Consider the reaction #2AgCl+Cu^(2+)rarr2AgCl+Cu#. Given that #DeltaG_("rxn")=-22.kJ#, How do you find the value for #E_(rxn)^@# ?

1 Answer
Aug 3, 2017

Answer:

You can do it like this:

Explanation:

The expression you need is :

#sf(DeltaG^@=-nFE_(rxn)^@)#

Where n is the no. of moles of electrons transferred which, in this case = 2.

#sf(F)# is the charge on 1 mole of electrons and is the Faraday Constant which = #sf(9.65xx10^(4)color(white)(x)"C/mol")#

#sf(DeltaG^@=-nFE_(rxn)^@)#

#:.##sf(-22.2xx10^(3)=-2xx9.65xx10^(4)xxE_(rxn)^@)#

#sf(E_(rxn)^@=(-22.2xx10^3)/(-2xx9.65xx10^4)=+0.115color(white)(x)V)#

We can check this by comparing standard electrode potentials which can be looked up:

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#

#sf(AgCl_((s))+erightleftharpoonsAg_((s))+Cl_((aq))^(-)" "E^@=+0.2223color(white)(x)V)#

#sf(Cu_((aq))^(2+)+2erightleftharpoonsCu_((s))" "E^@=+0.34color(white)(x)V)#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)#

The most +ve half cell will take in the electrons so the half - reactions will proceed in the directions shown by the arrows.

#sf(E_(rxn)^@)# is the arithmetic difference between the two 1/2 cells which you get by subtracting the least +ve #sf(E^@)# from the most +ve #rArr#

#sf(E_(rxn)^(@)=+0.34-(+0.2223)=+0.1177color(white)(x)V)#

This agrees ok with the calculated value.