
If \[{\rm{Z}}{{\rm{n}}^{2 + }}\] /Zn electrode is diluted 100 times, then the change of emf is
(A) increases of 59 mV
(B) Decrease of 59 mV
(C) Decrease of 29.5 mV
(D) Increase of 29.5 nv
Answer
164.1k+ views
Hint: EMF is the abbreviation for electromotive force. It equates to the potential difference in the state where no flow of current occurs. Potential difference and EMF both are measured in volts.
Formula used:
Nernst equation is to be used for the calculation of change of EMF.
\[E = {E^0} - \dfrac{{RT}}{{nF}}\ln Q\]
Here, E is for electrode potential, \[{E^0}\] is for standard electrode potential, R is for gas constant, T is temperature, n is for mole of electrons, F is Faraday's quotient.
Complete Step by Step Solution:
At temperature of 298 K, the Nernst 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,
\[E = {E^0} - \dfrac{{8.314 \times 298}}{{n \times 96485}} \times 2.303\log Q\]
\[E = {E^0} - \dfrac{{0.059}}{n}\log Q\]
\[Q = \dfrac{{\rm{1}}}{{\left[ {{\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}} \right]}}\]
So,
\[E = {E^0} - \dfrac{{0.059}}{n}\log \dfrac{1}{{\left[ {{\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}} \right]}}\]
So, \[\left[ {{\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}} \right]\] =100
\[{E^0} - E = \dfrac{{0.059}}{2}\log \dfrac{1}{{100}}\]
\[{E^0} - E = \dfrac{{0.059}}{2} \times \log {10^{ - 2}}\]
\[{E^0} - E = - 2 \times \dfrac{{0.059}}{2} \times 1\](log 10=1)
\[{E^0} - E = - 0.059\,{\rm{V}} = - {\rm{59}}\,{\rm{mV}}\]
Therefore, the change of EMF of the cell is 59 mV.
Hence, option B is right.
Additional Information:
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 mole of electrons involved and F is Faraday's constant.
Note: 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.
Formula used:
Nernst equation is to be used for the calculation of change of EMF.
\[E = {E^0} - \dfrac{{RT}}{{nF}}\ln Q\]
Here, E is for electrode potential, \[{E^0}\] is for standard electrode potential, R is for gas constant, T is temperature, n is for mole of electrons, F is Faraday's quotient.
Complete Step by Step Solution:
At temperature of 298 K, the Nernst 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,
\[E = {E^0} - \dfrac{{8.314 \times 298}}{{n \times 96485}} \times 2.303\log Q\]
\[E = {E^0} - \dfrac{{0.059}}{n}\log Q\]
\[Q = \dfrac{{\rm{1}}}{{\left[ {{\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}} \right]}}\]
So,
\[E = {E^0} - \dfrac{{0.059}}{n}\log \dfrac{1}{{\left[ {{\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}} \right]}}\]
So, \[\left[ {{\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}} \right]\] =100
\[{E^0} - E = \dfrac{{0.059}}{2}\log \dfrac{1}{{100}}\]
\[{E^0} - E = \dfrac{{0.059}}{2} \times \log {10^{ - 2}}\]
\[{E^0} - E = - 2 \times \dfrac{{0.059}}{2} \times 1\](log 10=1)
\[{E^0} - E = - 0.059\,{\rm{V}} = - {\rm{59}}\,{\rm{mV}}\]
Therefore, the change of EMF of the cell is 59 mV.
Hence, option B is right.
Additional Information:
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 mole of electrons involved and F is Faraday's constant.
Note: 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.
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