Calculate the standard cell potential (E∘) for the reaction? X(s)+Y+(aq)→X+(aq)+Y(s) if K = 5.14×10−3.

#E^@ = -"0.1354 V"#

This should therefore be nonspontaneous, which should make sense. Since #K < 1#, the reactants are favored in the reaction, and thus it is not spontaneous.

You would get the same conclusion with #DeltaG^@# at #"298.15 K"#:

#DeltaG^@ = -RTlnK#

#= - cdot overbrace((+))^(R) cdot overbrace((+))^(T) cdot overbrace((-))^(lnK) > 0#

which again means the reaction is nonspontaneous.

Well, recall that a change in #DeltaG# away from standard state is:

#DeltaG = DeltaG^@ + RTlnQ#

At chemical equilibrium though, #DeltaG = 0# and #Q = K#, so:

#DeltaG^@ = -RTlnK# #" "" "bb((1))#

As it turns out, with a little unit conversion, #"J" = "C" cdot "V"#, which means:

#DeltaG^@ = -nFE^@# #" "" "bb((2))#

• The #n# is the mols of electrons per mol of atom.
• #F = "96485 C/mol e"^(-)# is the Faraday constant.
• #E^@# is the overall cell potential in #"V"#.

So the units work out, and we have that #E^@ > 0# is spontaneous at #"298.15 K"#. Therefore, solving #(2)# for #E^@#, we get:

#E^@ = -(DeltaG^@)/(nF)#

Utilizing #(1)#, we get:

#barul(|stackrel(" ")(" "E^@ = (RT)/(nF)lnK" ")|)#

As a result, we have verified the true equation---your equation should have divided by #RT#, not multiplied.

Anyways, therefore, the #E^@# is:

#color(blue)(E^@) = ("8.314472 V"cdotcancel"C""/"cancel"mol atom"cdotcancel"K" cdot 298.15 cancel"K")/(cancel("1 mol e"^(-1))/cancel"1 mol atom" cdot 96485 cancel"C""/"cancel("mol e"^(-))) ln (5.14 xx 10^(-3))#

#=# #color(blue)ul(-"0.1354 V")#