Question

$E^{\circ }$ for $Fe/{Fe}^{2+}$ is +0.44 V and $E^{\circ }$ for $Cu/{Cu}^{2+}$ is -0.32V. Then, in the cell:

• A. $Cu$ oxidises ${Fe}^{2+}$ ion
• B. ${Cu}^{2+}$ ion oxidises $Fe$
• C. $Cu$ reduces ${Fe}^{2+}$ ion
• D. ${Cu}^{2+}$ ion reduces $Fe$

Answer 1

The correct option is B ${Cu}^{2+}$ ion oxidises $Fe$

Here, standard oxidation potential ( $E^0$) of $Fe$ is +0.44V and that of $Cu$ is -0.32V. Thus $Fe$ has greater value of $E^0$ than $Cu$. Thus, $Fe$ undergoes oxidation and $C{u}^{2+}$ undergoes reduction.

Hence, in the given cell, the $C{u}^{2+}$ undergoes oxidation and $Fe$ undergoes reduction. Thus here $C{u}^{2+}$ acts as oxidizing agent and oxidizes Fe atom.

$Fe+C{u}^{2+}\to F{e}^{2+}+Cu$