Question

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

Answer 1

I got `+"1.299 V"`.

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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")`