If the equilibrium constant for a two-electron redox reaction at $298 \mathrm{~K}\( is \)2.2 \times 10^{5},\( calculate the corresponding \)\Delta G^{\circ}\( and \)E^{\circ}$.

We can use the equation \(\Delta G^{\circ} = -RT \ln K\) to calculate the standard Gibbs free energy change: $$\Delta G^{\circ} = - (8.314\mathrm{~J~mol^{-1}K^{-1}}) (298\mathrm{~K}) \ln (2.2 \times 10^{5})$$ Computing this gives: $$\Delta G^{\circ} \approx -33922.3\mathrm{~J~mol^{-1}}$$

Now that we have the \(\Delta G^{\circ}\) value, we can use the Nernst equation to calculate \(E^{\circ}\): $$\Delta G^{\circ} = -nFE^{\circ}$$ Rearrange the equation to solve for \(E^{\circ}\): $$E^{\circ} = -\dfrac{\Delta G^{\circ}}{nF}$$ Plug in the values for \(\Delta G^{\circ}\), \(n\) (number of electrons transferred, two in this case), and \(F\): $$E^{\circ} = -\dfrac{-33922.3\mathrm{~J~mol^{-1}}}{(2)(96,485\mathrm{~C~mol^{-1})}}$$ Computing this gives: $$E^{\circ} \approx 0.176\mathrm{~V}$$ So, the standard Gibbs free energy change (\(\Delta G^{\circ}\)) for the reaction is approximately \(-33922.3\mathrm{~J~mol^{-1}}\) and the standard electrode potential (\(E^{\circ}\)) is approximately \(0.176\mathrm{~V}\).