# Find the equilibrium constant for a 3-electron transfer process, whose standard emf is #"0.59 V"# at #"298.15 K"#? #F = "96485.33 C/mol e"^(-)#

##
#a)# #10^15#

#b)# #10^20#

#c)# #10^25#

#d)# #10^30#

##### 1 Answer

Jun 28, 2016

The standard EMF (electromotive "force") is

#\mathbf(DeltaG = DeltaG^@ + RTlnQ)#

#\mathbf(DeltaG^@ = -nFE_"cell"^@)# where:

#DeltaG# is theGibbs' free energyfor the process.#DeltaG^@# is the Gibbs' free energy for the processat#mathbf(25^@ "C")# and#\mathbf("1 bar")# pressure.#R = "8.314472 J/mol"cdot"K"# is theuniversal gas constant.#T# is thetemperatureat which the process occurs, in#"K"# . For a standardthermodynamicprocess we assume it is#"298.15 K"# .#n# is the#"mol"# s of electrons transferred.#"F"# isFaraday's constant, which is approximately#"96485.33 C/mol e"^(-)# .#E_"cell"^@# is in#"V"# , and#"1 V"cdot"C" = "1 J"# .

Since we are finding

#DeltaG^@ = -RTlnK_"eq" = -nFE_"cell"^@#

#= e^(("3 mols e"^(-)"/1 mol atom"cdot"96485.33 C/mol e"^(-)cdot"0.59 V")"/"("8.314472 J/mol"cdot"K"cdot"298.15 K")#

#= e^68.89#

#~~ color(blue)(8.3xx10^(29))#

I found the actual answer to this elsewhere, and all they give is