Challenging Redox Problème

[SonhosDaVida]

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So I don't understand why choice B is not correct in Question 23. The answer key says, if choice B were true then tarnish could never occur because the reduction of silver would be carried out by the sulfide anion it binds.

What does this mean? Isn't choice D also having sulfide amount reduce silver?

The answer is D, but I don't understand why it can't be B?

 

[AdaptPrep]

Hi, SonhosDaVida--

Hopefully, this will help:

The original tarnishing of silver is silver (Ag) going to Ag+ in presence of sulfur (sulfur is reduced/oxidizing agent, silver is oxidized/reducing agent). Well, if sulfur oxidizes silver to tarnish it, how can it then turn around and reduce it back to Ag (which is happening in choice B)? To reduce silver from Ag+ back to Ag, we need an element that is a reducing agent (remember, sulfur was the oxidizing agent in the original tarnish process). Sulfur cannot be both an oxidizing and reducing agent in the presence of silver.

Good luck studying!

 

[SonhosDaVida]

@AdaptPrep Thanks for your explanation, but I am still a bit confused. So the tarnishing of a metal involves turning the metal into a cation?

Thus, silver is oxidized to Ag+. In choice B, the sulfide anion acts as a reducing agent, but in the reverse reaction, the sulfur is supposed to be the oxidizing agent, turning Ag to Ag+. This cannot happen because a compound or element cannot both be a reducing and oxidizing agent in a forward and its reverse reaction? This is a law for all chemical reactions? The same compound that is the oxidizing agent in the forward reaction cannot be the reducing agent in the reverse reaction?

So we turn to aluminum to reduce Ag+ to Ag in choice D. is this correct thinking? Thanks

 

[AdaptPrep]

Yes, your thinking is correct. However, technically, you can reverse the reaction, but you will have to hook up a battery or put energy into the system if it is not spontaneous. In answer B, they are not adding a battery or energy, so the reverse would not spontaneously occur.

Recall that Electrochemistry deals with charts that lists Standard Reduction Potentials listed with best oxidizing agents on top (+E values) and best reducing agents on the bottom (-E values). Be familiar with those trends and how to use those values (which this problem really does not require, but just FYI). For example, lithium loves to be Li+ so it readily donates an electron. This makes it a strong reducing agent (it is on the bottom of the lists with E = -3.05). Fluorine loves to be F- so it is readily accepts an electron. This makes it a strong oxidizing agent (it is on the top of the lists with E = +2.87).