Calculate $E_{c e l l}$ $E_(cell)$ for: $M g (s) ∣ ∣ M g^{2 +} (0.25 M) ∣ ∣ ∣ ∣ Z n^{2 +} (0.10 M) ∣ ∣ Z n (s)$ $Mg(s)|Mg^(2+)(0.25M)||Zn^(2+)(0.10M)|Zn(s)$ ?

Question

Calculate $E_{c e l l}$ $E_(cell)$ for: $M g (s) ∣ ∣ M g^{2 +} (0.25 M) ∣ ∣ ∣ ∣ Z n^{2 +} (0.10 M) ∣ ∣ Z n (s)$ $Mg(s)|Mg^(2+)(0.25M)||Zn^(2+)(0.10M)|Zn(s)$ ?

$M g (s) ∣ ∣ M g^{2 +} (0.25 M) ∣ ∣ ∣ ∣ Z n^{2 +} (0.10 M) ∣ ∣ Z n (s)$ $Mg(s)|Mg^(2+)(0.25M)||Zn^(2+)(0.10M)|Zn(s)$

Half-reactions
$M g (s) \to M g^{2 +} (a q) + 2 e^{-}$ $Mg(s)\toMg^(2+)(aq)+2e^(-)$
$Z n^{2 +} + 2 e^{-} \to Z n (s)$ $Zn^(2+)+2e^(-)\toZn(s)$

Combined
$M g (s) + Z n^{2 +} (a q) ⇌ M g^{2 +} (a q) + Z n (s)$ $Mg(s)+Zn^(2+)(aq)\rightleftharpoonsMg^(2+)(aq)+Zn(s)$

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.)

1 Answer

Impact of this question

6235 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-13T19:33:26

$E_(cell) = "1.598 V"$

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

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,

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

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

Science

Math

Humanities

... and beyond

Calculate $E_{c e l l}$ $E_(cell)$ for: $M g (s) ∣ ∣ M g^{2 +} (0.25 M) ∣ ∣ ∣ ∣ Z n^{2 +} (0.10 M) ∣ ∣ Z n (s)$ $Mg(s)|Mg^(2+)(0.25M)||Zn^(2+)(0.10M)|Zn(s)$ ?

My notes say this:

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Calculate E_(cell)EcellE_(cell) for: Mg(s)|Mg^(2+)(0.25M)||Zn^(2+)(0.10M)|Zn(s)Mg(s)∣∣Mg2+(0.25M)∣∣∣∣Zn2+(0.10M)∣∣Zn(s)Mg(s)|Mg^(2+)(0.25M)||Zn^(2+)(0.10M)|Zn(s)?

My notes say this:

1 Answer

Related questions

Impact of this question

Calculate $E_{c e l l}$ $E_(cell)$ for: $M g (s) ∣ ∣ M g^{2 +} (0.25 M) ∣ ∣ ∣ ∣ Z n^{2 +} (0.10 M) ∣ ∣ Z n (s)$ $Mg(s)|Mg^(2+)(0.25M)||Zn^(2+)(0.10M)|Zn(s)$ ?