Question

Consider the cell Cu/Cu^+2||Ag^+/Ag.If the concentration of Cu^+2 and Ag^+ ions becomes ten times, the emf of the cell is?

Answer 1

`sf(E_(cell)=+0.49color(white)(x)V)`

Explanation:

The cell diagram implies that we are under standard conditions where we have unit concentration which is then increased by a factor of 10.

You should be given the `sf(E^@)` values:

`sf(Cu^(2+)+2erightleftharpoonsCucolor(white)(xxxxxxx)E^@=+0.34color(white)(x)V)`

`sf(Ag^(+)+erightleftharpoonsAgcolor(white)(xxxxxxxx)E^@=+0.8color(white)(x)V)`

The `sf(Cu|Cu^(2+))` half cell has the least +ve value so this will shift right to left and give out electrons.

The cell reaction will be:

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

`sf(E_(cell)^@=+0.8-0.34=+0.46color(white)(x)V)`

The Nernst Equation shows the effect of concentration:

`sf(E_(cell)=E_(cell)^@-(RT)/(zF)lnQ)`

This can be simplified at `sf(25^@C)` to:

`sf(E_(cell)=E_(cell)^@-0.0591/zlogQ)`

This becomes:

`sf(E_(cell)=E_(cell)^@-0.0591/(2)logcolor(white)(x){[Cu^(2+)]]/[Ag^(+)]^2)`

Under standard conditions we have unit concentrations such that `sf(E_(cell)=E_(cell)^@)`.

Now we increase the concentrations by a factor of 10 `sf(rArr)`

`sf(E_(cell)=+0.46-0.0591/(2)logcolor(white)(x)[10/(10^2)])`

`sf(E_(cell)=+0.46+0.02955=+0.49color(white)(x)V)`

As you can see there is only a slight increase in the emf of the cell but, since you have increased the amount of materials, you will be able to draw more current from the cell.