Question

If the standard electrode potential for a cell is 2V at 300K, the equilibrium constant (K) for the reaction $Zn(s)+C{u}^{2+}(aq)\rightleftharpoons Z{n}^{2+}(aq)+Cu(s)$ at 300K is approximately
( R=$8J{K}^{-1}mo{l}^{-1}$, F=$96500Cmo{l}^{-1}$)
A. $e^{160}$
B. $e^{320}$
C. $e^{-160}$
D. $e^{-80}$

Answer 1

Hint: To solve this question, try to find two equations of the Gibbs energy change. One reaction should relate$△G$ with the equilibrium constant and the other should relate to $E^0$ of the cell. Now equate these two and find the answer.

Complete step-by-step reaction:
Gibbs free energy, denoted $△G$, combines enthalpy and entropy into one value. The change in free energy commonly called $△G$, is up to the sum of the enthalpy plus the merchandise of the temperature and entropy of the system. $\mathrm{\Delta }G$ can predict the direction of the chemical process under two conditions:
1. constant temperature and
2. constant pressure.
If $△G$ is positive, then the reaction is nonspontaneous (i.e., the input of external energy is critical for the reaction to occur) and if it's negative, then it's spontaneous (occurs without external energy input).
Spontaneous - could be a reaction that's fancy to be natural because it's a reaction that happens by itself with none external action towards it. Non spontaneous - needs constant external energy applied to that so as for the method to continue and once you stop the external action the method will cease. When solving for the equation, if change of G is negative, then we say that the reaction is spontaneous. If the change of G is positive, then it is non spontaneous. The symbol that's commonly used at no cost ENERGY is G. is more properly considered as "standard free energy change". Now, we have two equations for $△G$:
$△G=-nF{E^0}_{cell}$ and $△G=-RT\ln K_c$
Where F= Faraday, R=universal gas constant, T=temperature, $K_c$=equilibrium constant, n=number of moles
Equating both, we get
$nF{E^0}_{cell}=RT\ln K$
Also, we have $Zn->Z{n}^{2+}+2e^{-}$. Since there is an exchange of 2 electrons, we have n=2. By putting the values from the question, we have
$\ln K={\displaystyle \frac{2\times 96500\times 2}{300\times 8}}$, which is approximately equal to 160.

Hence the value of K will be $e^{160}$, which gives A as the correct answer option.

NOTE: You need to know the difference between the terms ‘ln’ and ‘log’. ‘ln’ is called the natural log and when it’s antilogarithm is taken, then the base of the power will be ‘e’. Whereas, the base of log is 10, so on taking antilogarithm, the base of the power becomes 10. That is why we obtain $e^{160}$ as the answer.