The first thing to do is to calculate the emf of the cell under standard conditions.
We can do this using standard electrode potentials (#sf(E^@)#):
List the 1/2 equations in order least positive to most positive:
# " " "E"^@("V")#
#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#
#sf(Sn^(4+)" "+" "2e" "rightleftharpoons" "Sn^(2+)" "+0.15)#
#sf(Au^(3+)" "+" "3e" "rightleftharpoons" "Au" "+1.52)#
#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)#
The more +ve half - cell will take in the electrons so the reactions in each half - cell will proceed in the direction shown by the arrows.
This gives the overall cell reaction:
#sf(2Au^(3+)+3Sn^(2+)rarr2Au+3Sn^(4+))#
To find #sf(E_(cell)^@)# subtract the least +ve electrode potential from the most positive:
#sf(E_(cell)^@=+1.52-(+0.15)=1.37color(white)(x)V)#
Since we are not under standard conditions we now need to use The Nernst Equation.
A useful form of this at #sf(25^@C)# is:
#sf(E_(cell)=E_(cell)^@-(0.05916)/(z)logQ)#
#sf(z)# is the number of moles of electrons transferred which, in this case = 6.
#sf(Q)# is the reaction quotient and is given by:
#sf(Q=([Sn^(4+)]^(3))/([Au^(3+)]^(2)[Sn^(2+)]^(3))#
#:.##sf(Q=((0.5)^(2))/((1.45)^(2)xx(0.87)^(3))=0.0902color(white)(x)"mol"^(-2).l^(2))#
#:.##sf(E_(cell)=1.37-(0.05916)/(6)log[0.0902])#
#sf(E_(cell)=1.37+0.103=+1.38color(white)(x)V)#
