Question

How many electrons are transferred in the following reaction?

Answer 1

Answer:

3Mg +2Al^3 ⇆ 3 Mg^2+ + 2Al

In this reaction 6 electrons are transferred

Explanation:

Step 1: The half reactions

Mg  -2e- ⇆ Mg^2+

Al^3+ +3e- ⇆ Al

Step 2: Balance both equations

3*(Mg  -2e- ⇆ Mg^2+)

2(Al^3+ +3e- ⇆ Al)

3Mg  -6e- ⇆ 3Mg^2+

2Al^3+ +6e- ⇆ 2Al

Step 3: The netto reaction

3Mg +2Al^3 ⇆ 3 Mg^2+ + 2Al

In this reaction 6 electrons are transferred

Answer 2

Answer:

6

Explanation:

Oxidation reaction is defined as the chemical 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 chemical 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:

$Mg(s)+Al^{3+}(aq.)\rightarrow Al(s)+Mg^{2+}(s)$

The half cell reactions for the above reaction follows:

Oxidation half reaction:  $Mg\rightarrow Mg^{2+}+2e^-$

Reduction half reaction:  $Al^{3+}+3e^-\rightarrow Al$

Magnesium is loosing 2 electrons to form the magnesium cation. Thus, it is getting oxidized. Aluminum anion is gaining 3 electrons to form Aluminum. Thus, it is getting reduced.

Thus, balancing the half-reactions as:-

Oxidation half reaction:  $3Mg\rightarrow 3Mg^{2+}+6e^-$

Reduction half reaction:  $2Al^{3+}+6e^-\rightarrow 2Al$

Thus, total number of electrons transferred = 6