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

#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")#