Question

What factors affect electrochemical cells?

Answer 1

Hint :In order to answer this question, to know the factors that affect electrochemical cells, we should go through the Gibbs free energy and the Nernst Equation to know the factors. And we will also prove them mathematically.

Complete Step By Step Answer:
According to the Gibbs Free energy, concentration and gas pressure, and temperature affect the electrochemical cells.
The Gibbs free energy measures how far a system is from equilibrium.
It therefore determines the voltage (driving force) of an electrochemical cell.
 $ \Delta G = - nFE\,or\,E = - \dfrac{{\Delta G}}{{nF}} $
where, $ n $ is the number of moles of electrons transferred and $ F $ is the Faraday Constant.
Now, we will discuss how Concentration and gas pressure affect electrochemical cells:-
 $ \Delta G = \Delta {G^\circ } - RT\,\ln Q $ , where $ Q $ is the reaction quotient.
For an equilibrium reaction such as $ A \rightleftharpoons B + C $ .
 $ Q = \dfrac{{[B][C]}}{{[A]}}\,or\,Q = \dfrac{{{P_B}{P_C}}}{{{P_A}}} $ if the substances are gases.
 $ E $ depends on $ \Delta G $ , $ \Delta G $ depends on $ Q $ , and $ Q $ depends on concentration and gas pressure.
Therefore both concentration and gas pressure affect the voltage of the cell.
Again, now we will discuss how temperature affects the electrochemical cells:-
According to the Nernst Equation,
 $ E = {E^\circ } - (\dfrac{{RT}}{{nF}})\ln Q $
The temperature term in this equation shows that temperature also affects the cell voltage.

Note :
The temperature dependence can be explained by the temperature dependence of work-functions. Any electrochemical reaction by definition is an electron transfer. The electron goes from the highest occupied $ MO $ of one species to the lowest unoccupied $ MO $ of the other. Both these levels are temperature dependent.