a) Consider the following reaction at 298K. 2 Cu 2+ (aq) + Hg (l) ------> 2 Cu +

Question

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?

Explanation / Answer

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-

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