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

Concept: The Nernst equation relates electrode potential with ionic concentration and reaction quotient. General form: \[ E = E^\circ - \frac{RT}{nF}\ln Q \] At \(298\,K\): \[ \frac{2.303RT}{F}=0.0591 \] Thus: \[ E = E^\circ - \frac{0.0591}{n}\log Q \]

Step 1: Understand the terms in equation.
Where:
• \(E^\circ\) = Standard electrode potential
• \(n\) = Number of electrons transferred
• \(Q\) = Reaction quotient The equation predicts how electrode potential changes with concentration.

Step 2: Check all options.
Option (A): \[ E=E^\circ-\dfrac{0.0591}{n}\log Q \] Correct standard form at \(298\,K\). Other options contain incorrect signs or incorrect placement of \(n\). Hence: \[ \boxed{(A)} \]