The pH can be calculated from molar concentration of hydronium ion, H 3 O + , or

Question

The pH can be calculated from molar concentration of hydronium ion, H3O+, or indirectly from molar concentration of hydroxide ion, OH-.

pH = -log[H3O+]

pH = 14 - pOH

pOH = -log[OH-]

Use stoichiometry and calculate the pH in the anode compartment after 5.00 minutes with a current of 0.0250amps, assuming a final solution volume of 80.0mL and a temperature of 20.0oC.

Calculate the pH in the cathode compartment after 5.00 minutes with a current of 0.0250amps, assuming a final solution volume of 80.0mL and a temperature of 20.0oC.

Explanation / Answer

Anode: 6H2O ------> 4H3O+ + O2 + 4e-

Cathode: 4H2O + 4e- -----> 2H2 + 4OH-

For 4 moles of electrons involved moles of H3O+ in anode = 4

and moles of OH- in cathode = 4

Total charge passed = 0.025 x 5 x 60 = 7.5 C

Charge of 1 e- = 1.602 x 10^-19 C

=> Number of e- involved = 7.5 / 1.602 x 10^-19 = 4.68 x 10^19

=> Moles of e- = 4.68 x 10^19 / 6.022 x 10^23 = 7.77 x 10^-5

=> Moles of H3O+ in anode = 7.77 x 10^-5

and Moles of OH- in cathode = 7.77 x 10^-5

Volume = 80 mL = 0.08 L

=> [H3O+] in anode = 7.77 x 10^-5 / 0.08 = 9.72 x 10^-4 M

[OH-] in cathode = 7.77 x 10^-5 / 0.08 = 9.72 x 10^-4 M

pH at anode = - log (9.72 x 10^-4) = 3.01

pOH at cathode = - log (9.72 x 10^-4) = 3.01

=> pH at cathode = 14 - 3.01 = 10.99

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