Question

Copper electrodes are placed into two aqueous solutions
of copper(II) sulfate at 25 ºC. One compartment contains
a 1.0 M solution while the other compartment contains a
0.10 M solution. The two compartments are connected
with a salt bridge and the electrodes are connected by a
wire passing through a voltmeter. In what direction do
the electrons flow through the wire, and what is the cell
potential read on the voltmeter?

Answer 1

Given :

The temperature at which the copper electrodes are placed into two aqueous solutions of copper(||) sulfate is = 25°C

Molarity of solution in 1st compartment = 1.0 M

Molarity of solution in 2nd compartment = 0.10 M

To Find :

The direction in which the electrons flow through the wire passing trough the voltmeter ?

The cell potential read on the voltmeter =?

Solution :

∴Since the 1st half cell (A) has concentration = 1 M

And the 2nd half cell (B) has concentration = 0.1 M

So, the electrode reduction potential for the A will be higher . Therefore reduction will take place at A .

So, now we can write reactions at anode and cathode half cells as :

At anode half cell i.e. B :

$Cu \rightarrow Cu^{2+}+ 2e^-$

At cathode half cell i.e. A :

$Cu^{2+} + 2e^- \rightarrow Cu$

So, electron will flow from cell B to A.

And the cell potential will be given  according to Nerst's equation as  :

$E_{cell } = E^0 _{cell} - \frac{0.059}{n} \times \log(\frac{Cu^{2+}_{anode}}{Cu^{2+}_{cathode}} )$

Here  n = no of electrons  transferred which is = 2

And since given cell is a concentration cell , so $E^0 _{cell} = 0$

So, on further putting the values the Nerst's equation becomes :

$E_{cell} =\frac{-0.059}{2} \log(\frac{0.1}{1} )$

       $= \frac{0.059}{2} V$

       = 0.0295 v

So, the value cell potential read on the voltmeter is 0.0295 V and the flow of electrons is from cell B to cell A.