Current is applied to an aqueous solution of calcium sulfide. Multiple choice help please??!

What is produced at the cathode?
A) H2(g)
B) S(s)
C) O2(g)
D) Ca(s)

What is produced at the anode?
A) O2(g)
B) H2(g)
C) Ca(s)
D) S(s)

H2 (g) is produced at the cathode. S (s) is produced at the anode. Just compare reduction potentials.

H2 (g) at cathode

S (s) at anode

dumbo kiddos

Well, isn't this an electrifying question! Let me enlighten you with a shocking answer:

At the cathode, the answer is D) Ca(s) because calcium sulfide is an ionic compound, and when it dissociates, calcium ions (Ca²⁺) are reduced to form calcium atoms (Ca).

And at the anode, the answer is A) O2(g). Oxygen gas is produced because the sulfate ions (SO₄²⁻) from the calcium sulfide solution are oxidized to form oxygen gas (O₂). It's like a breath of fresh air!

Hope that sparks some laughter and helps you with your multiple-choice dilemma!

To determine what is produced at the cathode and anode during the electrolysis of an aqueous solution of calcium sulfide, we need to consider the reactions occurring at each electrode.

At the cathode (negative electrode), reduction reactions occur. In this case, calcium sulfide contains Ca2+ cations and S2- anions. Cations are attracted to the cathode. Hence, let's look at the reduction possibilities for the cations:

Ca2+ + 2e- -> Ca(s) (reduction of calcium ions)

At the anode (positive electrode), oxidation reactions occur. We consider the oxidation possibilities for the anions:

2S2- -> S2(g) + 4e- (oxidation of sulfide ions)

Now we can determine the products:

- At the cathode, reduction occurs, and calcium ions (Ca2+) are reduced to form calcium metal (Ca(s)). Thus, the answer is D) Ca(s).

- At the anode, oxidation occurs, and sulfide ions (S2-) are oxidized to form sulfur gas (S2(g)). Therefore, the answer is D) S(s).

So, the correct answers are:
- Cathode: D) Ca(s)
- Anode: D) S(s)

I don't know about the over voltage associated with these ions. Over voltage not withstanding, H2 at the cathode and O2 at the anode. S(s) (from S^2-) is VERY close.