Question

For a spontaneous reaction $\Delta {G^0}$ and ${E^0}$ of cell will be respectively
A) \[-ve\] and \[ + ve\]
B) \[ + ve\] and \[-ve\]
C) \[ + ve\] and \[ + ve\]
D) \[-ve\] and \[-ve\]

Answer 1

Hint: $\Delta {G^0}$ and ${E^0}$ of cell will be respectively represent criterion for spontaneity of a cell reaction .A spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly.

Complete step by step answer:
According to the second law of thermodynamics, any spontaneous process must increase the entropy in the universe.
\[G{\text{ }} = {\text{ }}H{\text{ }}-{\text{ }}TS\]
Gibbs free energy is represented using the symbol G and typically has units of $kJ/mole$
if , for a reaction, $\Delta G$ is negative, then the reaction will be spontaneous
$\Delta G = \Delta H - T\Delta S$
\[\Delta G < 0\] , the process is exergonic and will proceed spontaneously in the forward direction to form more products
If ${E^0}cell$ is positive, cell reaction will be spontaneous.
Following reaction occur in the Daniel cell.
 (i)Zinc goes into solution as zinc ions in the vessel on the left-hand side.
\[Zn(s)\,\, \to \,\,Z{n^{2 + }}(aq) + 2{e^ - }\]
Since this reaction involves loss of electrons, it is known as oxidation half reaction. Since there are excess of electrons on the electrode, it is negatively charged and known as negative electrode or anode.
(ii) The electrons left behind on the anode (zinc electrode) flow through the external wire to the copper electrode where they react with the copper ions and form neutral copper atoms which are deposited on copper electrode on the right hand side.
\[C{u^{2 + }}(aq) + 2{e^ - }\,\, \to \,\,Cu(s)\]
Since this reaction involves gain of electrons, it is known as reduction half reaction. It is the cathode or positive electrode where reduction takes place.
\[Zn(s) + C{u^{2 + }}(aq)\,\,\, \to \,\,Z{n^{2 + }}(aq.) + Cu(s)\]
\[{E_{ox}} = E_{ox}^o - \dfrac{{2.303\;RT}}{{nF}}\log \;[{M^{n + }}]\]
\[{E_{red}} = E_{red}^o - \dfrac{{2.303\;RT}}{{nF}}\log \dfrac{1}{{[{M^{n + }}]}}\]
With the usual above conventions, the emf of the cell will have a positive value.
So,Option “A” is correct.

Note: Using a voltmeter to measure the electrical potential difference between two electrodes provides a quantitative indication of just how spontaneous a redox reaction is. The potential difference is measured in volts.