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

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#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.)

Half-reactions

Combined

Formula

**My notes say this:**

and

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

##### 1 Answer

#E_(cell) = "1.598 V"#

for these nonstandard concentrations. How do you change only the concentrations to force

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

The **Nernst equation** for the cell potential

#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

#-("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 **gets oxidized** in this reaction **more** favorably compared to zinc, as the

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 reactionafterit 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")#