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
582.9k+ views
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.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

