An electrochemical cell, consist of the following two redox couples, M$$^{x+}$$(aq)/M(s) [$$E_{red}^{\Theta} = +0.15$$ V] and Fe$$^{3+}$$(aq)/Fe(s) [$$E_{red}^{\Theta} = -0.036$$ V]. The cell EMF (E$$_{cell}$$) is recorded to be 0.2057 V. If the reaction quotient of the electrochemical reaction is found to be $$10^{-2}$$, then the value of $$x$$ is __________. (Nearest integer)

[Given: M is a p-block metal and $$\frac{2.303RT}{F} = 0.059$$ V]

To determine the value of $$x$$, we apply the Nernst equation to the given electrochemical cell.

The two redox couples are $$M^{x+}(aq)/M(s)$$ with $$E^\circ = 0.15\ \text{V}$$ and $$Fe^{3+}(aq)/Fe(s)$$ with $$E^\circ = -0.036\ \text{V}$$. Since $$0.15\ \text{V} > -0.036\ \text{V}$$, the $$M^{x+}/M$$ electrode acts as the cathode and the $$Fe^{3+}/Fe$$ electrode acts as the anode.

The balanced overall cell reaction is

$$3M^{x+}(aq) + xFe(s) \rightarrow 3M(s) + xFe^{3+}(aq).$$

Hence, the total number of electrons transferred is

$$n = 3x.$$

The standard cell potential is

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.15 - (-0.036) = 0.186\ \text{V}.$$

Using the Nernst equation,

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log Q,$$

where

$$E_{\text{cell}} = 0.2057\ \text{V}, \qquad E^\circ_{\text{cell}} = 0.186\ \text{V}, \qquad Q = 10^{-2}.$$

Substituting these values,

$$0.2057 = 0.186 - \frac{0.059}{3x}\log(10^{-2}).$$

Since

$$\log(10^{-2}) = -2,$$

the equation becomes

$$0.2057 = 0.186 + \frac{0.118}{3x}.$$

Rearranging,

$$0.2057 - 0.186 = \frac{0.118}{3x},$$

$$0.0197 = \frac{0.118}{3x}.$$

Therefore,

$$3x = \frac{0.118}{0.0197} \approx 6,$$

and hence

$$x = \frac{6}{3} = 2.$$

Therefore, the value of $$x$$ is 2.