Question

The voltage generated by the zinc concentration cell described by the line notation Zn ( s ) ∣ ∣ Zn 2 + ( aq , 0.100 M ) ∥ ∥ Zn 2 + ( aq , ? M ) ∣ ∣ Zn ( s ) is 25.0 mV at 25 °C . Calculate the concentration of the Zn 2 + ( aq ) ion at the cathode.

Answer 1

Answer: The concentration of $Zn^{2+}$ ion at cathode is 0.704 M

Explanation:

The half reactions for the cell is:

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, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

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

where,

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

$E_{cell}$ = 25.0 mV = 0.025 V   (Conversion factor: 1 V = 1000 mV)

$[Zn^{2+}_{cathode}]$ = ? M

$[Zn^{2+}_{anode}]$ = 0.100 M

Putting values in above equation, we get:

$0.025=0-\frac{0.0592}{2}\log \frac{0.100}{[Zn^{2+}_{anode}]}$

$[Zn^{2+}_{anode}]=0.704M$

Hence, the concentration of $Zn^{2+}$ ion at cathode is 0.704 M