Question

Calculate `E_(cell)` for: `Mg(s)|Mg^(2+)(0.25M)||Zn^(2+)(0.10M)|Zn(s)`?

`Mg(s)|Mg^(2+)(0.25M)||Zn^(2+)(0.10M)|Zn(s)`

Half-reactions
`Mg(s)\toMg^(2+)(aq)+2e^(-)`
`Zn^(2+)+2e^(-)\toZn(s)`

Combined
`Mg(s)+Zn^(2+)(aq)\rightleftharpoonsMg^(2+)(aq)+Zn(s)`

Formula
`E_(cell)=E°_(cell)-0.0592/nV\log(Q)`

My notes say this:

`E_(cell)=1.598V` (???)
`E°_(cell)=2.372V+(-0.762V)` (???)

and `Q=([Mg^(2+)])/([Zn^(2+)])=0.25/0.10` (this, I understand.)

Answer 1

`E_(cell) = "1.598 V"`

for these nonstandard concentrations. How do you change only the concentrations to force `E_(cell) = E_(cell)^@`?

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Typically (in the US, anyway), the anode is placed first and the cathode is placed second in cell notation so that electrons flow from left to right.

`overbrace("Mg"(s) | "Mg"^(2+)("0.25 M"))^"Anode"` `||` `overbrace("Zn"^(2+)("0.10 M")|"Zn"(s))^"Cathode"`

So the half-reactions are:

`-("Mg"^(2+)(aq) + cancel(2e^(-)) -> "Mg"(s)), " "-(E_(red)^@) = -(-"2.372 V")`
`ul("Zn"^(2+)(aq) + cancel(2e^(-)) -> "Zn"(s), " "E_(red)^@ = -"0.762 V")`
`"Mg"(s) + "Zn"^(2+)(aq) -> "Mg"^(2+)(aq) + "Zn"(s), " "E_(cell)^@ = ???`

These `E_(red)^@` were both negative, because they would react with `"HCl"` (i.e. their reduction potentials being more negative means the metals prefer to be oxidized compared to the standard hydrogen electrode half-reaction, which `"HCl"` is based on).

The Nernst equation for the cell potential `E_(cell)` is:

`E_(cell) = E_(cell)^@ - (RT)/(nF)lnQ`

where:

• `R = "8.314 V"cdot"C/mol"cdot"K"` is the universal gas constant.
• `T` is temperature in `"K"`.
• `n` is the mols of electrons PER mol of atom.
• `F = "96485 C/mol e"^(-)` is the Faraday constant.
• `Q` is the reaction quotient for when `E_(cell) ne 0` (i.e. non-equilibrium).

You have a different version because you seem to be using `T = "298.15 K"` by default.

`-("8.314 V"cdot"C/mol"cdot"K" cdot "298.15 K")/(n cdot "96485 C/mol e"^(-))cdot underbrace(lnQ)_(2.303logQ)`

`~~ -"0.0592 V"/nlogQ`

You have switched the sign on `E_(red)^@` for `"Mg"` because it gets oxidized in this reaction more favorably compared to zinc, as the `E_(red)^@` (`-"2.372 V"`) is more negative.

In general,

`E_(cell)^@ = E_(red)^@ + E_(o x)^@`

where `E_(o x)^@` is simply the opposite sign to the `E_(red)^@` value for the corresponding reduction reaction after it has been reversed.

A trick I do is take the more positive (less negative) value and subtract out the less positive (more negative) value to determine the spontaneous reaction direction.

Compare the two ways to do this:

`color(blue)(E_(cell)^@) = -"0.762 V" - (-"2.372 V")`

`= overbrace(E_("cathode")^@)^(-"0.762 V") - overbrace(E_("anode")^@)^(-"2.372 V")`

`= overbrace(E_(red)^@)^(-"0.762 V") + overbrace(E_(o x)^@)^(+"2.372 V")`

`= color(blue)(+"1.610 V")`

This suggests that our initial assumption was correct, that zinc gets reduced and magnesium gets oxidized. So, the reaction direction is right, and

`Q = (["Mg"^(2+)])/(["Zn"^(2+)]) = "0.25 M"/"0.10 M" = 2.5`

Therefore,

`color(blue)(E_(cell)) = "1.610 V" - "0.0592 V"/("2 mol e"^(-)"/1 mol Zn")log(2.5)`

`= color(blue)(+"1.598 V")`