# 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)#?

##
Ecell=0.14V

Q=2x106

Ecell=0.07785V

Ecell=0.14V

Q=2x106

Ecell=0.07785V

##### 1 Answer

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

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 **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.

#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

#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

#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