If one has the relevant half reaction, then the moles of any product can be dete

Question

If one has the relevant half reaction, then the moles of any product can be determined from stoichiometrically from the moles of electrons involved.

Also, the charge put into the chemical reaction, assuming 100% efficiency, is easily determined by:

charge = current x time                      or                    q = It

Calculate the volume of gas produced at the anode after 5.00 minutes of electrolysis with the current set to 0.0250amps (0.0250C/s). Assume a temperature of 20.0oC and atmospheric pressure of 1atm.

Calculate the volume of gas produced at the cathode after 15.00 minutes of electrolysis with the current set to 0.0250amps (0.0250C/s). Assume a temperature of 20.0oC and atmospheric pressure of 1atm.

Explanation / Answer

Anode: 6H2O -----> O2 (g) + 4H3O+ (aq) + 4e-

Cathode: 4H2O + 4e- -----> 2H2 (g) + 4OH- (aq)

4 moles of electrons involved produces 1 mole of O2 at anode and 2 moles of H2 at cathode

Charge = 0.025 x 5 x 60 = 7.5 C

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

=> Number of e- = 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

=> Moles of Gas (O2) produces at anode = 7.77 x 10^-5 / 4 = 1.94 x 10^-5 moles

Moles of Gas (H2) produces at cathode = 7.77 x 10^-5 / 2 = 3.89 x 10^-5 moles

We know that,

PV = nRT

=> 1 x Va = 1.94 x 10^-5 x 0.0821 x 293

=> Va = 4.67 x 10^-4 L gas is produced at anode

Since the current passes in the cathode is for 15 (5 x 3) minutes,

Moles of Gas (H2) produces at cathode = 3 x 3.89 x 10^-5 = 1.167 x 10^-4 moles

1 x Vc = 1.167 x 10^-4 x 0.0821 x 293

=> Vc = 2.81 x 10^-3 L gas produced at cathode

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