Question

A cell has a standard electrode potential of \(0.354\,V\) at \(298\,K\). If 2 electrons are transferred in the cell reaction, what is the equilibrium constant (\(K\)) of the reaction at \(298\,K\)?

• A. \(1\times10^{12}\)
• B. \(1\times10^{11}\)
• C. \(1\times10^{10}\)
• D. \(1\times10^{13}\)

Hint:

At \(298\,K\), \[ \boxed{ E^\circ=\frac{0.0591}{n}\log K } \] Remember: \[ \boxed{ \begin{aligned} E^\circ\gt 0 &\Rightarrow K\gt 1\\ E^\circ\lt 0 &\Rightarrow K\lt 1 \end{aligned} } \] A larger value of \(E^\circ\) corresponds to a larger equilibrium constant.

Answer 1

The Correct Option is A

Concept: The relationship between the standard cell potential and the equilibrium constant is \[ \boxed{ E^\circ=\frac{0.0591}{n}\log K } \] where

• \(E^\circ\) = Standard cell potential,

• \(n\) = Number of electrons transferred,

• \(K\) = Equilibrium constant.
A positive value of \(E^\circ\) indicates that the reaction is spontaneous and has a large equilibrium constant.

Step 1: Write the given data.
\[ E^\circ=0.354\,V \] \[ n=2 \]

Step 2: Apply the Nernst equation.
Using \[ E^\circ=\frac{0.0591}{n}\log K, \] we get \[ 0.354=\frac{0.0591}{2}\log K. \] Hence, \[ \log K=\frac{0.354\times2}{0.0591} \] \[ =\frac{0.708}{0.0591} \] \[ \approx11.98\approx12. \]

Step 3: Calculate the equilibrium constant.
\[ K=10^{12}. \] Therefore, \[ \boxed{K\approx1\times10^{12}.} \] Hence, \[ \boxed{\textbf{Option (A)}} \] is the correct answer.