Question

A dilute solution of sulfuric acid is electroylzed. Oxygen gas forms at the anode and hydrogen gas forms at the cathode. (The overall reaction is just the decomposition of water back to the elements). If the electrolysis is done at 4.80 volts and a current of 1.44 amps for 7.50 minutes, how many liters of oxygen gas (measured at STP) will be produced at the anode?

Answer 1

0.038 L

Step-by-step explanation:

During the electrolysis of dilute sulphuric acid; the half reactions are as follows;

Cathode;

4H^+(aq) + 4e ----> 2H2(g)

Anode:

4OH^-(aq) ----> 2H2O(l) + O2(g) + 4e

Overall reaction; 4H2O(l)------> 4H^+(aq) + 4OH^-(aq)

Hence 4 electrons are transferred in the process.

From the reaction equation;

4F liberates 22.4 dm^3 of oxygen

But 1F =96500C

Therefore

4(96500)C liberates 22.4 L of oxygen

Thus (1.44×7.50×60) will liberate 1.44×7.50×60 × 22.4/4(96500)

Volume of oxygen liberated= 14515.2 / 386000

Volume of oxygen liberated= 0.038 L