Question

Determine the mass of the metal at the cathode of the following galvanic cell when the cell is “dead” (at 298 K) assuming that the initial mass of both metal electrodes are 100 g and all solutions in the cell are exactly 1.0 L. Fe | Fe2+ (0.10 M) || Cd2+ (0.95 M) | Cd

Answer 1

Answer : The mass of the Cd at the cathode will be, 201.810 grams

Solution :

From the given cell we conclude that, the iron (Fe) undergoes oxidation by loss of electrons and thus act as anode. Cadmium (Cd) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode : $Fe\rightarrow Fe^{2+}+2e^-$     $E^0_{[Fe^{2+}/Fe]}=-0.44V$

Reaction at cathode : $Cd^{2+}+2e^-\rightarrow Cd$     $E^0_{[Cd^{2+}/Cd]}=-0.40V$

The balanced cell reaction will be,  

$Fe(s)+Cd^{2+}(aq)\rightarrow Fe^{2+}(aq)+Cd(s)$

First we have to calculate the standard electrode potential of the cell.

$E^0_{[Fe^{2+}/Fe]}=-0.44V$

$E^0_{[Cd^{2+}/Cd]}=-0.40V$

$E^0=E^0_{[Cd^{2+}/Cd]}-E^0_{[Fe^{2+}/Fe]}$

$E^0=(-0.40V)-(-0.44V)=0.04V$

Now we have to calculate the concentration of $Fe^{2+}\text{ and }Cd^{2+}$.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]}{[Cd^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = 0  (for dead cell)

Now put all the given values in the above equation, we get:

$0=0.04-\frac{0.0592}{2}\log \frac{[Fe^{2+}]}{[Cd^{2+}]}$

$\frac{[Fe^{2+}]}{[Cd^{2+}]}=22.695$

Let the concentration of $Cd^{2+}$ be, 'x'.

So, the concentration of $Fe^{2+}$ will be, $22.695\times [Cd^{2+}]=22.695x$

From this we conclude that,

$[Cd^{2+}]+[Fe^{2+}]=x+22.695x$

$0.95+.010=x+22.695x$

By solving the term 'x', we get:

$x=0.0443M$

Initial concentration of $Cd^{2+}$ = 0.95 M

Final concentration of $Cd^{2+}$ = 0.0443 M

The number of moles of $Cd^{2+}$ deposited = Initial concentration - final concentration

The number of moles of $Cd^{2+}$ deposited = 0.95 - 0.0443 = 0.9057 mole

Now we have to calculate the mass of Cd deposited.

$\text{Mass of Cd deposited}=\text{Moles of Cd deposited}\times \text{molar mass of Cd}=0.9057mole\times 112.411g/mole=101.810g$

Now we have to calculate the mass of the metal at the cathode.

The final mass of the Cd at the cathode = 100 + 101.810 = 201.810 g

Therefore, the mass of the Cd at the cathode will be, 201.810 grams