What are concentration cells? Give examples.

Concentration cells in general are electrochemical cells that have different concentrations of the same species in solution. For simplicity, let's stick to voltaic cells, because we'll make it more complex later.

EXAMPLE

Consider a copper cell with #"2.0 M"# on the left and #"0.05 M"# on the right.

#|" "" "" "ul|color(white)(/./)|" "color(white)(./)|" "" "" "" "" "|" "" "" "ul|color(white)(/./)|" "" "color(white)(.)|#
#ul|" "" ""2.0 M Cu"^(2+)" "|" "" "" "" "" "ul|" "" ""0.05 M Cu"^(2+)" "|#

Both beakers have a copper metal strip in them, and a salt bridge connects them (say, with #"KNO"_3#).

Big takeaways early on:

• If the concentrations were the same, then clearly, #E_(cell) = 0# (why?). Furthermore, #E_(cell)^@ = 0# already, as both species are identical.
• Since the concentration on the right is lower, the electron flow is such that concentration decreases on the left and increases on the right. This is a so-called "concentration gradient".

Due to the fact that concentration flows from high to low, and that these are cations,

• loss of #"2.0 M"# #"Cu"^(2+)# is due to taking in electrons to form #"Cu"(s)# on the left (decreasing #["Cu"^(2+)]# on the left). Thus, this is the reactant.
• gain in #"0.05 M"# #"Cu"^(2+)# is due to losing electrons from #"Cu"(s)# on the right (increasing #["Cu"^(2+)]# on the right). Thus, this is the product.

Hence, since electrons flow into the cathode (always), the electron flow is counterclockwise in this case, and the cathode is on the left, at the #2.0# #M# #Cu^(2+)# beaker.

This means the spontaneous direction for the reaction is:

#"Cu"^(2+)(aq)("2.0 M") -> "Cu"^(2+)(aq)("0.05 M")#

The cell potential at #"298.15 K"# in this case, based on the Nernst equation, is:

#color(blue)(E_(cell)) = cancel(E_(cell)^@)^(0) - (RT)/(nF)lnQ#

#= -("8.314472 V"cdotcancel"C""/"cancel"mol"cdotcancel"K" cdot 298.15 cancel"K")/((2 cancel("mol e"^(-)))/(1 cancel"mol Cu") cdot 96485 cancel"C""/"cancel("mol e"^(-))) ln \frac("0.05 M")("2.0 M")#

And it's good that this is going to be a positive value; since it is, we have correctly chosen the spontaneous reaction direction.

#=# #color(blue)("0.0474 V")#

MAKING IT COMPLEX?

Okay, what if the #"2.0 M"# was changed to #10^(-3) "M"#?

#|" "" "" "ul|color(white)(/./)|" "color(white)(./)|" "" "" "" "" "|" "" "" "ul|color(white)(/./)|" "" "color(white)(.)|#
#ul|" "color(white)(.)10^(-3)"M Cu"^(2+)" "|" "" "" "" "" "ul|" "" ""0.05 M Cu"^(2+)" "|#

I predict that the roles of cathode and anode get reversed. Now we have the right-hand beaker at the higher concentration, not lower.

Hence, the cathode is now on the right (at the #0.05# #M# beaker) and the anode is now on the left (at the #10^(-3)# #M# beaker).

This means the new spontaneous direction for the reaction is:

#"Cu"^(2+)(aq)(10^(-3)"M") larr "Cu"^(2+)(aq)("0.05 M")#

And now, the new cell potential becomes:

#color(blue)(E_(cell)') = cancel(E_(cell)^@)^(0) - (RT)/(nF)lnQ#

#= -("8.314472 V"cdotcancel"C""/"cancel"mol"cdotcancel"K" cdot 298.15 cancel"K")/((2 cancel("mol e"^(-)))/(1 cancel"mol Cu") cdot 96485 cancel"C""/"cancel("mol e"^(-))) ln \frac(10^(-3) "M")("0.05 M")#

#=# #color(blue)("0.0503 V")#

Notice that the #"0.05 M"#, which was previously on the top, is now on the bottom. This is because now the reaction flows the opposite direction to the original example.

In short, the smaller concentration goes on top, to guarantee that the concentration cell reaction is chosen in the spontaneous direction.