The apparatus depicted to the right is often used to demonstrate the electrolysis of water. Tubes A — Electrochemistry Chemistry Question
Problem Context
The apparatus depicted to the right is often used to demonstrate the electrolysis of water. Tubes A and B are initially filled with an aqueous solution of H2SO4 or Na2SO4.
Describe the purpose of adding the H2SO4 or Na2SO4 rather than using pure water.
Model Answer
Because pure water is a poor conductor of electricity, the H2SO4 or Na2SO4 is added to provide electrolyte (so that the solution will conduct).
Give the formula of the gas produced in;
i. tube A
Model Answer
i) tube A is the cathode, therefore it is the site of reduction where H2 is produced
ii. tube B
Model Answer
while (ii) tube B is the anode, where oxidation occurs, therefore O2 is produced.
Describe a chemical test that could be used to identify the gas collected in tube A. Include the procedure and expected observation.
Model Answer
Because H2 is flammable, a burning splint can be inserted into the products from Tube A. If there is a “pop” associated with the reaction, it confirms that the gas is H2.
Calculate the number of moles of gas expected to be collected in tube B when a 600. milliamp current is applied for 40.0 minutes. (Assume no side reactions occur.)
Model Answer
Charge = current × time: 0.600 C·s-1 × 2400 s = 1440 C
and: 1440 C × = 3.73×10–3 mol O2
Calculate the volume of the gas produced in part d. for a temperature is 20 ˚C and a pressure in the laboratory of 735 mmHg. (The vapor pressure of water is 17.5 mmHg.)
Model Answer
First correct for vapor pressure of water:
Ptotal = PO2 + PH2O so PO2 = 735-17.5 = 717.5 mmHg
= 0.095 L = 95 mL
If H2O2 is formed in a side reaction the quantity of only one of the products is affected. Identify the product affected and state how its quantity compares with that produced with no side reaction. Explain your answer.
Model Answer
The quantity of O2 would be affected, but the quantity of H2 would not. The yield of O2 would be decreased because some of the electricity would oxidize H2O into peroxide (H2O2) instead of O2.