Question

The electrode potential for the Daniell cell given is $1.1V$
$Zn(s)/Z{n^{2 + }}(aq)\parallel C{u^{2 + }}(aq)/Cu(s)$
Write overall cell reaction and calculate the standard Gibbs energy for the reaction [$F{\text{ }} = {\text{ }}96487{\text{ }}c/mol$]

Answer 1

Gibbs free energy – It is an energy required to do work. The decrease of Gibb’s (-$\Delta G$) energy (-) of the cell reaction provides a measure of electrical work.

Complete step by step answer:
From the given cell representation, we can see that zinc is the anode and copper is the cathode. At anode oxidation takes place and at cathode reduction takes place.
Therefore, the oxidation half reaction is :
$Zn(s) \to Z{n^{2 + }}(aq) + 2{e^ - }$
The reduction half reaction is :
$C{u^{2 + }}(aq) + 2{e^ - } \to Cu(s)$
Now , in order to obtain overall cell reaction we add these two reactions . Therefore the overall cell reaction is :
$Zn(s) + C{u^{2 + }}(aq) \to Z{n^{2 + }}(aq) + Cu(s)$
The value of standard Gibb’s energy change is:
$\Delta {G^o} = - nF{E^O}cell$
n = Number of electron exchanged in the given reaction
F = Faraday constant its value is 96487 C
${E^0}cell$ = Standard EMF of the cell
$\Delta {G^0} = - 2 \times 96487 \times 1.1$
$= - 212,271.4J$
$= 212.27KJ$

Note: (1) Gibb’s free energy – It is energy required to do work . The decrease of Gibb’s energy (-$\Delta G$) of the cell reaction provides a measure of electrical work (${W_{elec}}$) .
(2) The difference in the electrode potential of the two electrodes of the cell is termed as electromotive force (or EMF) or cell voltage$({E_{cell}})$.
$EMF = {E_{red}}(Cathode) - {E_{red}}(Anode)$
Or
${E_{cell}} = {E_{Cathode}} - {E_{Anode}}$
(3) (i) Numerical values of reduction potential and oxidation potential are equals . But they are opposite in sign. The reduction potential is defined by tendency to gain electrons, whereas oxidation potential is described by tendency to lose electrons.
(ii) Reduction potential increases when the concentration of ions increases. It decreases when the concentration of the ions in solution decreases.