Tarnish on Silver - TBR Gen Chem

[arc5005]

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Tarnish on silver is attributed to Ag2S. Placing silver in an aluminum pan and adding aqueous baking soda (NaHCO3) can remove the tarnish. Which of the following equations shows the reaction that occurs?

A) Ag2S (s) + 2 NaHCO3 (aq) → H2 (g) + 2 Ag (s) + Na2S (s)
B) Ag2S (s) + 2 NaHCO3 (aq) → CO2 (g) + 2 Ag (s) + S (s)
C) Ag2S (s) + 2 Al (s) → 2 Ag (s) + Al2S (s)
D) 3 Ag2S (s) + 2 Al (s) → 6 Ag (s) + Al2S3 (s)

D) 3 Ag2S (s) + 2 Al (s) → 6 Ag (s) + Al2S3 (s)

Choice A can be eliminated, because both the H+ and the Ag+ are reduced in the reaction as given, and nothing is oxidized. An oxidation-reduction reaction must have both an oxidation half-reaction and reduction half-reaction. 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. Choice C is eliminated, because Al cannot exist in a +1 oxidation state. The correct answer is choice D, because the oxidation state of aluminum (which was oxidized) is +3.

I am still having a hard time understanding this, despite the reasoning they provide.

 

[Lawpy]

The question asks for a chemical equation that describes the reaction of removing tarnish from silver. The question states that silver tarnish can be removed by using aluminum pan and adding aqueous baking soda. Removing tarnish means you want to get pure silver from silver sulfide. Another way to think of this is Ag2S is impure/dirty and Ag is pure/clean.

Removing tarnish means Ag2S is the reactant and Ag is the product. We can see that this implies a redox reaction since there’s a change in oxidation state of silver from Ag2S (where Ag here has +1 oxidation state) to Ag (where Ag here has 0 oxidation state). Since there is a decrease in oxidation state in silver, Ag is being reduced.

For redox reactions to run, there needs to be an accompanying oxidation reaction. In other words, there is a reducing agent that reduces Ag2S into Ag (and the reducing agent gets oxidized in the process).

The question says aluminum pan and aqueous baking soda are used in the reaction. Aluminum is a possible reducing agent since its oxidation state is 0 as a reactant and can be oxidized to +3 (+1 and +2 oxidation states are very rare). For NaHCO3, Na has +1 oxidation state, H has +1 oxidation state, C has +4 oxidation state and O has -2 oxidation state. You can’t increase the oxidation states of Na, H, or C any further since they are already in its stable electron configuration with octets filled. You could increase oxidation state of O by making O2 gas but NaHCO3 usually decomposes to form CO2 and H2O, which isn’t a redox reaction since oxidation states aren’t changing. So the reducing agent used is Al and not NaHCO3.

A simpler way is to use the process of elimination for this question to find what is the reducing agent. Something has to be oxidized if silver is being reduced.

In choice A, hydrogen in NaHCO3 has +1 oxidation state and H2 has 0 oxidation state, so hydrogen is being reduced. But silver is also being reduced. There isn’t an oxidation reaction, so it’s incorrect.

In choice B, Ag2S decomposes into Ag and S. Now that is a redox reaction since S in sulfide has -2 oxidation state that increases to 0 in elemental sulfur. But the question says aluminum and NaHCO3 are needed to remove the silver tarnish. If Ag2S can just spontaneously decompose to provide pure silver, there isn’t a need to have any of those extra stuff since the tarnish can simply provide pure silver on its own via sulfide anion! So choice B is incorrect.

Choices C and D both have aluminum oxidizing and are thus redox reactions. But choice C has aluminum in +1 oxidation state, which is very rare and unstable. The reason had to do with its electron configuration: Al = 1s2 2s2 2p6 3s2 3p1. Aluminum really prefers to get rid off its electrons in its highest energy orbitals (so 3s2 and 3p1), since it wants to have a stable octet as fast as possible (aluminum certainly doesn’t want to have 5 electrons added to its 3p orbital to form a larger octet, since negatively charged electrons repel each other and it requires more energy to add more electrons, so removing electrons is simply better). Removing 1 electron does form a stable state since the 3s orbital is still filled (this is why Al can exist in +1 oxidation state), but aluminum still wants to get rid off high energy electrons in 3s orbital to form an octet, which is why Al most commonly has an oxidation state of +3.

So choice D is a better answer than choice C, since it has Al in its +3 oxidation state.

 

[arc5005]

The question asks for a chemical equation that describes the reaction of removing tarnish from silver. The question states that silver tarnish can be removed by using aluminum pan and adding aqueous baking soda. Removing tarnish means you want to get pure silver from silver sulfide. Another way to think of this is Ag2S is impure/dirty and Ag is pure/clean.

Removing tarnish means Ag2S is the reactant and Ag is the product. We can see that this implies a redox reaction since there’s a change in oxidation state of silver from Ag2S (where Ag here has +1 oxidation state) to Ag (where Ag here has 0 oxidation state). Since there is a decrease in oxidation state in silver, Ag is being reduced.

For redox reactions to run, there needs to be an accompanying oxidation reaction. In other words, there is a reducing agent that reduces Ag2S into Ag (and the reducing agent gets oxidized in the process).

The question says aluminum pan and aqueous baking soda are used in the reaction. Aluminum is a possible reducing agent since its oxidation state is 0 as a reactant and can be oxidized to +3 (+1 and +2 oxidation states are very rare). For NaHCO3, Na has +1 oxidation state, H has +1 oxidation state, C has +4 oxidation state and O has -2 oxidation state. You can’t increase the oxidation states of Na, H, or C any further since they are already in its stable electron configuration with octets filled. You could increase oxidation state of O by making O2 gas but NaHCO3 usually decomposes to form CO2 and H2O, which isn’t a redox reaction since oxidation states aren’t changing. So the reducing agent used is Al and not NaHCO3.

A simpler way is to use the process of elimination for this question to find what is the reducing agent. Something has to be oxidized if silver is being reduced.

In choice A, hydrogen in NaHCO3 has +1 oxidation state and H2 has 0 oxidation state, so hydrogen is being reduced. But silver is also being reduced. There isn’t an oxidation reaction, so it’s incorrect.

In choice B, Ag2S decomposes into Ag and S. Now that is a redox reaction since S in sulfide has -2 oxidation state that increases to 0 in elemental sulfur. But the question says aluminum and NaHCO3 are needed to remove the silver tarnish. If Ag2S can just spontaneously decompose to provide pure silver, there isn’t a need to have any of those extra stuff since the tarnish can simply provide pure silver on its own via sulfide anion! So choice B is incorrect.

Choices C and D both have aluminum oxidizing and are thus redox reactions. But choice C has aluminum in +1 oxidation state, which is very rare and unstable. The reason had to do with its electron configuration: Al = 1s2 2s2 2p6 3s2 3p1. Aluminum really prefers to get rid off its electrons in its highest energy orbitals (so 3s2 and 3p1), since it wants to have a stable octet as fast as possible (aluminum certainly doesn’t want to have 5 electrons added to its 3p orbital to form a larger octet, since negatively charged electrons repel each other and it requires more energy to add more electrons, so removing electrons is simply better). Removing 1 electron does form a stable state since the 3s orbital is still filled (this is why Al can exist in +1 oxidation state), but aluminum still wants to get rid off high energy electrons in 3s orbital to form an octet, which is why Al most commonly has an oxidation state of +3.

So choice D is a better answer than choice C, since it has Al in its +3 oxidation state.

thank you so much!!