and the Co2+
concentration is 5.94×10-4M ?
2H+(aq)
+ Co(s)------>
H2(g) +
Co2+(aq)
Answer: ? V
The cell reaction as written above is spontaneous for the
concentrations given.. t/f
d)
When the following equation is balanced properly under
acidic conditions, what are the coefficients of
the species shown?
Pb2++ Cu+
-----> Pb
+ Cu2+
Water appears in the balanced equation as
a (reactant, product, neither) with a
coefficient of ?. (Enter 0 for neither.)
How many electrons are transferred in this reaction?
Part a
2 Cu2+(aq) + Hg (l) ------>2 Cu+(aq)+ Hg2+(aq)
Oxidation reaction
Hg (l) ------>Hg2+ + 2e-
Eox = - 0.855 V
Reduction reaction
2 Cu2+(aq) + 2e- ------>2 Cu+(aq)
Ered = 0.15 V
Standard reduction potential
E°cell = Eox + Ered
= - 0.855 + 0.15
= - 0.705 V
E° < 0
n = 2 mol e-
G° = - nFE°cell
= - 2 x 96500 x (-0.705)
= 136065 J/mol
G° > 0
K = exp (-G°/RT) = exp ( - 136065/8.314*298) = 1.4*10^-24
K < 1
The reaction is reactant favored
Part b
3 Hg2+(aq) + 2 Cr (s) ------> 3 Hg (l) + 2 Cr3+(aq)
Oxidation reaction
2 Cr (s) ------> 2 Cr3+(aq) + 6e-
Eox = 0.73 V
Reduction reaction
3 Hg2+(aq) + 6e- -----> 3 Hg (l)
Ered = 0.855 V
Standard reduction potential
E°cell = Eox + Ered
= 0.73 + 0.855
= 1.585 V
E° > 0
n = 6 mol e-
G° = - nFE°cell
= - 6 x 96500 x (1.585)
= - 917715 J/mol
G° < 0
K = exp (-G°/RT) = exp ( 917715/8.314*298) = 7.3*10^160
K > 1
The reaction is product favored
Part c
2H+(aq) + Co(s)------> H2(g) + Co2+(aq)
Oxidation reaction
Co(s)------> Co2+(aq) + 2e-
Eox = 0.28 V
Reduction reaction
2H+(aq) + 2e- ------> H2(g)
Ered = 0 V
Standard reduction potential
E°cell = 0.28 V
From the Nernst equation
E = E°cell - (0.0592/2) log[Co2+](PH2) / [H+]2
= 0.28 - (0.0592/2) log[5.94*10^-4] (6.47*10^-3)/ [1.43]2
= 0.45 V
Part d
Pb2+ + Cu+ -----> Pb + Cu2+
Oxidation reaction
2Cu+ = 2Cu2+ + 2e-
Reduction reaction
Pb2+ + 2e- = Pb
The balanced reaction
2Cu+ + Pb2+ ? 2Cu2+ + Pb
Water does not appear with 0 coefficient
2 mol e-