Silver will spontaneously reduce which of the following?
Reaction E (volts)
Na+ + e- ---> Na -2.71
Co2+ + 2e- ---> Co -0.28
Ag+ + e- ---> Ag .80
A) Na
B) Co2+
C) Na+
D)Co
E) None of these
6 answers
none of these.
Please explain why
Here is a link for the activity series (they are written as oxidations and not reductions). Any metal ABOVE an ion will displace that ion from solution but will not displace any ion ABOVE IT.
http://www.files.chem.vt.edu/RVGS/ACT/notes/activity_series.html
Therefore, Na + Ag^+ ==> Ag + Na^+ BUT
Na^+ + Ag ==> No reaction.
You can do it another way by writing half reactions and see which will give you a positive voltage.
Ag ==> Ag^+ + e Eo = -0.8
Na^+ + e ==> Na Eo = -2.71
---------------------------
Ag + Na^+ ==> Ag^+ + Na
and Eocell = -3.something and negative voltages are not spontaneous.
Ag ==> Ag^+ + e Eo = -0.8
Co^2+ + 2e ==> Co Eo = -.28
------------------
2Ag + Co^2+ ==> 2Ag^+ + Co
and Eocell = -1.something. Negative. not spontaneous.
Ag ==> Ag^+ + e Eo = -0.80
Na ==> Na^+ + e Eo = 2.71
BUT this is not a reaction because BOTH reactions are oxidations. You must have one oxidation half cell and one reduction half cell to get a cell. Both can't be oxidations and both can't be reductions.
http://www.files.chem.vt.edu/RVGS/ACT/notes/activity_series.html
Therefore, Na + Ag^+ ==> Ag + Na^+ BUT
Na^+ + Ag ==> No reaction.
You can do it another way by writing half reactions and see which will give you a positive voltage.
Ag ==> Ag^+ + e Eo = -0.8
Na^+ + e ==> Na Eo = -2.71
---------------------------
Ag + Na^+ ==> Ag^+ + Na
and Eocell = -3.something and negative voltages are not spontaneous.
Ag ==> Ag^+ + e Eo = -0.8
Co^2+ + 2e ==> Co Eo = -.28
------------------
2Ag + Co^2+ ==> 2Ag^+ + Co
and Eocell = -1.something. Negative. not spontaneous.
Ag ==> Ag^+ + e Eo = -0.80
Na ==> Na^+ + e Eo = 2.71
BUT this is not a reaction because BOTH reactions are oxidations. You must have one oxidation half cell and one reduction half cell to get a cell. Both can't be oxidations and both can't be reductions.
compare the e values with silver and i think you will see why...
But the given Ecell for Ag was positive .80 why did you use -.80? And how do you know that Na+ is a reduction?
Ag^+ + e = Ag is given in the table as 0.80 but I wrote it as
Ag ==> Ag^+ +e therefore, I changed the sign.
For the second part of your question, answer A is Na and answer C is Na^+. I used both of them. If you start with Na it MUST go to Na^+ and if you start with Na^+ it MUST go to Na.
Ag ==> Ag^+ +e therefore, I changed the sign.
For the second part of your question, answer A is Na and answer C is Na^+. I used both of them. If you start with Na it MUST go to Na^+ and if you start with Na^+ it MUST go to Na.
0 answers