To solve this problem, we will use the Nernst equation at equilibrium.$\\$At equilibrium, the cell potential $E_{cell}$ is zero, and the Nernst equation is given by:$\\ E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln K \\$Where:$\\$• $E^\circ_{cell} = 0.59$ V (standard cell potential)$\\$• $R = 8.314$ J mol$^{-1}$ K$^{-1}$ (universal gas constant)$\\$• $T = 298$ K (temperature)$\\$• $n = 1$ (number of electrons involved in the reaction)$\\$• $F = 96485$ C mol$^{-1}$ (Faraday's constant)$\\$• $K$ is the equilibrium constant$\\$At equilibrium, $E_{cell} = 0,$ so the equation becomes:$\\ 0 = E^\circ_{cell} - \frac{RT}{nF} \ln K \\$Rearrange to solve for $\ln K: \\ \ln K = \frac{nF}{RT} E^\circ_{cell} \\$Substitute the known values:$\\ \ln K = \frac{1 \times 96485}{8.314 \times 298} \times 0.59 \\$Calculate the value:$\\ \ln K = \frac{96485 \times 0.59}{2477.172} \\ \ln K \approx \frac{56926.15}{2477.172} \\ \ln K \approx 22.98 \\$To find $K,$ we take the exponential of both sides:$\\ K = e^{22.98} \\ K \approx 1.0 \times 10^{10} \\$Therefore, the equilibrium constant for the cell reaction is $1.0 \times 10^{10}. \\$This corresponds to Option 3.