Question

A substance that will oxidize aqueous iron (II) ions into iron (III) ions:
A. chlorine
B. bromine
C. platinum
D. oxygen

Answer 1

The redox reactions occur such that one atom gets oxidized while the other atom gets reduced. The oxidized atom loses electrons to become a cation while the reduced atom gains electrons to become anion. The overall electrons transferred remain the same.

Complete step by step answer:
-There are two types of agents. One is the oxidizing agents and the other is the reducing agents. Oxidizing agents are those which oxidize the other atom by reducing themselves. Reducing agents are those that reduce the other atom by oxidizing themselves.
-Here we see that iron is getting oxidized from iron(II) to iron(III). It means that some atom is oxidizing iron by reducing itself. That atom is gaining electrons to do so. The atom with the higher value of electronegativity will be able to gain electrons easily.
-From the given atoms, we see that chlorine gas is a molecule of group 17 which belongs to the halogen family. It can gain 1 electron to attain noble gas configuration. Oxygen atom belongs to group 16 and can gain 2 electrons to become stable.
-In the question, we see that iron ions are showing a change in one electron only. So we need to select an atom that can gain only one electron easily. So, we prefer chlorine over oxygen though oxygen is more electronegative than chlorine.
-Platinum is a metal and it gets oxidized and not reduced during the redox reactions. Bromine atoms also belong to group 17 of the periodic table but it has lesser tendency to reduce itself as it is less electronegative than chlorine. So chlorine is used to oxidize iron.

-The balanced chemical reaction for the oxidation of iron and reduction of chlorine can be written as
$\begin{aligned} & C{{l}_{2}}+2{{e}^{-}}\to 2C{{l}^{-}} \\\ & \text{2F}{{\text{e}}^{2+}}-2{{e}^{-}}\to 2F{{e}^{3+}} \\\ \end{aligned}$
The complete redox reaction can be written as
$C{{l}_{2}}\left( g \right)+2F{{e}^{2+}}\left( aq \right)\to 2C{{l}^{-}}\left( aq \right)+2F{{e}^{3+}}\left( aq \right)$
So, the correct answer is “Option A”.

Note: If there are metals which have variable valencies like copper and iron, they have different strengths of their reducing capabilities. Example - Copper is a better reducing agent when it is in (II) form rather than (I) form. So always take care of such metals while observing the reducing power.