How do I find the reaction potential of forming "MgO" and "Zn" from "ZnO" and "Mg"?
I have been given the task of finding the reaction potential of
Mg + ZnO -> MgO + Zn
(I have the potential value, I'm supposed to go through the steps).
The hint I was given was to start by finding the potentials of the reactions:
2Mg + O2 -> 2MgO
2ZnO -> 2Zn + O2.
I have split these further into:
oxidation: 2Mg -> 2Mg2+ +4e- (e=2.37V)
reduction: O2 + 4e- -> 2O2- (e=???)
oxidation: 2O2- -> O2 + 4e- (e=???)
reduction: 2ZnO+ + 4e- -> 2Zn (e=-0.76V)
I can't find the reaction potential for the oxygen and any chart, and I have no idea how I would find it myself. I've been looking for 3 days with no luck. Any help our suggestions for resources would be helpful
I have been given the task of finding the reaction potential of
(I have the potential value, I'm supposed to go through the steps).
The hint I was given was to start by finding the potentials of the reactions:
I have split these further into:
oxidation:
reduction:
oxidation:
reduction:
I can't find the reaction potential for the oxygen and any chart, and I have no idea how I would find it myself. I've been looking for 3 days with no luck. Any help our suggestions for resources would be helpful
1 Answer
I got
Let's just use what we are familiar with. You want
"Mg"(s) + cancel(1/2"O"_2(g)) -> "MgO"(s) " "bb((1))
ul("ZnO"(s) -> "Zn"(s) + cancel(1/2"O"_2(g))) " "bb((2))
"Mg"(s) + "ZnO"(s) -> "MgO"(s) + "Zn"(s) and I have divided the coefficients by
2 so that we involve only"1 mol" of the oxides. That is on purpose, as that is how formation reactions are defined.
Keep it simple. You have
For
DeltaG_(rxn)^@(1) = DeltaG_(f,MgO(s, "periclase"))^@ = -"569.024 kJ/mol"
The second reaction is also for the formation of
DeltaG_(rxn)^@(2) = -DeltaG_(f,ZnO(s))^@ = +"318.32 kJ/mol"
(reversed from the original, negative formation value)
Therefore, the net
DeltaG_(rxn)^@ = DeltaG_(rxn)^@(1) + DeltaG_(rxn)^@(2)
= -"569.024 kJ/mol" + ("318.32 kJ/mol") = -"250.704 kJ/mol"
for the reaction:
"Mg"(s) + "ZnO"(s) -> "MgO"(s) + "Zn"(s)
As a result, since this reaction involves two-electron transfers,
Therefore, we can now relate
DeltaG_(rxn)^@ = -nFE_(cell)^@
As a result:
=> color(blue)(E_(cell)^@) = -(DeltaG_(rxn)^@)/(nF)
= -(-250.704 cancel"kJ""/"cancel"mol atoms" xx "1000 J"/cancel"1 kJ")/((2 cancel("mol e"^(-)))/(cancel"1 mol atom") cdot "96485 C"/cancel("mol e"^(-)))
= color(blue)(+"1.299 V")