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

1 Answer
Mar 6, 2018

I got +"1.299 V".


Let's just use what we are familiar with. You want E_(cell)^@ for:

"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 DeltaG_f^@ from your textbook appendix, so use those! I'm looking them up from here, but use your textbook values.

For (1), that is the formation reaction of "MgO"(s) (formation Gibbs' free energies for elements in their elemental state are just 0), so look that up to be

DeltaG_(rxn)^@(1) = DeltaG_(f,MgO(s, "periclase"))^@ = -"569.024 kJ/mol"

The second reaction is also for the formation of "ZnO"(s) (well, the reverse, actually), so

DeltaG_(rxn)^@(2) = -DeltaG_(f,ZnO(s))^@ = +"318.32 kJ/mol"
(reversed from the original, negative formation value)

Therefore, the net DeltaG_(rxn)^@ is:

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, n = "2 mol e"^(-)"/mol atoms". Also, "1 mol" of atom goes to "1 mol" of oxide.

Therefore, we can now relate DeltaG_(rxn)^@ to E_(cell)^@:

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")