Question

What is the voltage of the voltaic cell Zn|Zn2+||Cu2+|Cu at 298 K if [Zn2+] = 0.2 M and [Cu2+] = 4.0 M? Cu2+ + 2e- → Cu Eo = +0.34V.

Answer 1

Answer : The voltage of the voltaic cell is 1.14 V

Explanation :

From the given cell representation, we conclude that

The copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction:  $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$E^o_{cell}=(+0.34V)-(-0.76V)$

$E^o_{cell}=1.1V$

Now we have to calculate the emf or voltage of the cell.

Using Nernest equation at 298 K :

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

where,

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

$E_{cell}$ = ?

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

$E_{cell}=1.1-\frac{0.0592}{2}\log \frac{0.2}{4.0}$

$E_{cell}=1.14V$

Therefore, the voltage of the voltaic cell is 1.14 V