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Question: Write Nernst equation for single electrode potential at 298 K....

Write Nernst equation for single electrode potential at 298 K.

Explanation

Solution

For finding the electrode potential of a substance we use the Nernst equation which is written as:
E=E∘−RTnFln⁡QE={{E}^{\circ }}-\dfrac{RT}{nF}\ln Q
Where E is the electrode potential of the substance, E∘{{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∘−RTnFln⁡QE={{E}^{\circ }}-\dfrac{RT}{nF}\ln Q
Where E is the electrode potential of the substance, E∘{{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→Xn++ne−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=[Xn+]Q=[{{X}^{n+}}]
So, putting the values in the formula, we get:
E=E∘−R x 298nFln⁡[Xn+]E={{E}^{\circ }}-\dfrac{R\text{ x 298}}{nF}\ln [{{X}^{n+}}]
Converting it into natural log,
E=E∘−2.303 x R x 298nFlog⁡[Xn+]E={{E}^{\circ }}-\dfrac{2.303\text{ x }R\text{ x 298}}{nF}\log [{{X}^{n+}}]

Note:
Suppose, the reaction is Zn+Cu2+→Zn2++CuZn+C{{u}^{2+}}\to Z{{n}^{2+}}+Cu, then the Nernst equation will be:
E=E∘−2.303 x R x 2982 x 96500 log⁡[Zn2+Cu2+]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.