Given \[{E^0}_{C{r^{3 + }}/Cr} = - 0.72,{E^0}_{F{e^{2 + }}/Fe} = - 0.42V\].
The potential for the cell, \[Cr|C{r^{3 + }}(0.1M)||F{e^{2 + }}(0.01M)|Fe\]is:
(a) - 0.26 V
(b) 0.26 V
(c) 0.339 V
(d) -0.339 V
Answer
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Hint: We know that Nernst Equation can be used to determine cell potential in non-standard situations. It connects the measured cell potential to the reaction's quotient, allowing for the precise determination of equilibrium constants (including constants of solubility).
Complete answer: The Nernst equation is a relationship between an electrochemical reaction's reduction potential (half-cell or full-cell reaction) and the normal electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.
\[E = {E^0} - \dfrac{{RT}}{{zF}}\log Q\]
Where,
\[E\]= reduction rate of the cell
\[{E^0}\]= standard potential of the cell
R = universal gas constant
T= temperature in kelvin
z= ion charge(moles of electrons)
F= Faraday constant
Q=reaction quotient
We need to find the value of E°, that is
\[{E^0}_{cell} = {E^0}_{cathode} - {E^0}_{anode}\]
\[{E^0} = - 0.42 - ( - 0.72)\]
\[ = + 0.30V\]
Here we need to balance the equation,
therefore,
\[2Cr(s) + 3F{e^{2 + }}(0.01M) \rightleftharpoons 2C{r^{3 + }}(0.1M) + 3Fe(s)\]
Substituting the values to Nernst equation,
\[E = {E^0} - \dfrac{{0.059}}{n}{\log _{10}}Q\]
\[ = 0.30 - \dfrac{{0.059}}{6}\log {10^4}\]
\[ = 0.261V\]
Therefore,
\[E = 0.26V\]
Hence the correct answer is option (B)- 0.26V.
Additional Information:
In an electrochemical cell, the cell potential,\[{E_{cell}}\], is the difference in potential between two half cells. The tendency of electrons to flow from one half cell to the other causes the potential difference. During reduction, on the other hand, the material loses electrons and becomes negatively charged.
Note:
Remember the Nernst equation, \[E = {E^0} - \dfrac{{RT}}{{zF}}\log Q\]. The Nernst equation connects an electrochemical cell's cell potential, the normal cell potential, temperature, and the reaction quotient. The Nernst equation is often used to measure an electrochemical cell's cell potential at any temperature, strain, or reactant concentration.
Complete answer: The Nernst equation is a relationship between an electrochemical reaction's reduction potential (half-cell or full-cell reaction) and the normal electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.
\[E = {E^0} - \dfrac{{RT}}{{zF}}\log Q\]
Where,
\[E\]= reduction rate of the cell
\[{E^0}\]= standard potential of the cell
R = universal gas constant
T= temperature in kelvin
z= ion charge(moles of electrons)
F= Faraday constant
Q=reaction quotient
We need to find the value of E°, that is
\[{E^0}_{cell} = {E^0}_{cathode} - {E^0}_{anode}\]
\[{E^0} = - 0.42 - ( - 0.72)\]
\[ = + 0.30V\]
Here we need to balance the equation,
therefore,
\[2Cr(s) + 3F{e^{2 + }}(0.01M) \rightleftharpoons 2C{r^{3 + }}(0.1M) + 3Fe(s)\]
Substituting the values to Nernst equation,
\[E = {E^0} - \dfrac{{0.059}}{n}{\log _{10}}Q\]
\[ = 0.30 - \dfrac{{0.059}}{6}\log {10^4}\]
\[ = 0.261V\]
Therefore,
\[E = 0.26V\]
Hence the correct answer is option (B)- 0.26V.
Additional Information:
In an electrochemical cell, the cell potential,\[{E_{cell}}\], is the difference in potential between two half cells. The tendency of electrons to flow from one half cell to the other causes the potential difference. During reduction, on the other hand, the material loses electrons and becomes negatively charged.
Note:
Remember the Nernst equation, \[E = {E^0} - \dfrac{{RT}}{{zF}}\log Q\]. The Nernst equation connects an electrochemical cell's cell potential, the normal cell potential, temperature, and the reaction quotient. The Nernst equation is often used to measure an electrochemical cell's cell potential at any temperature, strain, or reactant concentration.
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