Question

How does the concentration of electrolyte affect an electrochemical cell?

Answer 1

The concentration effect in an electrochemical cell is described by The Nernst Equation.

I will show how this works by this example:

`Zn_((s))+Cu_((aq))^(2+)rarrZn_((aq))^(2+)+Cu_((s))`

`E_(cell)^(0)=-1.1V`

This is the emf of the cell when operating under standard conditions I.e 1 Atmosphere, 298K and unit concentration. If we alter these conditions, in this case, concentration the Nernst Equation can be used to find the new emf.

The equation is:

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

R= Gas Constant = 8.31JK/mol.
T = absolute temperature.
F = Faraday Constant = `9.64xx10^4C.mol.^(-1)`
Z = no. moles electrons transferred.
Q = Reaction Quotient. This = `K_c` when the reaction has reached equilibrium.

Let's say `[Zn^(2+)] = 1mol.l^(-1)` and `[Cu^(2+)] = 2.0 mol.l^(-1)`

`K_c =([Zn^(2+)]_e)/([Cu^(2+)]_e)`

The concentrations are those at equilibrium. At this point current has been drawn from the cell such that the potential difference between the 2 half cells is zero so `DeltaG=0` and Q =`K_c`.
However cell emf's are measured such that no current is drawn from the cell and the concentrations at equilibrium can be assumed to be the same as those at the start.

So we can use those initial concentrations in the Nernst Equation:

`E_(cell)=E_(cell)^(0)-(RT)/(zF)` `ln([[Zn^(2+)]]/[[Cu^(2+)]))`

`E_(cell)= -1.1-8.31xx298/(2xx9.64xx10^(4)` `ln(1/2)`

`E_(cell)=-1.1-(-0.0089) = -1.09V`