Consider the 2 half - cells : #sf(Cu^(2+)+2erightleftharpoonsCu)# #sf(E^(@)=+0.34"V")# #sf(Ag^(+)+erightleftharpoonsAg)# #sf(E^(@)=+0.80"V")# What is the solution to the points below?

What is the standard emf of the cell formed from the two 1/2 cells ?
What is the emf of the cell formed from the two half - cells if #[Cu^(2+)]=0.130M# and #[Ag^(+)]=0.0001M# ?

1 Answer
Nov 25, 2016

Answer:

#sf(E_(cell)^@=+0.46color(white)(x)V)#

#sf(E_(cell)=+0.25color(white)(x)V)#

Explanation:

List the 1/2 equations in order least positive to most positive:

#sf(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx)E^(@)color(white)(xx)(V))#

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

#sf(Cu^(2+)" "+" "2e" "rightleftharpoons" "Cucolor(white)(xxxxxxxxxxxxxxxxx)+0.34)#

#sf(Ag^(+)" "+" "e" "rightleftharpoons" "Agcolor(white)(xxxxxxxxxxxxxxxxxx)+0.80)#

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

Note we use the #rightleftharpoons# symbol to show that the 1/2 cells can go in either direction depending on what they are coupled with.

The 1/2 cell with the most +ve #"E"^@# is the one which will take in the electrons.

From this we can see that the 2nd 1/2 cell will be driven left to right and the 1st 1/2 cell right to left in accordance with the arrows.

So the cell reaction when it is working is:

#sf(Cu+2Ag^+rarrCu^(2+)+2Ag)#

Because #sf(E_(cell)^@)# is an empirically measured quantity it always has a +ve value. So to calculate it just subtract the least +ve value from the most +ve:

#sf(E_(cell)^@=+0.80-(+0.34)=+0.46color(white)(x)V)#

To find the emf of the cell under non - standard condition we need to use the Nernst Equation. At #sf(25^@C)# this simplifies down to:

#sf(E_(cell)=E_(cell)^@-(0.05916)/(z)logQ)#

#sf(Q)# is the reaction quotient which, in this case is given by:

#sf(Q=([Cu^(2+)])/([Ag^(+)]^2))#

#sf(z)# is the no. moles of electrons transferred which, in this case = 2.

Putting in the numbers:

#sf(E_(cell)=+0.46-(0.05916)/(2)log[[0.130]/[(10^(-4))^(2)]]" "V)#

#sf(E_(cell)=+0.46-(0.02958xx7.1139)" "V)#

#sf(E_(cell)=+0.46-0.2104=0.25color(white)(x)V)#