Question

a) Calculate the potential in the following cell (no current is flowing)

Zn|Zn²⁺(1.00M)|| UO2 ²⁺(1.00M), U⁴⁺ (0.1M), H3O ⁺ (1.00M) |Pt

b) The cathode has a volume of 50 mL. What is the cell potential (no current) after adding 10 mL, 25 mL, 50 mL, and 100 mL 1M NaOH to the cathode?

c) What is the Gibbs Free Energy for the 5 cells in a) and b). What is the spontaneous reaction in each case?

Answer 1

We will use the Nernst equation,

Ecell = E^ocell - 0.059/n log K

where, K = equlibrium constant = [products]/[reactants]

E^ocell = Ecathode - Eanode

a) E^ocell = Ecathode - Eanode

              = 0.2682 - (-0.7621)

              = 1.03 V

Use Nernst equation,

Ecell = E^ocell - 0.059/n log Q

Q = [Zn2+][U4+]/[Zn][UO2^2+][H+]

Fed the values we have,

Ecell = 1.03 - 0.059/2[log(0.1/1.0)]

         = 1.06 V

b) Now when 10 mL of NaOH is added

we have, moles of NaOH = M x L = 1 x 0.01 = 0.01 moles

Initial moles of H+ = 1 x 0.05 = 0.05 moles

Remaining moles of H+ = 0.05 - 0.01 = 0.04 moles

Molarity of H+ = 0.04/(0.05 + 0.01) = 0.67 M

Feed this value in Nernst equation,

Ecell = 1.03 - 0.059/2[log(0.1/0.67)]

         = 1.05 V

Similarly when, 25 mL NaOH is added.

we have, moles of NaOH = M x L = 1 x 0.025 = 0.025 moles

Initial moles of H+ = 1 x 0.05 = 0.05 moles

Remaining moles of H+ = 0.05 - 0.025 = 0.025 moles

Molarity of H+ = 0.025/(0.05 + 0.01) = 0.42 M

Feed this value in Nernst equation,

Ecell = 1.03 - 0.059/2[log(0.1/0.42)]

         = 1.048 V

Similarly, when 50 mL of NaOH is added,          

we have, moles of NaOH = M x L = 1 x 0.05 = 0.05 moles

Initial moles of H+ = 1 x 0.05 = 0.05 moles

Remaining moles of H+ = 0.05 - 0.05 = 0.0 moles

Feed this value in Nernst equation,

Ecell = 1.03 V

Now, when 100 mL of NaOH is added, we have 50 mL of NaOH excess,

The reaction would be non-spontaneous

c) Free energy is calculated as,

i) delta G = -nFE

            = -(2 x 96485 x 1.06) = -204.55 kJ spontaneous reaction

ii) delta G = -(2 x 96485 x 1.05) = -202.62 kJ spontaneous reaction

iii) delta G = -(2 x 96485 x 1.048) = -202.23 kJ spontaneous reaction

iv) delta G = -(2 x 96485 x 1.03) = -198.76 kJ At equlibrium

iv) delta G = non-spontaneous reaction