The half-reactions for the oxidation-reduction reaction between Al(s) and Zn₂ (aq) are represented above. Based on the half-reactions, what is the coefficient for Al(s) if the equation for the oxidation-reduction reaction is balanced with the smallest whole-number coefficients?

Respuesta :

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[tex]Al(s)\rightarrow Al^{3+}(aq)+3e^-[/tex]

[tex]Zn^{2+}(aq)+2e^-\rightarrow Zn(s)[/tex]

The half-reactions for the oxidation-reduction reaction between Al(s) and Zn²⁺(aq) are represented above. Based on the half-reactions, what is the coefficient for Al(s) if the equation for the oxidation-reduction reaction is balanced with the smallest whole-number coefficients?

Answer :  The coefficient for Al(s) is, 2

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The given oxidation-reduction half reaction are :

Oxidation : [tex]Al(s)\rightarrow Al^{3+}(aq)+3e^-[/tex]

Reduction : [tex]Zn^{2+}(aq)+2e^-\rightarrow Zn(s)[/tex]

Now balance the charge.

In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 3 and then added both equation, we get the balanced redox reaction.

Oxidation : [tex]2Al(s)\rightarrow 2Al^{3+}(aq)+6e^-[/tex]

Reduction : [tex]3Zn^{2+}(aq)+6e^-\rightarrow 3Zn(s)[/tex]

The balanced chemical equation will be,

[tex]2Al(s)+3Zn^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Zn(s)[/tex]

Thus, the coefficient for Al(s) is, 2

Answer:

  • The coefficient for [tex]Al(s) = 2[/tex]

Explanation:

The half reactions are

[tex]AI(s) --> AI^{3+}(aq) + 3e^-\\\\Zn^{2+}(aq) + 2e^- --> Zn(s)[/tex]

Therefore,

[tex]AI --> AI^3+ + 3e^-\\\\2AI --> 2AI^{3+} + 6e^-[/tex]

[tex]Zn^2+ + 2e^- --> Zn\\\\3Zn^{2+} + 6e --> 3Zn[/tex]

Now add both half reaction

[tex]2AI --> 2AI^3+ + 6e^-\\\\3Zn^2^+ + 6e^- --> 3Zn\\\\= 2AI(s) + 3Zn^{2+}_{(aq)} --> 2AI^{3+}_{(aq)} + 3Zn(s)[/tex]

The coeffiecient of [tex]AI = 2[/tex]

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