#E_(react)=-2.18"V"#
The proposed reaction is:
#2Au_((s))+3Zn_((aq))^(2+)rarr2Au_((aq))^(3+)+3Zn_((s))#
So the 1/2 equations are :
#Zn_((aq))^(2+)+2erarrZn_((s))# #E_1=-0.76"V"#
#Au_((s))rarrAu_((aq))^(3+)+3e# #E_2=-1.42"V"#
So total potential = #E_1+E_2=-1.42+(-0.76)=-2.18"V"#
#DeltaG=-nFE_(react)=-nF(-2.18)#
This means #DeltaG# is +ve so the reaction is not feasible.
A preferred approach is to use standard electrode potentials which should be listed negative to positive:
#Zn^(2+)+2erightleftharpoons Zn# #E^(0)=-0.76"V"#
#Au^(3+)+3erightleftharpoonsAu# #E^(0)=+1.42"V"#
Use the rule "Bottom left oxidises top right" or "top right reduces bottom left".
From this we can see that #Au# is not able to reduce #Zn^(2+)# to #Zn# as the Au3+/Au half cell #E^(0)# value needs to be < -0.76V.
