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

The standard Gibbs free energy change is given by $\Delta G° = -nFE°$, where $n = 2$ electrons, $F = 96500$ C/mol, and $E° = 0.354$ V. Substituting: $\Delta G° = -2 \times 96500 \times 0.354 \approx -68{,}322$ J $\approx -68.3$ kJ; the negative value confirms the reaction is spontaneous under standard conditions.