Question

A cell was set up having the following reaction Mg(s) + Cd2+ (aq) → Mg2+ (aq) + Cd (s) E°cell = 1.97 V The Magnesium electrode was dipped in a 1.00 M solution of MgSO4 and the Cadmium electrode was dipped in a solution of unknown Cd2+ concentration. The cell potential was measured to be 1.80 V. What is the unknown Cd2+ concentration?

Answer 1

Answer : The concentration of unknown $Cd^{2+}$ will be, $1.807\times 10^{-6}M$

Solution :

The balanced cell reaction will be,  

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

Here magnesium (Mg) undergoes oxidation by loss of electrons, thus act as anode. Cadmium (Cd) undergoes reduction by gain of electrons and thus act as cathode.

Now we have to calculate the concentration of unknown $Cd^{2+}$.

Using Nernest equation :

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

where,

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

$E_{cell}$ = emf of the cell = 1.80 V

$E^o_{cell}$ = standard cell potential = 1.97 V

$[Mg^{2+}]$ = concentration of magnesium ion = 1.00 M

$[Cd^{2+}]$ = concentration of cadmium ion = ?

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

concentration of unknown $Cd^{2+}$.

$1.80=1.97-\frac{0.0592}{2}\log \frac{(1.00)}{[Cd^{2+}]}$

$[Cd^{2+}]=1.807\times 10^{-6}M$

Therefore, the concentration of unknown $Cd^{2+}$ will be, $1.807\times 10^{-6}M$