Question

The Nernst equation for a half-cell reaction at \(298\,K\) is:

• A. \(E=E^\circ-\dfrac{0.0591}{n}\log Q\)
• B. \(E=E^\circ+\dfrac{0.0591}{n}\log Q\)
• C. \(E=E^\circ-\dfrac{n}{0.0591}\log Q\)
• D. \(E=E^\circ+\dfrac{n}{0.0591}\log Q\)

Hint:

At \(298\,K\): \[ E=E^\circ-\frac{0.0591}{n}\log Q \] This formula is extremely important for electrochemistry numericals.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The Nernst equation relates the reduction potential of an electrochemical cell or a half-cell to the standard electrode potential, temperature, and the activities (or concentrations) of the chemical species undergoing oxidation and reduction. It allows the calculation of cell potentials under non-standard state conditions.
Step 2: Key Formula or Approach:
The fundamental thermodynamic form of the Nernst equation is: $$ E = E^\circ - \frac{RT}{nF} \ln Q $$ Where: - $E$ is the reduction potential under given conditions. - $E^\circ$ is the standard reduction potential. - $R$ is the universal gas constant ($8.314 \text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$). - $T$ is the absolute temperature in Kelvin. - $n$ is the number of moles of electrons transferred in the balanced half-reaction. - $F$ is Faraday's constant ($96485 \text{ C}\cdot\text{mol}^{-1}$). - $Q$ is the reaction quotient, and $\ln Q$ is the natural logarithm ($2.303 \log_{10} Q$).
Step 3: Detailed Explanation:
Let's substitute the constant parameters for a standard temperature of $T = 298 \text{ K}$ and convert the natural logarithm to a base-10 logarithm: $$ E = E^\circ - \frac{2.303 \times R \times T}{nF} \log Q $$ $$ E = E^\circ - \frac{2.303 \times 8.314 \times 298}{n \times 96485} \log Q $$ Multiplying and dividing the numerical constants in the numerator and denominator gives: $$ \frac{2.303 \times 8.314 \times 298}{96485} \approx 0.05915 \text{ V} $$ Substituting this combined constant value back into our equation yields: $$ E = E^\circ - \frac{0.0591}{n} \log Q $$ This matches the structure given in option (A).
Step 4: Final Answer:
The correct form of the Nernst equation at 298 K is $E = E^\circ - \frac{0.0591}{n} \log Q$.