For the reaction of #"Au"^(3+)# with #"Cl"_2#, if #["Cl"^(-)]# and #["Au"^(3+)]# were initially #"0.10 M"# and #"0.25 M"# respectively, while #P_("Cl"_2) = "0.5 bar"# initially, what is #E_(cell)#?

Well, what I would do is draw out the problem to see what you got...

The Nernst equation is:

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

• #E_(cell)# is the cell potential at nonstandard #T# and #P#. #""^@# signifies #"1 bar"# and #25^@ "C"#.
• #R# and #T# are known from the ideal gas law, and are in #"J/mol"cdot"K"# and #"K"#, respectively.
• #n# is the mols of electrons transferred in the overall reaction.
• #F = "96485 C/mol e"^(-)# is the Faraday constant.
• #Q# is the reaction quotient.

From the looks of it, you are supposed to calculate #E_(cell)# at this nonstandard, nonequilibrium state.

First, write out the half-reactions. You need to know how to look up reduction reactions in a standard reduction potential table, which is mandatory at any chemistry exam with electrochemistry questions.

#"Au"^(3+)(aq) + 3e^(-) -> "Au"(s)#, #" "E_(red)^@ = "1.50 V"#

#"Cl"_2(g) + 2e^(-) -> 2"Cl"^(-)(aq)#, #" "E_(red)^@ = "1.36 V"#

Since #E_(red)^@("Au") > E_(red)^@("Cl"_2)#, gold would rather be reduced than oxidized. Thus, the overall reaction at standard conditions is:

#2("Au"^(3+)(aq) + cancel(3e^(-)) -> "Au"(s))#, #" "E_(red)^@ = "1.50 V"#
#3(2"Cl"^(-)(aq) -> "Cl"_2(g) + cancel(2e^(-)))#, #" "E_(o x)^@ = -"1.36 V"#
#"-------------------------------------------"#
#2"Au"^(3+)(aq) + 6"Cl"^(-)(aq) -> 3"Cl"_2(g) + 2"Au"(s)#

And its standard cell potential, i.e. #E_(cell)^@# (and NOT #E_(cell)# as you had stated!) is:

#color(blue)(ul(E_(cell)^@)) = E_(red)^@ + E_(o x)^@#

#= "1.50 V" + (-"1.36 V") = ul(color(blue)(+"0.14 V"))#

For this reaction, we have the reaction quotient, which is written just like #K# but with current concentrations/partial pressures,

#Q = ((P_("Cl"_2(g))//P^@)^3)/((["Au"^(3+)(aq)]//c^@)^2(["Cl"^(-)]//c^@)^6)#

where #P^@ = "1 bar"# and #c^@ = "1 M"# such that #Q# is unitless. It must be, or else it cannot be operated on by #ln#.

We can immediately obtain the reaction quotient with the given data:

#color(blue)(ulQ) = ("0.5 bar"//"1 bar")^3/(("0.25 M"//"1 M")^2("0.10 M"//"1 M")^6)#

#= color(blue)(ul(2 xx 10^6))#

And that allows us to finally get the nonstandard cell potential. You didn't tell us the temperature, but I am guessing it is #"298.15 K"#.

#color(blue)(ul(E_(cell))) = "0.14 V" - (("8.314472 V"cdot"C/mol"cdot"K")("298.15 K"))/(("6 mol e"^(-)/("1 mol atom"))("96485 C/mol e"^(-)))ln(2 xx 10^6)#

#=# #"0.07787 V"#

#~~# #color(blue)(ul("0.07785 V"))#

And indeed it is at #"298.15 K"#. The answer wasn't exact because we were either using slightly different Faraday constants or universal gas constants, or your temperature was more like #"298 K"#.