Question #2eac3

1 Answer
Feb 6, 2016

Answer:

#(a)#

#E_(cell)^@=+0.32"V"#

#(b)#

#E_(cell)=+0.37"V"#

Explanation:

(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.