Question

Consider the redox reaction below.

Answer 1

Answer: The half reaction that describes the reduction reaction is $Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)$

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)$

The half reactions follows:

Oxidation half reaction:  $Mg(s)\rightarrow Mg^{2+}(aq.)+2e^-$

Reduction half reaction:  $Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)$

Hence, the half reaction that describes the reduction reaction is $Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)$

Answer 2

Answer:

1)

Explanation:

Fe+2 + 2e -> Fe

The element that is reduced gains electrons. In this case Fe is gaining electrons