Question

Using the following portion of the activity series for oxidation half-reactions, determine which combination of reactants will result in a reaction. Li(s) Al3 (aq) eCr(s) Al3 (aq) 3e75) A) Li(s) with Al(s) B) Li(s) with Al3 (aq) C) Li (aq) with Al(s) D) Li (aq) with Al3 (aq)

Answer 1

The question is incomplete, the complete question is:

Using the following portion of the activity series for oxidation half-reactions, determine which combination of reactants will result in a reaction.

$Li(s)\rightarrow Li^+(aq)+e^-$

$Al(s)\rightarrow Al^{3+}(aq)+3e^-$

A) Li(s) with Al(s)

B) Li(s) with $Al^{3+}$ (aq)

C) $Li^+$ (aq) with Al(s)

D) $Li^+$ (aq) with $Al^{3+}$ (aq)

Answer: The correct option is B): Li(s) with $Al^{3+}$ (aq)

Explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction.  

The chemical species will undergo a reduction reaction if the value of standard reduction potential is more positive or less negative.

For the given half-reactions:

$Li(s)\rightarrow Li^+(aq)+e^-;E^o_{Li^+/Li}=-3.04V$

$Al(s)\rightarrow Al^{3+}(aq)+3e^-;E^o_{Al^{3+}/Al}=-1.662V$

As the value of standard reduction potential of aluminium is less negative. Thus, it undergoes reduction reaction and lithium will undergo oxidation reaction.

The half-reaction follows:

Oxidation half-reaction:  $Li(s)\rightarrow Li^+(aq)+e^-$            ( × 3)

Reduction half-reaction:  $Al^{3+}(aq)+3e^-\rightarrow Al(s)$

Overall cell-reaction:  $3Li(s)+Al^{3+}(aq)\rightarrow 3Li^+(aq)+Al(s)$

Hence, the correct option is B): Li(s) with $Al^{3+}$ (aq)