Question

Zn|Zn2+(1.00M)|| UO22+(1.00M), U4+(0.1M), H3O+(1.00M) |Pt

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

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

a.

UO₂²⁺ + 4H⁺ + 2e^- = U⁴⁺ + 2H₂O                          E^o = + 0.334 V

Zn²⁺ + 2e^- = Zn                                                     E^o = - 0.760 V

The spontaneous reaction is

Zn + UO₂²⁺ + 4H⁺  = Zn²⁺ + U⁴⁺ + 2H₂O            n=2

E^o_cell = 0.334 – (- 0.760) = 1.094 V    (use standard reduction potentials)

E_cell = E^o_cell – (0.0592 V /2) log Q     

where Q = (reaction quotient) is

Q = [Zn²⁺ ] [ U⁴⁺ ] / ( [UO₂²⁺][ H⁺]⁴ )

     = 1x0.1/(1x1⁴) = 0.1

logQ = -1

E_cell = 1.094 V + 0.0296 V = 1.124 V

b.

Q = [Zn²⁺ ] [ U⁴⁺ ] / ( [UO₂²⁺][ H⁺]⁴ ) = 0.1/[ H⁺]⁴         

logQ = -1 + 4·pH

E_cell = E^o_cell – (0.0592 V /2) log Q

E_cell = 1.094 V - 0.0296 V (4·pH -1)

Calculate pH in the cathode after each addition.

Initial [H+] = 1 M , pH = 0    The half cell contains 0.050 L x 0.1 mol/L = 0.005 mol H⁺

Addition, mL NaOH	[H+], mol/L	pH	Ecell = 1.094 V - 0.0296 V (4·pH -1)
0	1	0	1.094 V
10	0.66	0.180	…
25	0.33	0.481	…
50	1.10-7	7.00	….
100	[HO-] = 0.5 [H+] = 2x10-14	13.7	…

Note: [H⁺] is corrected for dilution

Calculate E_cell for each line.

c.

Use the general equation:

deltaG = - n F E_cell    (use the E_cell values calculated at a and b.)

Ex:

dG = - (2 mol) x 96458 C/mol x 1.094 V = 211.0 kJ