An electrochemical cell, consist of the following two redox...

Given standard reduction potentials:
$E^{\ominus}_{M^{x+}/M} = +0.15$ V
$E^{\ominus}_{Fe^{3+}/Fe} = -0.036$ V

Since $E^{\ominus}_{M^{x+}/M} > E^{\ominus}_{Fe^{3+}/Fe}$, reduction occurs at the $M^{x+}/M$ electrode (cathode) and oxidation occurs at the $Fe^{3+}/Fe$ electrode (anode).

Cathode reaction: $(M^{x+} + x e^{-} \rightarrow M) \times 3$
Anode reaction: $(Fe \rightarrow Fe^{3+} + 3 e^{-}) \times x$

Overall balanced cell reaction:
$3 M^{x+} + x Fe \rightarrow 3 M + x Fe^{3+}$

The number of electrons transferred in the balanced reaction is $n = 3x$.

Standard cell potential $E^{\ominus}_{cell}$ is:
$E^{\ominus}_{cell} = E^{\ominus}_{cathode} - E^{\ominus}_{anode}$
$E^{\ominus}_{cell} = 0.15 - (-0.036) = 0.186$ V

Using the Nernst equation:
$E_{cell} = E^{\ominus}_{cell} - \dfrac{2.303RT}{nF} \log Q$

Substitute the given values ($E_{cell} = 0.2057$ V, $Q = 10^{-2}$, and $\dfrac{2.303RT}{F} = 0.059$ V):
$0.2057 = 0.186 - \dfrac{0.059}{3x} \log(10^{-2})$

$0.2057 - 0.186 = -\dfrac{0.059}{3x} \times (-2)$

$0.0197 = \dfrac{0.118}{3x}$

$3x = \dfrac{0.118}{0.0197}$

$3x \approx 6$

$x = 2$

Answer: $2$