Chemistry, 17.06.2021 23:30 kkingstone1231

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)

Answers

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)

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)

Answer from: Quest

The answer options is d

Answer from: Quest

Variables,control,conclusion and finally scientific theory

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Using the following portion of the activity series for oxidation half-reactions, determine which com...