The equilibrium constant for the reaction Zn(s) + Sn$$^{2+}$$(aq) $$\rightleftharpoons$$ Zn$$^{2+}$$(aq) + Sn(s) is $$1 \times 10^{20}$$ at 298 K. The magnitude of standard electrode potential of Sn/Sn$$^{2+}$$ if E$$^0_{\text{Zn}^{2+}/\text{Zn}} = -0.76$$ V is ______ $$\times 10^{-2}$$ V. (Nearest integer)
Given: $$\frac{2.303 RT}{F} = 0.059$$ V

We need to find the standard electrode potential of Sn/Sn$$^{2+}$$ given the equilibrium constant and $$E^0_{\text{Zn}^{2+}/\text{Zn}}$$.

The reaction is: Zn(s) + Sn$$^{2+}$$(aq) $$\rightleftharpoons$$ Zn$$^{2+}$$(aq) + Sn(s)

At equilibrium:

$$E^0_{\text{cell}} = \frac{0.059}{n}\log K$$

Here $$n = 2$$ (two electrons transferred).

$$E^0_{\text{cell}} = \frac{0.059}{2}\log(1 \times 10^{20}) = \frac{0.059}{2} \times 20 = 0.59 \text{ V}$$

$$E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}}$$

Zn is oxidized (anode), Sn$$^{2+}$$ is reduced (cathode):

$$E^0_{\text{cell}} = E^0_{\text{Sn}^{2+}/\text{Sn}} - E^0_{\text{Zn}^{2+}/\text{Zn}}$$

$$0.59 = E^0_{\text{Sn}^{2+}/\text{Sn}} - (-0.76)$$

$$E^0_{\text{Sn}^{2+}/\text{Sn}} = 0.59 - 0.76 = -0.17 \text{ V}$$

The question asks for the magnitude of $$E^0_{\text{Sn/Sn}^{2+}}$$.

Note: $$E^0_{\text{Sn/Sn}^{2+}}$$ (oxidation potential) = $$-E^0_{\text{Sn}^{2+}/\text{Sn}}$$ (reduction potential) = $$-(-0.17) = 0.17$$ V.

The magnitude = $$|{-0.17}| = 0.17$$ V $$= 17 \times 10^{-2}$$ V.

The correct answer is 17.