How do I find the reaction potential of forming #"MgO"# and #"Zn"# from #"ZnO"# and #"Mg"#?

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")#