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stira [4]
2 years ago
8

Balance each of the following equations according to the half- reaction method:

Chemistry
1 answer:
lukranit [14]2 years ago
5 0

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

(a) Sn^{2+}+2Cu^{2+}\rightarrow Sn^{4+}+2Cu^+

(b) H_2S+Hg_2^{2+}\rightarrow 2Hg+S+2H^+

(c) 5CN^-+2ClO_2+H_2O\rightarrow 5CNO^-+2Cl^-+2H^+

(d) Fe^{2+}+Ce^{4+}\rightarrow Fe^{3+}+Ce^{3+}

(e) 2HBrO\rightarrow 3Br^-+2Br+2O_2+H_2O+3H^+

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.

<u>(a) The given chemical reaction is,</u>

Sn^{2+}+Cu^{2+}\rightarrow Sn^{4+}+Cu^+

The oxidation-reduction half reaction will be :

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

Reduction : Cu^{2+}+1e^-\rightarrow Cu^+

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

The balanced chemical equation will be,

Sn^{2+}+2Cu^{2+}\rightarrow Sn^{4+}+2Cu^+

<u>(b) The given chemical reaction is,</u>

H_2S+Hg_2^{2+}\rightarrow Hg+S

The oxidation-reduction half reaction will be :

Oxidation : H_2S\rightarrow S+2H^++2e^-

Reduction : Hg_2^{2+}+2e^-\rightarrow 2Hg

The electrons in oxidation and reduction reaction are same. Now add both the equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

H_2S+Hg_2^{2+}\rightarrow 2Hg+S+2H^+

<u>(c) The given chemical reaction is,</u>

CN^-+ClO_2\rightarrow CNO^-+Cl^-

The oxidation-reduction half reaction will be :

Oxidation : CN^-+H_2O\rightarrow CNO^-+2H^++2e^-

Reduction : ClO_2+4H^++5e^-\rightarrow Cl^-+2H_2O

In order to balance the electrons, we multiply the oxidation reaction by 5 and reduction reaction by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

5CN^-+2ClO_2+H_2O\rightarrow 5CNO^-+2Cl^-+2H^+

<u>(d) The given chemical reaction is,</u>

Fe^{2+}+Ce^{4+}\rightarrow Fe^{3+}+Ce^{3+}

The oxidation-reduction half reaction will be :

Oxidation : Fe^{2+}\rightarrow Fe^{3+}+1e^-

Reduction : Ce^{4+}+1e^-\rightarrow Ce^{3+}

The electrons in oxidation and reduction reaction are same. Now add both the equation, we get the balanced redox reaction.

The balanced chemical equation will be,

Fe^{2+}+Ce^{4+}\rightarrow Fe^{3+}+Ce^{3+}

<u>(e) The given chemical reaction is,</u>

HBrO\rightarrow Br^-+O_2

The oxidation-reduction half reaction will be :

Oxidation : HBrO+H_2O\rightarrow O_2+Br+3H^++3e^-

Reduction : HBrO+H^++2e^-\rightarrow Br^-+H_2O

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.

The balanced chemical equation in a acidic solution will be,

2HBrO\rightarrow 3Br^-+2Br+2O_2+H_2O+3H^+

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