This question is really making me upset i just can't get the answer right. i've tried so many times.
If someone could help it would be muchly appeciated!
Experiments were performed with four metals and solutions of their ions: A-A, B-B2+, C-C2+, D-D2+. the results of the experiments were:
D2 + C = No reactions
B2 + C = B + C2
A + B = A + B2
a) write the equation for each reduction half reaction.
b) Arrange them in the relative reduction potential table with the easiest to reduce at the top and the hardest at the bottom.
c) write the equation for each ocidation half reaction.
d) Arrange them in the relative oxidation potential table with the easiest to oxidixe at the top and the hardest at the bottom.
e) Which is the best oxidixing agent? which is the best reducing agent??
I think you have made typos or just didn't write some of the problem?
Is A-A really A-A^+ or A-A^2+ or something similar to the other pairs?
In the reactions, is the first one of
D2 + C==>NR really D^2+ + C ==> NR?
Is the second one of
B2 + C ==> B + C2 really B^2+ + C ==>B + C^2+?
etc.
A-A^+
D^2+ + C = No Reactions
B^2+ + C = B C^2
A^+ + B = A + B^2+
Arrange these couples in pairs something like the following where the first column is the metal(element) and the second column is the ion:
.....M.......ion
.....A.......A^+
.....B.......B^2+
.....C.......C^2+
.....D.......D^2+
Now remember the activity series. Here is a link to remind you of what it looks like.
http://www.files.chem.vt.edu/RVGS/ACT/notes/activity_series.html
In the activity series, any ELEMENT will replace any ION BELOW IT IN THE SERIES but will NOT replace an ion ABOVE it in the series. For example,
2Na + HCl ==> 2NaCl + H2. Notice that Na is ABOVE H^+ so the reaction goes as written. But Cu will not replace H^+ as
Cu + 2HCl ==> CuCl2 + H2 IS NOT CORRECT. WHY? Because Cu is BELOW H^+ in the series (find them so you can see what is going on) so the correct equation to write is
Cu + HCl ==> No reaction.
Now to your post.
D^2+ + C ==> No reaction which means that C (the element) is BELOW D^2+; therefore, in a new table you can write them like this.
.........M..............ion
.........D..............D^2+
.........C..............C^2+
From rxn 2 (B^2+ + C==> B + C^2+) you know C(the element) must be ABOVE B and you can add it to the table just below C like this.
.........M..............ion
.........D..............D^2+
.........C..............C^2+
.........B..............B^2+
The last equation of A^+ + B ==> A + B^2+ means B must be ABOVE A so it goes at the bottom of the table. The final table should look like this
D
C
B
A.
Now that they are lined up right, you can answer the other questions.
a) To write the reduction half reactions, we need to identify the species being reduced in each case. The reduction half reaction is always the species gaining electrons. Based on the given information:
D2 + C = No reactions (No reduction of D2 or C)
B2 + C = B + C2 (B2+ is reduced to B)
A + B = A + B2 (B is reduced to B2+)
The reduction half reactions are:
D2+ + 2e- -> D (no reaction)
B2+ + 2e- -> B
A -> A+ + e-
b) To arrange the metals in the relative reduction potential table, we need to compare their reduction potentials. The easier it is for a metal to get reduced, the higher its reduction potential.
From the given information, we can deduce the relative reduction potentials:
E(D2+/D) > E(B2+/B) > E(A+/A)
c) To write the oxidation half reactions, we need to identify the species being oxidized in each case. The oxidation half reaction is always the species losing electrons.
Based on the given information, we can write the oxidation half reactions:
D2 -> D2+ + 2e- (no reaction)
B -> B2+ + 2e-
A+ + e- -> A
d) To arrange the metals in the relative oxidation potential table, we need to compare their oxidation potentials. The easier it is for a metal to get oxidized, the higher its oxidation potential.
From the given information, we can deduce the relative oxidation potentials:
E(D/D2+) < E(B/B2+) < E(A/A+)
e) The best oxidizing agent is the species with the highest oxidation potential. As per our previous calculations:
E(D/D2+) < E(B/B2+) < E(A/A+)
Therefore, D/D2+ is the best oxidizing agent.
The best reducing agent is the species with the highest reduction potential. As per our previous calculations:
E(D2+/D) > E(B2+/B) > E(A+/A)
Therefore, A/A+ is the best reducing agent.