1. Calculate the standard free-energy change (in kilojoules) at 25 °C for the re

ID: 961945 • Letter: 1

Question

1. Calculate the standard free-energy change (in kilojoules) at 25 °C for the reaction:

O2(g) + 4H+(aq) + 4Cl-(aq) ---> 2H2O(l) + 2Cl2(g)

a. -12.5 kJ b. -50.2 kJ c. -25.1 kJ d. +12.5 kJ e. +50.2 kJ

2. At 25oC, calculate the equilibrium constant for the reaction:

Sn(s) + 2 H+ (aq) à Sn2+ (aq) + H2(g).

a. 5.4 x 104 b. 2.3 x 102 c. 1.1 x 102 d. 1.0 x 10-3 e. 1.8 x 10-5

3. What is the correct shorthand notation for the voltaic cell that utilizes the reaction below.

Fe(s) + Cu2+ (aq) --> Fe2+(aq) + Cu(s)

a. Fe(s) / Fe2+(aq) // Cu2+(aq) / Cu(s)

b. Cu(s) / Cu2+(aq) // Fe2+(aq) / Fe(s)

c. Cu2+(aq) / Cu(s) // Fe2+(aq) / Fe(s)

d. Fe2+(aq) / Fe(s) // Cu2+(aq) / Cu(s)

e. Fe2+(aq) / Fe(s) // Cu(s)/ Cu2+(aq)

4. The oxidation half-reaction for the standard hydrogen electrode is:

a. H2 (g) 2H+ (aq) + 2e-

b. 2H+ (aq) + 2OH- (aq) H2O (l)

c. Pt2+(aq) + 2e- à Pt(s)

d. 2H+(aq) + 2e- H2 (g)

e. 2H+ + Cl2 (g) 2HCl (aq)

1. Ecell = Eanode - Ecathode = 1.358 - 1.229 = 0.129 V

dG = -nFEcell

= -4 x 96485 x 0.129

= -50.2 kJ

So the answer for free energy is,

b. -50.2 kJ

2. Ecell = Eanode - Ecathode = -0.14

-nFEcell = -RTlnK

-2 x 96485 x -0.14 = -8.314 x 298 ln K

K = 1.8 x 10^-5

So the equilibrium constant for the reaction is,

e. 1.8 x 10^-5

3. The correct shorthand notation for the cell would be,

a. Fe(s)|Fe2+(aq)||Cu2+(aq)|Cu(s)

4. Th oxidation half reaction for the standard hydrogen electrode is:

a. H2(g) ---> 2H+(aq) + 2e-

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