Following question I need help with part b) Please verify my answer for part a) too if possible.
Consider the following reaction for silver tarnishing:
3Ag2S(s) + 2Al(s) -> 6Ag(s) + Al2S3(s)
a. State the oxidation number for each element in the reaction.
3Ag2S(s) + 2Al(s) -> 6Ag(s) + Al2S3(s)
(+1)+(-2)+(+3) -> (+1)+(+3)+(-2)
verify these please
b. Identify the oxidized reactant and the reduced reactant.
Please help me with b) above....
All of the oxidation numbers are right except for Al(s). All elements in the free state have an oxidation number of zero. (Note: I don't consider the reaction written as the reason silver tarnishes. Rather it is a way of removing the sulfide from silver that is tarnished.
For the part b, oxidation is the loss of electrons. Reduction is the gain of electrons. Which element/ion loses and which gains electrons.
Yes that's right, free states have oxidation number of zero. Ok will work out reduction and oxidation, made it a bit clearer. Thanks for your help, very valuable :)
To identify the oxidized reactant and the reduced reactant in this reaction, we need to determine which elements are undergoing a change in oxidation number.
In this reaction, silver (Ag) changes from an oxidation state of +1 in silver sulfide (Ag2S) to 0 in metallic silver (Ag). Aluminum (Al) changes from an oxidation state of 0 to +3 in aluminum sulfide (Al2S3).
For oxidation, the oxidation number of an element increases, indicating the loss of electrons. For reduction, the oxidation number decreases, indicating the gain of electrons.
Based on this information, we can identify the oxidized reactant and the reduced reactant as follows:
Oxidized reactant: Aluminum (Al)
Reason: Since the oxidation number of Al increases from 0 to +3, it is being oxidized.
Reduced reactant: Silver sulfide (Ag2S)
Reason: Since the oxidation number of Ag decreases from +1 to 0, it is being reduced.
Therefore, the answer for part b) is:
The oxidized reactant is Aluminum (Al) and the reduced reactant is Silver sulfide (Ag2S).
For part a), let's verify the oxidation numbers you provided:
3Ag2S(s) + 2Al(s) -> 6Ag(s) + Al2S3(s)
Ag: (+1) * 2 = +2 on the left side, and (+1) * 6 = +6 on the right side.
S: (-2) * 2 = -4 on the left side, and (-2) * 3 = -6 on the right side.
Al: 0 on the left side, and (+3) * 2 = +6 on the right side.
Therefore, the oxidation numbers you provided are correct for part a).