Question

A zinc rod is dipped in 0.1M solution of $${\text{ZnS}}{{\text{O}}_{\text{4}}}$$. The salt is 95% dissociated at this dilution 298K. calculate the electrode potential. [$${{\text{E}}^{\text{o}}}_{{\text{Z}}{{\text{n}}^{{\text{ + + }}}}{\text{/Zn}}}{\text{ = - 0}}{\text{.76eV}}$$]

Answer 1

Hint: The standard electrode potential is used for measuring the potential for equilibrium. The potential of the electrode is defined as the potential difference present between the electrode and electrolyte. When the concentration of all the species is equal to unity involved in a semi cell the potential of electrodes is known as standard electrode potential.

Complete step by step answer:
The concentration of the zinc(II) ion in aqueous will be the following:
Concentration of zinc (II) = $$0.1 + \dfrac{{95}}{{100}} = 0.095$$
Now look at the following equation :
$${\text{Z}}{{\text{n}}^{{\text{2 + }}}}{\text{(aq) + 2}}{{\text{e}}^{\text{ - }}} \to {\text{Zn}}$$
So the formula for the Nernst equation is the following:
$${\text{E = }}{{\text{E}}^{\text{o}}}{\text{ + }}\dfrac{{{\text{0}}{\text{.0591}}}}{{\text{n}}}{\text{log}}\dfrac{{{\text{[Z}}{{\text{n}}^{{\text{2 + }}}}{\text{(aq)]}}}}{{{\text{[Zn]}}}}$$
The value of $${{\text{E}}^{\text{o}}}$$is -0.76eV
The value of n is 2 as two electrons are needed to convert zinc (II) into zinc.
The value for $${\text{[Z}}{{\text{n}}^{{\text{2 + }}}}{\text{(aq)]}}$$is 0.095
The value for [Zn] is 1
Now substituting the value in the formula is :
$${\text{E = - 0}}{\text{.76 + }}\dfrac{{{\text{0}}{\text{.0591}}}}{2}{\text{log}}\dfrac{{0.095}}{1}$$
$${\text{E = - 0}}{\text{.76 + 0}}{\text{.02953}} \times {\text{( - 1}}{\text{.0223)}}$$
$${\text{E = - 0}}{\text{.76 - 0}}{\text{.03021}}$$
$${\text{E = - 0}}{\text{.79V}}$$
The electrochemical cells are based on the concept of redox reactions which is made up of the two half reactions. The one of the half reactions is oxidation which usually occurs at anode and it is done by losing the electrons. The reduction reaction occurs at the cathode which occurs by gaining the electrons.

So, the value for the electrode potential is -0.79V.

Note: The substances which have high standard reduction potential are considered as good oxidising agents. Whereas the substances which have low standard reduction potential are considered as good reducing agents. The oxidation potential of the electrode is considered to be the negative of the reduction potential.