Question

Calculate the half-cell potential at $298K$ for the reaction,
$Z{n^{2 + }} + 2{e^ - } \to Zn$
If $\left[ {Z{n^{2 + }}} \right] = 0.1M$ and ${E^o} = - 0.76volt$
A. $ - 0.847volt$
B. $ - 0.789volt$
C. $ - 0.475volt$
D. $ - 0.366volt$

Answer 1

Hint: If standard electrode potential is known to us, we can easily calculate electrode potentials at different concentrations and temperatures using the Nernst equation. In this question, Nernst's equation for a single electrode can be applied.

Formulae used: ${E_{{m^{n + }}/m}} = {E^o}_{{m^{n + }}/m} - \dfrac{{0.0591}}{n}log\dfrac{1}{{\left[ {{m^{n + }}} \right]}}$

Complete step by step answer:
In this question, we are directly given the standard electrode potential of the zinc ion and its concentration. Looking at the reaction, we can easily conclude that it is a reduction reaction occurring at a single electrode. Thus, in this case, we can use the Nernst equation for a single electrode to obtain the cell potential at the given temperature with the specified concentration:
${E_{{m^{n + }}/m}} = {E^o}_{{m^{n + }}/m} - \dfrac{{0.0591}}{n}log\dfrac{1}{{\left[ {{m^{n + }}} \right]}}$
Where ${E_{{m^{n + }}/m}}$ is the cell potential at the given conditions, ${E^o}_{{m^{n + }}/m}$ is the standard electrode potential, $n$ is the number of electrons involved in oxidation/reduction during the reaction and $[{m^{n + }}]$ is the concentration of the ionic species.
Here, we have $n = 2$ and $[{m^{n + }}] = [Z{n^{2 + }}] = 0.1$
Substituting the values we have into the above equation, we get:
${E_{{m^{n + }}/m}} = - 0.76 - \dfrac{{0.0591}}{2}\log \dfrac{1}{{0.1}}$
Solving and simplifying the above equation, we get:
${E_{{m^{n + }}/m}} = - 0.76 - \dfrac{{0.0591}}{2}$ (since $\log \dfrac{1}{{0.1}} = \log 10 = 1$)
Therefore, the electrode potential at $298K$ is:
${E_{{m^{n + }}/m}} = - 0.789volt$

Hence, the correct option is option B.

Additional Information: Nernst equation, although very useful, has some drawbacks. The Nernst Equation is only applicable when there is no current flow through the electrode. When there is current flow, the ion activity at the surface of the electrode changes and we encounter errors while measuring cell potential as the Nernst equation doesn’t account for some terms generated due to current flow.

Note: The value of $0.0591$ taken in the above form of the Nernst equation is actually a generalization, based on the fact that reaction is happening at the standard temperature, that is, $298K$. For calculating cell potentials at other temperatures, we should replace the $0.0591$ term with $\dfrac{{2.303RT}}{F}$ where $R$ is the universal gas constant, $T$ is the absolute temperature and $F$ stands for Faraday. The Nernst equation for a single electrode, which is used here, is not the same as the general form of the Nernst equation. In the equation for a single electrode as above, the 1 in the numerator is caused due to the fact that we take the concentration of solid compounds to be unity.