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

## $M g \left(s\right) | M {g}^{2 +} \left(0.25 M\right) | | Z {n}^{2 +} \left(0.10 M\right) | Z n \left(s\right)$ Half-reactions $M g \left(s\right) \setminus \to M {g}^{2 +} \left(a q\right) + 2 {e}^{-}$ $Z {n}^{2 +} + 2 {e}^{-} \setminus \to Z n \left(s\right)$ Combined $M g \left(s\right) + Z {n}^{2 +} \left(a q\right) \setminus r i g h t \le f t h a r p \infty n s M {g}^{2 +} \left(a q\right) + Z n \left(s\right)$ Formula E_(cell)=E°_(cell)-0.0592/nV\log(Q) My notes say this: ${E}_{c e l l} = 1.598 V$ (???) E°_(cell)=2.372V+(-0.762V) (???) and $Q = \frac{\left[M {g}^{2 +}\right]}{\left[Z {n}^{2 +}\right]} = \frac{0.25}{0.10}$ (this, I understand.)

Jul 13, 2018

${E}_{c e l l} = \text{1.598 V}$

for these nonstandard concentrations. How do you change only the concentrations to force ${E}_{c e l l} = {E}_{c e l l}^{\circ}$?

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:

$- \left(\text{Mg"^(2+)(aq) + cancel(2e^(-)) -> "Mg"(s)), " "-(E_(red)^@) = -(-"2.372 V}\right)$
$\underline{\text{Zn"^(2+)(aq) + cancel(2e^(-)) -> "Zn"(s), " "E_(red)^@ = -"0.762 V}}$
"Mg"(s) + "Zn"^(2+)(aq) -> "Mg"^(2+)(aq) + "Zn"(s), " "E_(cell)^@ = ???

These ${E}_{red}^{\circ}$ were both negative, because they would react with $\text{HCl}$ (i.e. their reduction potentials being more negative means the metals prefer to be oxidized compared to the standard hydrogen electrode half-reaction, which $\text{HCl}$ is based on).

The Nernst equation for the cell potential ${E}_{c e l l}$ is:

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \ln Q$

where:

• $R = \text{8.314 V"cdot"C/mol"cdot"K}$ is the universal gas constant.
• $T$ is temperature in $\text{K}$.
• $n$ is the mols of electrons PER mol of atom.
• $F = {\text{96485 C/mol e}}^{-}$ is the Faraday constant.
• $Q$ is the reaction quotient for when ${E}_{c e l l} \ne 0$ (i.e. non-equilibrium).

You have a different version because you seem to be using $T = \text{298.15 K}$ by default.

$- \left({\text{8.314 V"cdot"C/mol"cdot"K" cdot "298.15 K")/(n cdot "96485 C/mol e}}^{-}\right) \cdot {\underbrace{\ln Q}}_{2.303 \log Q}$

$\approx - \frac{\text{0.0592 V}}{n} \log Q$

You have switched the sign on ${E}_{red}^{\circ}$ for $\text{Mg}$ because it gets oxidized in this reaction more favorably compared to zinc, as the ${E}_{red}^{\circ}$ ($- \text{2.372 V}$) is more negative.

In general,

${E}_{c e l l}^{\circ} = {E}_{red}^{\circ} + {E}_{o x}^{\circ}$

where ${E}_{o x}^{\circ}$ is simply the opposite sign to the ${E}_{red}^{\circ}$ 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")

$= {\overbrace{{E}_{red}^{\circ}}}^{- \text{0.762 V") + overbrace(E_(o x)^@)^(+"2.372 V}}$

$= \textcolor{b l u e}{+ \text{1.610 V}}$

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)

$= \textcolor{b l u e}{+ \text{1.598 V}}$