Use the following reactions to arrange the elements A, B, C, and D in order of their redox activity, from greatest to least.
2 A + B2+ --> 2 A+ + B
B + D2+--> B2+ + D
A+ + C -->No reaction
2 C + B2+ --> 2 C+ + B
Can someone explain?
thanks
I think the easy way to look at this is to study the activity series. Here is a link.
http://www.files.chem.vt.edu/RVGS/ACT/notes/activity_series.html
An example or two. Any metal will oxidize any ION below it in the series; therefore, note that Li will oxidize H^+ or Cu^+2 or Ag^+. Therefore, Li goes ABOVE H and Cu. Cu will not react with HCl, for example, to give H2 gas; therefore, Cu is BELOW H so we would place the metals as Li, H, Cu.
It works the same way with A, B, C, and D EXCEPT that instead of giving the placement in the table the problem gives you the reactions.
2A + B^2+ ==> 2A^+ + B
Note that A is oxidized and B^2+ is reduced. Therefore the abbreviated table is
A
B
B + D^2+ ==> B^2+ + D
B is oxidized, D^2+ is reduced.
So
B
D is the placement.
A^+ + C ==> NR (you may see it better if we keep it consistent and write it as C + A^+ ==> NR) which means C is below A but last equation says it is above B. So I write it as
A
C
B
D
Check my thinking.
Well, let's look at the reactions and start making jokes about them!
First, we have 2 A + B2+ --> 2 A+ + B. It's like A is calling B and saying, "Hey, I need some positive vibes, can you lend me some electrons?" And B is like, "Sure thing, A, I got your back!"
Now, we have B + D2+ --> B2+ + D. B is now asking D, "Buddy, can you spare some electrons?" And D says, "Sorry, B, I don't have any to spare, but I can take some of yours!" And suddenly, B and D become B2+ and D.
Next, we have A+ + C --> No reaction. This one is just sad. A is feeling lonely and goes to C for some chemical bonding, but C is like, "Nah, I'm not feeling it today, sorry!"
Lastly, we have 2 C + B2+ --> 2 C+ + B. C takes matters into its own hands, being proactive and saying to B, "I don't need your electrons, I'll take them from you myself!" And just like that, C becomes C+ and B is left without.
So, let's rank their redox activity from greatest to least based on these reactions:
1. B
2. C
3. A
4. D
B is the most active, always ready to give or take electrons. C comes next, able to take electrons from B but not from A. A is less active, unable to react with C or D. And poor D is the least active, D2+ stays as it is.
Hope this explanation brought a smile to your face!
To determine the order of redox activity for elements A, B, C, and D, we need to examine the given reactions and identify which elements are oxidized and which ones are reduced.
Let's analyze each reaction step by step:
1. 2A + B2+ → 2A+ + B
In this reaction, element A is oxidized from its elemental state (A) to its cationic state (A+), while element B is reduced from its cationic state (B2+) to its elemental state (B). This suggests that A is a stronger reducing agent than B.
2. B + D2+ → B2+ + D
In this reaction, element B is oxidized from its elemental state (B) to its cationic state (B2+), while element D is reduced from its cationic state (D2+) to its elemental state (D). This tells us that B is a stronger reducing agent than D.
3. A+ + C → No reaction
This reaction shows that element A+ is not capable of reducing element C. Therefore, we cannot conclude the relative redox activity between A+ and C based on this reaction alone.
4. 2C + B2+ → 2C+ + B
In this reaction, element C is oxidized from its elemental state (C) to its cationic state (C+), while element B is reduced from its cationic state (B2+) to its elemental state (B). This indicates that C is a stronger reducing agent than B.
Based on the above analysis, we can order the elements from greatest to least redox activity as follows:
A > B > C > D
Therefore, element A is the most active (strongest reducing agent) while element D is the least active (weakest reducing agent) in terms of redox activity.
To determine the order of redox activity for the elements A, B, C, and D, we need to analyze the given reactions and identify the elements that are undergoing reduction (gaining electrons) or oxidation (losing electrons). The general rule is that an element undergoing oxidation is less reactive, while an element undergoing reduction is more reactive.
Let's go through the reactions one by one:
1) 2A + B2+ -> 2A+ + B: In this reaction, A is getting oxidized (loses electrons) from A to A+, while B2+ is getting reduced (gains electrons) from B2+ to B. Therefore, A is less reactive than B.
2) B + D2+ -> B2+ + D: Here, B is being oxidized from B to B2+, while D2+ is being reduced from D2+ to D. This suggests that B is less reactive than D.
3) A+ + C -> No reaction: In this case, A+ doesn't undergo any changes, so we can't determine its reactivity relative to the other elements based on this reaction.
4) 2C + B2+ -> 2C+ + B: In this reaction, C is being oxidized from C to C+, and B2+ is getting reduced from B2+ to B. This indicates that C is less reactive than B.
Based on the information we gathered from the given reactions, we can conclude that the order of redox activity for the elements A, B, C, and D, from greatest to least, is: D > B > A > C.