(a)
List the standard electrode potentials:
#color(white)(xxxxxxxxxxxxxxxxxxx) E^@"V"#
#stackrelcolor(blue)(larr)color(white)(xxxxxxxxxxxxxx)#
#Zn^(2+)+2erightleftharpoonsZn" "-0.76#
#Fe^(2+)+2erightleftharpoonsFe" "-0.44"#
#stackrelcolor(red)(rarr)color(white)(xxxxxxxxxxxxxx)#
Overall reaction:
#Zn_((s))+Fe_((aq))^(2+)rarrZn_((aq))^(2+)+Fe_((s))#
To find #E_(cell)^@# subtract the least +ve #E^@# value from the most +ve:
#E_(cell)^@=-0.44-(-0.76)=+0.32"V"#
(b)
Now that the conditions are non - standard we need to use the Nernst Equation:
#E_(cell)=E_(cell)^@-(RT)/(nF)lnQ#
Where #Q# is the reaction quotient.
#n# is the number of moles of electrons transferred.
At 298K this simplifies down to:
#E_(cell)=E_(cell)^@-(0.05916)/(2)logQ#
#:.E_(cell)=E_(cell)^@-(0.05916)/(2)log([[Zn^(2+)]]/([Fe^(2+)]))#
#:.E_(cell)=+0.32-(0.05916)/(2)log((0.01)/(0.5))#
#E_(cell)=+0.32-(0.5916)/(2)xx(-1.698)#
#E_(cell)=0.32+0.05=+0.37"V"#
You can see that this has increased the value of #E_(cell)# indicating that the reaction has been driven to the right.
This is in accordance with Le Chatelier as #[Zn^(2+)]# has been decreased relative to #[Fe^(2+)]# so shifting the position of equilibrium to the right.
