Question

If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(1.5 \times 10^{-4}\), calculate the corresponding \(\Delta G^{\circ}\) and \(E_{\text {cell }}^{\circ}\).

Short answer

Using the given equilibrium constant (K=1.5 x 10⁻⁴) and temperature (298 K), we calculate the corresponding Gibbs free energy change (ΔG°) using the formula ΔG° = -RT ln(K), which yields ΔG° = 32,839.57 J/mol. We then use the formula ΔG° = -nFE°₈cell to find the standard cell potential (E°₈cell), resulting in E°₈cell = -0.17 V.

Step-by-step solution

Calculate ΔG° using the equilibrium constant

We can use the given equilibrium constant (K=1.5 x 10⁻⁴) and the formula ΔG° = -RT ln(K) to find the Gibbs free energy change. First, plug in the values for R (8.314 J/mol·K) and T (298 K) into the formula: ΔG° = - (8.314 J/mol·K) x (298 K) x ln(1.5 x 10⁻⁴) Now, calculate the natural logarithm of K: ln(1.5 x 10⁻⁴) = -13.41 So, we can plug this value back into the formula for ΔG° : ΔG° = - (8.314 J/mol·K) x (298 K) x (-13.41) Finally, calculate the value of ΔG° : ΔG° = 32839.57 J/mol

Calculate E°₈cell using the ΔG° value

Now that we have the value for ΔG°, we can use the formula ΔG° = -nFE°₈cell to find the standard cell potential. First, plug in the values for n (2) and F (96485 C/mol) into the formula: 32839.57 J/mol = - (2) x (96485 C/mol) x E°₈cell Now, isolate E°₈cell by dividing both sides by -2F: E°₈cell = (32839.57 J/mol) / (-2 x 96485 C/mol) Finally, calculate the value of E°₈cell : E°₈cell = -0.17 V So, the corresponding Gibbs free energy change (ΔG°) is 32,839.57 J/mol, and the standard cell potential (E°₈cell) is -0.17 V.