Question

A galvanic cell has two solutions: solution one is 0.3 M FeCl2, solution two is 1.7 M MgCl2. The electrode in solution one is Fe and the electrode in solution two is Mg. The temperature is 85 °C. What is the spontaneous cell potential in this setup?

Answer 1

The cell potential is +1.77 V.

First of all work out the emf of the cell under standard conditions:

I've looked up the standard electrode potentials for each half - cell and these are:

`Mg_((aq))^(2+)+2e^(-)rightleftharpoonsMg_((s))` `E^0=-2.36 V`

`Fe_((aq))^(2+)+2e^(-)rightleftharpoonsFe_((s))` `E^0=-0.44V`

To work out `E_(cell)^(0)` we subtract the least positive from the most positive so:

`E_(cell)^(0)=-0.44-(-2.36)=1.92V`

We can see that the `Mg` half - cell has the more negative value so electrons will flow from the `Mg` to the `Fe` half - cell.

So the cell reaction is :

`Mg_((s))+Fe_((aq))^(2+)rarrMg_((aq))^(2+)+Fe_((s))`

The question is concerned with non - standard conditions which are much more prevalent in the real world. To help us here we can use The Nernst Equation. A useful form is :

`E_(cell)=E_(cell)^(0)-((2.303RT)/(nF))logQ`

Here are their definitions:

R is the Gas Constant and = 8.31J/K/mol
T is the absolute temperature in Kelvin
n is the number of moles transferred in the reaction, in this case 2
F is the Faraday Constant and is the amount of charge carried by a mole of electrons. It has the value of `9.65`x`10^(4)C.mol^(-1)`
`Q` is the reaction quotient which is analogous to `K_c` and in this case is given by

`([Mg_((aq))^(2+)])/([Fe_((aq))^(2+)]`.

The concentrations used are initial concentrations and not equilibrium ones as are normally used . We can make this approximation as we are measuring the emf of the cell, so no current is being drawn from it and so the initial concentrations don't differ from those at equilibrium.

Now we put in the values from the question:

T = 85 deg C = 383K

`E_(cell)=1.92-(((2.303).(8.31)(383))/(2.(9.65x10^4)))log(1.7/(1.3))`

`E_(cell)=1.92-0.152=1.77V`