Chemistry: Balance the following oxidation-reduction reactions, which occur in acidic solution, using the half-reaction method. (use the lowest possible coefficients. include states-of-matter under the given conditions in your answer.) o2(g) + pb(s) → h2o(l) + pb2+(aq) (b) no3−(aq) + sn(s) → no(g) + sn2+(aq) (c) cl2(g) + cr3+(aq) → cl −(aq) + cr2o72−(aq) (d) f2(g) + mn2+(aq) → f −(aq) + mno4−(aq)

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Balance the following oxidation-reduction reactions,

Ответ:

jalalblal

09.01.2022

Chemistry

Answer : The balanced chemical equation in a acidic solution are,

(a) $2Pb+O_2+4H^+\rightarrow 2Pb^{2+}+2H_2O$

(b) $2Sn+NO_3^-+4H^+\rightarrow 2Sn^{2+}+NO+2H_2O$

(d) $2Mn^{2+}+8H_2O+5F_2\rightarrow 2MnO_4^-+16H^++10F^-$

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,

$O_2(g)+Pb(s)\rightarrow H_2O(l)+Pb^{2+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Pb\rightarrow Pb^{2+}$

Reduction : $O_2\rightarrow H_2O$

Now balance oxygen atom on both side.

Oxidation : $Pb\rightarrow Pb^{2+}$

Reduction : $O_2\rightarrow 2H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Pb\rightarrow Pb^{2+}$

Reduction : $O_2+4H^+\rightarrow 2H_2O$

Now balance the charge.

Oxidation : $Pb\rightarrow Pb^{2+}+2e^-$

Reduction : $O_2+4H^++4e^-\rightarrow 2H_2O$

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 : $2Pb\rightarrow 2Pb^{2+}+4e^-$

Reduction : $O_2+4H^++4e^-\rightarrow 2H_2O$

The balanced chemical equation will be,

$2Pb+O_2+4H^+\rightarrow 2Pb^{2+}+2H_2O$

(b) The given chemical reaction is,

$NO_3^-(aq)+Sn(s)\rightarrow NO(g)+Sn^{2+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sn\rightarrow Sn^{2+}$

Reduction : $NO_3^-\rightarrow NO$

Now balance oxygen atom on both side.

Oxidation : $Sn\rightarrow Sn^{2+}$

Reduction : $NO_3^-\rightarrow NO+2H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sn\rightarrow Sn^{2+}$

Reduction : $NO_3^-+4H^+\rightarrow NO+2H_2O$

Now balance the charge.

Oxidation : $Sn\rightarrow Sn^{2+}+2e^-$

Reduction : $NO_3^-+4H^++4e^-\rightarrow NO+2H_2O$

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 : $2Sn\rightarrow 2Sn^{2+}+4e^-$

Reduction : $NO_3^-+4H^++4e^-\rightarrow NO+2H_2O$

The balanced chemical equation will be,

$2Sn+NO_3^-+4H^+\rightarrow 2Sn^{2+}+NO+2H_2O$

$Cl_2(g)+Cr^{3+}(aq)\rightarrow Cl^-(aq)+Cr_2O_7^{2-}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $2Cr^{3+}\rightarrow Cr_2O_7^{2-}$

Reduction : $Cl_2\rightarrow 2Cl^-$

Now balance oxygen atom on both side.

Oxidation : $2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}$

Reduction : $Cl_2\rightarrow 2Cl^-$

Now balance hydrogen atom on both side.

Oxidation : $2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}+14H^+$

Reduction : $Cl_2\rightarrow 2Cl^-$

Now balance the charge.

Oxidation : $2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}+14H^++6e^-$

Reduction : $Cl_2+2e^-\rightarrow 2Cl^-$

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 : $2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}+14H^++6e^-$

Reduction : $3Cl_2+6e^-\rightarrow 6Cl^-$

The balanced chemical equation will be,

$2Cr^{3+}+7H_2O+3Cl_2\rightarrow Cr_2O_7^{2-}+14H^++6Cl^-$

(d) The given chemical reaction is,

$F_2(g)+Mn^{2+}(aq)\rightarrow F^-(aq)+MnO_4^-(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Mn^{2+}\rightarrow MnO_4^-$

Reduction : $F_2\rightarrow 2F^-$

Now balance oxygen atom on both side.

Oxidation : $Mn^{2+}+4H_2O\rightarrow MnO_4^-$

Reduction : $F_2\rightarrow 2F^-$

Now balance hydrogen atom on both side.

Oxidation : $Mn^{2+}+4H_2O\rightarrow MnO_4^-+8H^+$

Reduction : $F_2\rightarrow 2F^-$

Now balance the charge.

Oxidation : $Mn^{2+}+4H_2O\rightarrow MnO_4^-+8H^+5e^-$

Reduction : $F_2+2e^-\rightarrow 2F^-$

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 : $2Mn^{2+}+8H_2O\rightarrow 2MnO_4^-+16H^+10e^-$

Reduction : $5F_2+10e^-\rightarrow 10F^-$

The balanced chemical equation will be,

$2Mn^{2+}+8H_2O+5F_2\rightarrow 2MnO_4^-+16H^++10F^-$

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