Question

The voltage generated by the zinc concentration cell described by the line notation zn(s) ∣∣ zn2+(aq,0.100 m) ∥∥ zn2+(aq,? m) ∣∣ zn(s) is 14.0 mv at 25 °c. calculate the concentration of the zn2+(aq) ion at the cathode.

Answer 1

Answer: The concentration of $Zn^{2+}$ ion at cathode is 0.295 M

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):  $Zn(s)\rightarrow Zn^{2+}(0.100M,aq.)+2e^-$

Reduction half reaction (cathode):  $Zn^{2+}(?M,aq.)+2e^-\rightarrow Zn(s)$

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

$E_{cell}$ = 14.0 mV = 0.014 V    (Conversion factor:  1 V = 1000 mV)

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

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

Putting values in above equation, we get:

$0.014=0-\frac{0.0592}{2}\log \frac{0.100M}{[Zn^{2+}]_{cathode}}$

$[Zn^{2+}]_{cathode}=0.295M$

Hence, the concentration of $Zn^{2+}$ ion at cathode is 0.295 M

Answer 2

Reaction involved in present electrochemical cell,
At Anode: Zn     →    Zn^2+      +       2e^2-
At cathode:      Zn^2+      +       2e^2-      →      Zn
Net Reaction: Zn + Zn^2+ ('x' m)  →   Zn^2+(0.1 m) + Zn
Number of electrons involved in present electrochemical cell = n = 2

According to Nernst equation for electrochemical cell,
Ecell = -2.303 $\frac{RT}{nT} log \frac{[Zn^2+]R}{[Zn^2+]L}$ = 0.014

Given: T = $25^{0}C = 298 K$, F = 96500 C, R = gas constant = 8.314J/K.mol, [Zn^2+]R = 0.1 m , Ecell = 0.014 v

∴ 0.014 = - 2.303 $\frac{8.314X298}{2X96500}log \frac{0.1}{x}$
∴ log$\frac{0.1}{x}$ = $\frac{-2.303X8.314X298}{2X96500X0.014}$ = -2.1117
∴ log x = log(0.1) + 2.1117
∴x = 13.09 m