Question

3. A 265 mL solution of NaCl was electrolyzed for 6.90 min. If the pH of the final solution was 12.02, calculate the average current used? = __________________A

4. What is the standard emf of a galvanic cell made of a Pb electrode in a 1.0 M Pb(NO3)2 solution and a Al electrode in a 1.0 M Al(NO3)3 solution at 25C?

Ecell= _____________________V

Answer 1

3.

NaCl --> Na+ + Cl -

So the electrolytic reactions will be:

2Cl- => Cl2 + 2e

2H2O + 2e- => H₂ + 2OH^-

From each mole of Cl- electrolysed one mole of OH- is produced

As pH = 12.02

so pOH = 1.98

pOH= -log[OH-] = 1.98

so [OH-] = 0.0104 M

So [Cl-] = 0.0104 M

so concentration of NaCl = .0104 M

Volume = 265 mL

so moles of NaCl electrolysed = 0.0104 X 265 / 1000 = 0.00275 moles

For one gram equivalent of eletrolyte electrolyzed the charge will be 96,485 Coulombs

Charge reletaed to 0.00275 moles or gm equivalent = 96,485 X 0.00275 C = 265.33 Coloumbs

Current = charge / time = 265.33 Coloumbs / 6.9 X 60 seconds = 0.6409 Amperes