Respuesta :
Answer : The balanced chemical equation in a acidic solution are,
(a) [tex]2Pb+O_2+4H^+\rightarrow 2Pb^{2+}+2H_2O[/tex]
(b) [tex]2Sn+NO_3^-+4H^+\rightarrow 2Sn^{2+}+NO+2H_2O[/tex]
(c) [tex]2Cr^{3+}+7H_2O+3Cl_2\rightarrow Cr_2O_7^{2-}+14H^++6Cl^-[/tex]
(d) [tex]2Mn^{2+}+8H_2O+5F_2\rightarrow 2MnO_4^-+16H^++10F^-[/tex]
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.
(a) The given chemical reaction is,
[tex]O_2(g)+Pb(s)\rightarrow H_2O(l)+Pb^{2+}(aq)[/tex]
The oxidation-reduction half reaction will be :
Oxidation : [tex]Pb\rightarrow Pb^{2+}[/tex]
Reduction : [tex]O_2\rightarrow H_2O[/tex]
- Now balance oxygen atom on both side.
Oxidation : [tex]Pb\rightarrow Pb^{2+}[/tex]
Reduction : [tex]O_2\rightarrow 2H_2O[/tex]
- Now balance hydrogen atom on both side.
Oxidation : [tex]Pb\rightarrow Pb^{2+}[/tex]
Reduction : [tex]O_2+4H^+\rightarrow 2H_2O[/tex]
- Now balance the charge.
Oxidation : [tex]Pb\rightarrow Pb^{2+}+2e^-[/tex]
Reduction : [tex]O_2+4H^++4e^-\rightarrow 2H_2O[/tex]
In order to balance the electrons, we multiply the oxidation reaction by 2 and then added both equation, we get the balanced redox reaction.
Oxidation : [tex]2Pb\rightarrow 2Pb^{2+}+4e^-[/tex]
Reduction : [tex]O_2+4H^++4e^-\rightarrow 2H_2O[/tex]
The balanced chemical equation will be,
[tex]2Pb+O_2+4H^+\rightarrow 2Pb^{2+}+2H_2O[/tex]
(b) The given chemical reaction is,
[tex]NO_3^-(aq)+Sn(s)\rightarrow NO(g)+Sn^{2+}(aq)[/tex]
The oxidation-reduction half reaction will be :
Oxidation : [tex]Sn\rightarrow Sn^{2+}[/tex]
Reduction : [tex]NO_3^-\rightarrow NO[/tex]
- Now balance oxygen atom on both side.
Oxidation : [tex]Sn\rightarrow Sn^{2+}[/tex]
Reduction : [tex]NO_3^-\rightarrow NO+2H_2O[/tex]
- Now balance hydrogen atom on both side.
Oxidation : [tex]Sn\rightarrow Sn^{2+}[/tex]
Reduction : [tex]NO_3^-+4H^+\rightarrow NO+2H_2O[/tex]
- Now balance the charge.
Oxidation : [tex]Sn\rightarrow Sn^{2+}+2e^-[/tex]
Reduction : [tex]NO_3^-+4H^++4e^-\rightarrow NO+2H_2O[/tex]
In order to balance the electrons, we multiply the oxidation reaction by 2 and then added both equation, we get the balanced redox reaction.
Oxidation : [tex]2Sn\rightarrow 2Sn^{2+}+4e^-[/tex]
Reduction : [tex]NO_3^-+4H^++4e^-\rightarrow NO+2H_2O[/tex]
The balanced chemical equation will be,
[tex]2Sn+NO_3^-+4H^+\rightarrow 2Sn^{2+}+NO+2H_2O[/tex]
(c) The given chemical reaction is,
[tex]Cl_2(g)+Cr^{3+}(aq)\rightarrow Cl^-(aq)+Cr_2O_7^{2-}(aq)[/tex]
The oxidation-reduction half reaction will be :
Oxidation : [tex]2Cr^{3+}\rightarrow Cr_2O_7^{2-}[/tex]
Reduction : [tex]Cl_2\rightarrow 2Cl^-[/tex]
- Now balance oxygen atom on both side.
Oxidation : [tex]2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}[/tex]
Reduction : [tex]Cl_2\rightarrow 2Cl^-[/tex]
- Now balance hydrogen atom on both side.
Oxidation : [tex]2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}+14H^+[/tex]
Reduction : [tex]Cl_2\rightarrow 2Cl^-[/tex]
- Now balance the charge.
Oxidation : [tex]2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}+14H^++6e^-[/tex]
Reduction : [tex]Cl_2+2e^-\rightarrow 2Cl^-[/tex]
In order to balance the electrons, we multiply the reduction reaction by 3 and then added both equation, we get the balanced redox reaction.
Oxidation : [tex]2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}+14H^++6e^-[/tex]
Reduction : [tex]3Cl_2+6e^-\rightarrow 6Cl^-[/tex]
The balanced chemical equation will be,
[tex]2Cr^{3+}+7H_2O+3Cl_2\rightarrow Cr_2O_7^{2-}+14H^++6Cl^-[/tex]
(d) The given chemical reaction is,
[tex]F_2(g)+Mn^{2+}(aq)\rightarrow F^-(aq)+MnO_4^-(aq)[/tex]
The oxidation-reduction half reaction will be :
Oxidation : [tex]Mn^{2+}\rightarrow MnO_4^-[/tex]
Reduction : [tex]F_2\rightarrow 2F^-[/tex]
- Now balance oxygen atom on both side.
Oxidation : [tex]Mn^{2+}+4H_2O\rightarrow MnO_4^-[/tex]
Reduction : [tex]F_2\rightarrow 2F^-[/tex]
- Now balance hydrogen atom on both side.
Oxidation : [tex]Mn^{2+}+4H_2O\rightarrow MnO_4^-+8H^+[/tex]
Reduction : [tex]F_2\rightarrow 2F^-[/tex]
- Now balance the charge.
Oxidation : [tex]Mn^{2+}+4H_2O\rightarrow MnO_4^-+8H^+5e^-[/tex]
Reduction : [tex]F_2+2e^-\rightarrow 2F^-[/tex]
In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 5 and then added both equation, we get the balanced redox reaction.
Oxidation : [tex]2Mn^{2+}+8H_2O\rightarrow 2MnO_4^-+16H^+10e^-[/tex]
Reduction : [tex]5F_2+10e^-\rightarrow 10F^-[/tex]
The balanced chemical equation will be,
[tex]2Mn^{2+}+8H_2O+5F_2\rightarrow 2MnO_4^-+16H^++10F^-[/tex]
