The cell potential for $$Zn | Zn^{2+}(aq) || Sn^{x+} | Sn$$ is $$0.801 \text{ V}$$ at $$298 \text{ K}$$. The reaction quotient for the above reaction is $$10^{-2}$$. The number of electrons involved in the given electrochemical cell reaction is ______.
(Given $$E^0_{Zn^{2+}|Zn} = -0.763 \text{ V}$$, $$E^0_{Sn^{x+}|Sn} = +0.008 \text{ V}$$ and $$\dfrac{2.303RT}{F} = 0.06 \text{ V}$$)

We need to find the number of electrons involved in the electrochemical cell reaction.

The cell is: $$Zn | Zn^{2+}(aq) || Sn^{x+} | Sn$$

Anode (oxidation): $$Zn \rightarrow Zn^{2+} + 2e^-$$

Cathode (reduction): $$Sn^{x+} + xe^- \rightarrow Sn$$

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

$$E^0_{cell} = E^0_{Sn^{x+}|Sn} - E^0_{Zn^{2+}|Zn}$$

$$= 0.008 - (-0.763) = 0.771 \text{ V}$$

$$E_{cell} = E^0_{cell} - \dfrac{0.06}{n} \log Q$$

where n is the number of electrons transferred and Q is the reaction quotient.

$$0.801 = 0.771 - \dfrac{0.06}{n} \log(10^{-2})$$

$$0.801 = 0.771 - \dfrac{0.06}{n} \times (-2)$$

$$0.801 = 0.771 + \dfrac{0.12}{n}$$

$$0.801 - 0.771 = \dfrac{0.12}{n}$$

$$0.03 = \dfrac{0.12}{n}$$

$$n = \dfrac{0.12}{0.03} = 4$$

With $$n = 4$$, the cathode half-reaction is: $$Sn^{4+} + 4e^- \rightarrow Sn$$, so $$x = 4$$.

The overall balanced reaction: $$2Zn + Sn^{4+} \rightarrow 2Zn^{2+} + Sn$$ (4 electrons transferred).

Hence, the answer is 4.