Question

Balance the following oxidation-reduction reactions, which occur in acidic solution, using the half-reaction method.

Answer 1

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

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

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

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

(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,

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

(c) The given chemical reaction is,

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

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,

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