Write Nernst equation for single electrode potential at 298 K.
Answer
550.8k+ views
Hint: For finding the electrode potential of a substance we use the Nernst equation which is written as:
$E={{E}^{\circ }}-\dfrac{RT}{nF}\ln Q$
Where E is the electrode potential of the substance, ${{E}^{\circ }}$ is the standard electrode potential, R is the gas constant, T is the temperature, n is the number of electrons involved in the reaction, F is the Faraday's constant and Q is the ratio of the concentration of the ions involved in the reaction. Usually, the reaction involves an oxidant and a reductant, so for a single electrode potential, only one part of the reaction is present.
Complete answer:
When a metal electrode is dipped in its solution, then the electrode has the tendency to either lose an electron or gain electrons, so this tendency is known as Electrode potential.
For finding the electrode potential of a substance we use the Nernst equation which is written as:
$E={{E}^{\circ }}-\dfrac{RT}{nF}\ln Q$
Where E is the electrode potential of the substance, ${{E}^{\circ }}$ is the standard electrode potential, R is the gas constant, T is the temperature, n is the number of electrons involved in the reaction, F is the Faraday’s constant and Q is the ratio of the concentration of the ions involved in the reaction.
So, the single reaction for example when the oxidation takes place of X, the reaction will be:
$X\to {{X}^{n+}}+n{{e}^{-}}$
And the temperature is also mentioned in the question, i.e., 289 K and the value of Q will be:
$Q=[{{X}^{n+}}]$
So, putting the values in the formula, we get:
$E={{E}^{\circ }}-\dfrac{R\text{ x 298}}{nF}\ln [{{X}^{n+}}]$
Converting it into natural log,
$E={{E}^{\circ }}-\dfrac{2.303\text{ x }R\text{ x 298}}{nF}\log [{{X}^{n+}}]$
Note:
Suppose, the reaction is $Zn+C{{u}^{2+}}\to Z{{n}^{2+}}+Cu$, then the Nernst equation will be:
$E={{E}^{\circ }}-\dfrac{2.303\text{ x }R\text{ x 298}}{2\text{ x 96500 }}\log \left[ \dfrac{Z{{n}^{2+}}}{C{{u}^{2+}}} \right]$
The concentration of the solids is not taken, only the aqueous solution is taken.
$E={{E}^{\circ }}-\dfrac{RT}{nF}\ln Q$
Where E is the electrode potential of the substance, ${{E}^{\circ }}$ is the standard electrode potential, R is the gas constant, T is the temperature, n is the number of electrons involved in the reaction, F is the Faraday's constant and Q is the ratio of the concentration of the ions involved in the reaction. Usually, the reaction involves an oxidant and a reductant, so for a single electrode potential, only one part of the reaction is present.
Complete answer:
When a metal electrode is dipped in its solution, then the electrode has the tendency to either lose an electron or gain electrons, so this tendency is known as Electrode potential.
For finding the electrode potential of a substance we use the Nernst equation which is written as:
$E={{E}^{\circ }}-\dfrac{RT}{nF}\ln Q$
Where E is the electrode potential of the substance, ${{E}^{\circ }}$ is the standard electrode potential, R is the gas constant, T is the temperature, n is the number of electrons involved in the reaction, F is the Faraday’s constant and Q is the ratio of the concentration of the ions involved in the reaction.
So, the single reaction for example when the oxidation takes place of X, the reaction will be:
$X\to {{X}^{n+}}+n{{e}^{-}}$
And the temperature is also mentioned in the question, i.e., 289 K and the value of Q will be:
$Q=[{{X}^{n+}}]$
So, putting the values in the formula, we get:
$E={{E}^{\circ }}-\dfrac{R\text{ x 298}}{nF}\ln [{{X}^{n+}}]$
Converting it into natural log,
$E={{E}^{\circ }}-\dfrac{2.303\text{ x }R\text{ x 298}}{nF}\log [{{X}^{n+}}]$
Note:
Suppose, the reaction is $Zn+C{{u}^{2+}}\to Z{{n}^{2+}}+Cu$, then the Nernst equation will be:
$E={{E}^{\circ }}-\dfrac{2.303\text{ x }R\text{ x 298}}{2\text{ x 96500 }}\log \left[ \dfrac{Z{{n}^{2+}}}{C{{u}^{2+}}} \right]$
The concentration of the solids is not taken, only the aqueous solution is taken.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

