Chemistry, 02.09.2019 21:30, ntrann22

Balance each of the following equations according to the half- reaction method:
(a) sn2+ + cu2+ --> sn4+ + cu+
(b) h2s + hg2 2+ --> hg + s (in acid)
(c) cn- + clo2 --> cno- + cl- (in acid)
(d) fe2+ + ce 4+ --> fe3+ +ce3+
(e) hbro --> br- + o2 (in acid)

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answered: hiene

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

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

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

(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.

(a) The given chemical reaction is,

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

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,

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

(b) The given chemical reaction is,

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

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,

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

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

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,

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

(d) The given chemical reaction is,

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

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,

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

(e) The given chemical reaction is,

$HBrO\rightarrow Br^-+O_2$

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,

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

answered: Guest

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answered: Guest

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