Question

1. Which of the following occurred during the electrolysis of aqueous potassium sulfate? One or more answers may be correct. You will receive negative points for incorrect answers.

brown color disappears at the other electrode

twice as much gas was formed at one electrode that the other

a brown color formed at one electrode

copper was plated onto one of the electrodes

the indicator on one side turned yellow and the other side turned blue

the indicator turned pink at one electrode

gas bubbles at both platinum electrodes

gas bubbles were visible only at one electrode

2. During electrolysis of an aqueous solution of copper sulfate, what products are produced at the anode (both electrodes are stainless)? One or more answers may be correct. You will receive negative points for incorrect answers.

oxygen gas

electrons

Cu²⁺

OH^-

H₃O⁺

K⁺

3.Which of the following occurred during the electrolysis of aqueous potassium sulfate? One or more answers may be correct. You will receive negative points for incorrect answers.

brown color disappears at the other electrode

twice as much gas was formed at one electrode that the other

a brown color formed at one electrode

copper was plated onto one of the electrodes

the indicator on one side turned yellow and the other side turned blue

the indicator turned pink at one electrode

gas bubbles at both platinum electrodes

gas bubbles were visible only at one electrode

Answer 1

During electrolysis of electrolysis of aqueous potassium sulfate solution:

H⁺ (H2O) and K⁺ compete at cathode for reduction, out of which  H⁺ (H2O) reduces to Hydrogen gas.

water and sulfate ion compete at anode, out of which water is oxidized to form oxygen gas.

Since gas is formed at both electrodes, gas bubbles also form at both platinum electrodes (Answer)

Due to the release of H2(g) at cathode it becomes more basic, and anode becomes acidic.

Hence the indicator on one side turned yellow and the other side turned blue (Answer)

The reaction at cathode is:

4H⁺ + 4e^- ----> 2H₂(g)

The reaction at anode is:

2H₂O(l) -----> 4H⁺(aq) + O_2(g) +  4e^-

Since 2H₂(g) at cathode and 1 O₂(g) formed at anode, twice as much gas was formed at one electrode that the other (Answer)

Hence the correct options are:

twice as much gas was formed at one electrode that the other

the indicator on one side turned yellow and the other side turned blue

gas bubbles at both platinum electrodes

#2: The chemical reaction at anode is:

2H₂O(l) -----> 4H⁺(aq) + O_2(g) +  4e^-   

Oxygen gas, electron and H₃O⁺ are formed at anode. (Answer)