Question

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` ?

Answer 1

`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)`