An electrochemical cell consists of the follo | vedclass

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(B) The cell reaction involves the oxidation of $M$ and reduction of $Fe^{3+}$.$M \rightarrow M^{x+} + xe^-$ (Anode)$Fe^{3+} + 3e^- \rightarrow Fe$ (C

Vedclass · Accepted Answer

$2$

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(B) The cell reaction involves the oxidation of $M$ and reduction of $Fe^{3+}$.$M ightarrow M^{x+} + xe^-$ (Anode)$Fe^{3+} + 3e^- ightarrow Fe$ (Cathode)Balanced cell reaction: $3M + xFe^{3+} ightarrow 3M^{x+} + xFe$.The number of electrons transferred is $n = 3x$.Standard cell potential: $E^{\ominus}_{cell} = E^{\ominus}_{cathode} - E^{\ominus}_{anode} = 0.15 - (-0.036) = 0.186 	ext{ V}$.Using the Nernst equation: $E_{cell} = E^{\ominus}_{cell} - \frac{0.059}{n} \log Q$.$0.2057 = 0.186 - \frac{0.059}{3x} \log(10^{-2})$.$0.2057 - 0.186 = -\frac{0.059}{3x} 	imes (-2)$.$0.0197 = \frac{0.118}{3x}$.$3x = \frac{0.118}{0.0197} \approx 6$.$x = 2$.

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