Question

A voltaic cell is constructed with two Zn2+-Zn electrodes, where the half-reaction is Zn2+ + 2e− → Zn (s) E° = -0.763 V The concentrations of zinc ion in the two compartments are 4.50 M and 1.11 ⋅ 10−2 M, respectively. The cell emf is ________ V.

Answer 1

Answer : The cell emf for this cell is 0.077 V

Solution :

The balanced cell reaction will be,  

Oxidation half reaction (anode):  $Zn(s)\rightarrow Zn^{2+}+2e^-$

Reduction half reaction (cathode):  $Zn^{2+}+2e^-\rightarrow Zn(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}{diluted}}{[Zn^{2+}{concentrated}]}$

where,

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

$E_{cell}$ = ?

$[Zn^{2+}{diluted}]$ = 0.0111 M

$[Zn^{2+}{concentrated}]$ = 4.50 M

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

$E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0111M}{4.50M}$

$E_{cell}=0.077V$

Therefore, the cell emf for this cell is 0.077 V