At 298 K, the standard reduction potential for Cu$$^{2+}$$/Cu electrode is 0.34 V. Given :
K$$_{sp}$$Cu(OH)$$_2$$ $$= 1 \times 10^{-20}$$. Take $$\frac{2.303RT}{F} = 0.059$$ V. The reduction potential at pH = 14 for the above couple is $$(-) x \times 10^{-2}$$ V. The value of x is _____.

At pH = 14, [OH⁻] = 1 M. Using K$$_{sp}$$ of Cu(OH)₂:

$$K_{sp} = [Cu^{2+}][OH^-]^2$$

$$1 \times 10^{-20} = [Cu^{2+}] \times (1)^2$$

$$[Cu^{2+}] = 10^{-20} \text{ M}$$

Using the Nernst equation for Cu²⁺/Cu:

$$E = E° - \frac{0.059}{n}\log\frac{1}{[Cu^{2+}]}$$

$$E = 0.34 - \frac{0.059}{2}\log\frac{1}{10^{-20}}$$

$$E = 0.34 - \frac{0.059}{2} \times 20$$

$$E = 0.34 - 0.59 = -0.25 \text{ V}$$

$$E = -25 \times 10^{-2} \text{ V}$$

Therefore $$x = \mathbf{25}$$.