
The relationship between standard reduction potential of cell and equilibrium constant is shown by:
A. \[{E^0}_{cell} = \dfrac{n}{{0.059}}\log {K_c}\]
B. \[{E^0}_{cell} = \dfrac{{0.059}}{n}\log {K_c}\]
C. \[{E^0}_{cell} = 0.059\log {K_c}\]
D. \[{E^0}_{cell} = \dfrac{{\log {K_c}}}{n}\]
Answer
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Hint: The term Standard reduction potential defines a chemical species' tendency to get reduced. The process of gaining electrons by a chemical species is called reduction. In normal conditions, this is measured in volts.
Complete Step by Step Solution:
Let's derive the relationship between equilibrium constant and cell potential. We know that change of standard free energy relates to the equilibrium constant.
\[\Delta {G^0} = - RT\ln {K_C}\] --- (1)
Where, R is for gas constant, T is for temperature and K is constant.
Also, the change of free energy equates to an electrochemical cell's work done.
\[\Delta {G^0} = - nF{E^0}_{cell}\] ---- (2)
From equations (1) and (2),
\[nF{E^0}_{cell} = RT\ln {K_C}\]
The above equation rearranges to get
\[{E^0}_{cell} = \dfrac{{RT\ln {K_C}}}{{nF}}\]
At temperature of 298 K, the following equation becomes,
The value of R is 8.314 \[{\rm{J/mol}}\,{\rm{K}}\] and F is Faraday constant whose value is 96485 J/(V.mol).
So,
\[{E^0}_{cell} = \dfrac{{8.314 \times 298}}{{n \times 96485}} \times 2.303\log {K_C}\]
\[{E^0}_{cell} = \dfrac{{0.059}}{n}\log {K_C}\]
Therefore, option B is right.
Additional Information: Let's understand the process of electrolysis. A process in which the passing of electricity takes place through electrolytes and a chemical reaction occurs is termed electrolysis. And a device that generates electricity through the electrolysis process is termed a galvanic cell or electrochemical cell. In the process of electrolysis in a galvanic cell, the conversion of chemical energy to electrical energy takes place.
Note: In electrochemical cells, there is a relation between the work done by an electrochemical cell and the Gibbs free energy. The change of free energy equates to an electrochemical cell's work done. Formula is, \[\Delta {G^0} = - nF{E^0}_{cell}\]. Here, \[\Delta {G^0}\] is for change of free energy, n is for the mole of electrons involved and F is Faraday's constant.
Complete Step by Step Solution:
Let's derive the relationship between equilibrium constant and cell potential. We know that change of standard free energy relates to the equilibrium constant.
\[\Delta {G^0} = - RT\ln {K_C}\] --- (1)
Where, R is for gas constant, T is for temperature and K is constant.
Also, the change of free energy equates to an electrochemical cell's work done.
\[\Delta {G^0} = - nF{E^0}_{cell}\] ---- (2)
From equations (1) and (2),
\[nF{E^0}_{cell} = RT\ln {K_C}\]
The above equation rearranges to get
\[{E^0}_{cell} = \dfrac{{RT\ln {K_C}}}{{nF}}\]
At temperature of 298 K, the following equation becomes,
The value of R is 8.314 \[{\rm{J/mol}}\,{\rm{K}}\] and F is Faraday constant whose value is 96485 J/(V.mol).
So,
\[{E^0}_{cell} = \dfrac{{8.314 \times 298}}{{n \times 96485}} \times 2.303\log {K_C}\]
\[{E^0}_{cell} = \dfrac{{0.059}}{n}\log {K_C}\]
Therefore, option B is right.
Additional Information: Let's understand the process of electrolysis. A process in which the passing of electricity takes place through electrolytes and a chemical reaction occurs is termed electrolysis. And a device that generates electricity through the electrolysis process is termed a galvanic cell or electrochemical cell. In the process of electrolysis in a galvanic cell, the conversion of chemical energy to electrical energy takes place.
Note: In electrochemical cells, there is a relation between the work done by an electrochemical cell and the Gibbs free energy. The change of free energy equates to an electrochemical cell's work done. Formula is, \[\Delta {G^0} = - nF{E^0}_{cell}\]. Here, \[\Delta {G^0}\] is for change of free energy, n is for the mole of electrons involved and F is Faraday's constant.
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