Galvanic cell run at 25° C: calculate #\DeltaG# when #K=2.79xx10^7#?

#DeltaG = -"29.10 kJ/mol"#

This is just a way to get you to work more with the shift in #DeltaG# from standard conditions:

#DeltaG = DeltaG^@ + RTlnQ#

where #@# indicates standard conditions, like #"1 M"# concentrations, #"1 atm"# partial pressures, etc. for a given temperature.

If we were to have been at equilibrium (which we're not), then

#cancel(DeltaG)^(0) = DeltaG^@ + RTlncancel(Q)^(K)#

Thus, assuming this is #K_c#,

#DeltaG^@ = -RTlnK_c#

#= -8.314 cancel"J""/mol"cdotcancel"K" xx "1 kJ"/(1000 cancel"J") cdot 298.15 cancel"K" cdot ln(2.79 xx 10^7)#

#= -"42.50 kJ/mol"#

Next, the reaction quotient #Q# is for the spontaneous reaction:

#2("Cr"^(2+)(aq) -> "Cr"^(3+)(aq) + e^(-))#, #E_(red)^@ = -"0.41 V"#
#ul("Co"^(2+)(aq) + 2e^(-) -> "Co"(s))#, #E_(red)^@ = -"0.277 V"#
#2"Cr"^(2+)(aq) + "Co"^(2+)(aq) -> "Co"(s) + 2"Cr"^(3+)(aq)#

We don't need #E_(cell)^@#, but we did need to figure out what direction was spontaneous. Since #E_(red)^@# for #"Cr"^(2+ ->3+)# was more negative, that oxidation is spontaneous and we chose correctly.

(Also, #"Co"^(2+)# is more likely than #"Co"^+#, so that has the #"0.20 M"# concentration...)

#Q = (["Cr"^(3+)]^2)/(["Cr"^(2+)]^2["Co"^(2+)])#

#= ("2.0 M"//"1 M")^2/(("0.30 M"//"1 M")^2("0.20 M"//"1 M"))#

#= 222.22#

Finally,

#color(blue)(DeltaG) = overbrace(-"42.50 kJ/mol")^(DeltaG^@) + overbrace("0.008314 kJ/mol"cdotcancel"K" cdot 298.15 cancel"K" cdot ln(222.22))^(RTlnQ)#

#=# #color(blue)(-"29.10 kJ/mol")#

So even though this reaction is less spontaneous with these concentrations at this temperature, it is still spontaneous.