Consider the redox reaction below.

Upper M g (s) plus Pper f e superscript 2 plus (a q) right arrow upper M g superscript 2 plus (a q) plus upper F e (s).

Which equation is a half reaction that describes the reduction that is taking place?

Upper F e superscript 2 plus (a q) plus 2 e superscript minus right arrow upper F e (s).

Upper M g (s) right arrow upper M g superscript 2 plus (a q) plus 2 e superscript minus.

Upper Fe superscript 2 plus (a q) right arrow upper F e (s) plus 2 e superscript minus.

Upper M g (s) plus 2 e superscript minus right arrow upper Mg superscript 2 plus (a q).

See explanation below

Explanation:

To answer this better, let's write the overall equation:

Mg(s) + Fe²⁺(aq) > Mg²⁺(aq) + Fe(s)

Now the question shows us options of which equation is the half reaction that describes reduction. To do this, we need to know what is happening here, and what a reduction means here.

A reduction in a reaction takes place when an atom or compound gains electrons. This can be easily identified in this reaction because when an atom has a higher positive number, means that it's losing electron, and if it has a lower positive number and even negative numbers, means that is gaining the electrons.

In other words, if you have a number line, and put the 0 in the center, any movement that goes from 1 position to the right of the line, means that it's losing electrons (oxidizing) and if it goes to the left, it's gaining electrons (Reduction).

With this explained, let's write the half reactions taking place:

Mg(s) > Mg²⁺

Fe²⁺ > Fe(s)

When an atom it's in solid state, the oxidation number of the atom it's always zero, therefore, in the case of Mg, it goes from 0 to 2+, and Fe goes from 2+ to 0. According to the explanation above, if we go to the left we are reducting and if we go to the right, we are oxidizing. Therefore, as Fe goes from 2+ to 0, this means that is gaining 2 electrons, and it's reducting while Mg, it's oxidizing and losing electrons.

The half reactions then:

Mg(s) > Mg²⁺ + 2e⁻    (Oxidizing)

Fe²⁺ + 2e⁻> Fe(s)       (Reducting)

In this case, it would be the option 1.

fe 55.85 g/mol and h2o 18.02 g/mol.