To determine how many electrons will be lost in the redox reaction [tex]\( Al + Mn^{2+} \longrightarrow Al^{3+} + Mn \)[/tex], let's go through the steps involved in balancing the reaction using the half-reaction method.
### Step-by-Step Solution:
1. Identify the Oxidation and Reduction Half-Reactions:
- Oxidation Half-Reaction: This involves the aluminum (Al) atom losing electrons to become [tex]\( Al^{3+} \)[/tex].
[tex]\[
Al \longrightarrow Al^{3+} + 3e^-
\][/tex]
- Reduction Half-Reaction: This involves the manganese ion ([tex]\( Mn^{2+} \)[/tex]) gaining electrons to become manganese (Mn).
[tex]\[
Mn^{2+} + 2e^- \longrightarrow Mn
\][/tex]
2. Balance the Electrons in the Half-Reactions:
- The oxidation half-reaction shows that [tex]\( Al \)[/tex] loses 3 electrons.
- The reduction half-reaction shows that [tex]\( Mn^{2+} \)[/tex] gains 2 electrons.
3. Equalize the Number of Electrons:
- To balance the number of electrons, we need to find a common multiple for 3 and 2, which is 6.
- Multiply the oxidation half-reaction by 2:
[tex]\[
2(Al \longrightarrow Al^{3+} + 3e^-) \quad \Rightarrow \quad 2Al \longrightarrow 2Al^{3+} + 6e^-
\][/tex]
- Multiply the reduction half-reaction by 3:
[tex]\[
3(Mn^{2+} + 2e^- \longrightarrow Mn) \quad \Rightarrow \quad 3Mn^{2+} + 6e^- \longrightarrow 3Mn
\][/tex]
4. Combine the Balanced Half-Reactions:
- Adding these balanced half-reactions together gives us the overall balanced redox reaction:
[tex]\[
2Al + 3Mn^{2+} \longrightarrow 2Al^{3+} + 3Mn
\][/tex]
5. Determine the Total Number of Electrons Lost:
- Here, 2 moles of [tex]\( Al \)[/tex] each lose 3 electrons.
- Therefore, the total number of electrons lost is:
[tex]\[
2 \times 3 = 6
\][/tex]
Thus, in the balanced redox reaction, 6 electrons will be lost in all.
So the correct answer is: 6.
