Chemistry, 09.12.2019 18:31 mansoorahmedd67

What are the coefficients of the six species in the balanced equation below? remember to include coefficients for h2o(l) and h+(aq) in the appropriate blanks. bro3_(aq) + sn2+(aq) + → br-(aq) + sn4+(aq) +

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Answer from: bbb3771

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

$BrO_3^-(aq)+6H^+(aq)+3Sn^{2+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sn^{4+}(aq)$

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.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion (H^+) at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$BrO_3^-(aq)+Sn^{2+}(aq)\rightarrow Br^-(aq)+Sn^{4+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $BrO_3^-\rightarrow Br^-$

Reduction : $Sn^{2+}\rightarrow Sn^{4+}$

First balance the main element in the reaction.

Oxidation : $BrO_3^-\rightarrow Br^-$

Reduction : $Sn^{2+}\rightarrow Sn^{4+}$

Now balance oxygen atom on both side.

Oxidation : $BrO_3^-\rightarrow Br^-+3H_2O$

Reduction : $Sn^{2+}\rightarrow Sn^{4+}$

Now balance hydrogen atom on both side.

Oxidation : $BrO_3^-+6H^+\rightarrow Br^-+3H_2O$

Reduction : $Sn^{2+}\rightarrow Sn^{4+}$

Now balance the charge.

Oxidation : $BrO_3^-+6H^+6e^-\rightarrow Br^-+3H_2O$

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

The charges are not balanced. Now we are multiplying reduction reaction by 3 and adding both equation, we get the balanced redox reaction.

Oxidation : $BrO_3^-+6H^+6e^-\rightarrow Br^-+3H_2O$

Reduction : $3Sn^{2+}\rightarrow 3Sn^{4+}+6e^-$

The balanced chemical equation in acidic medium will be,

$BrO_3^-(aq)+6H^+(aq)+3Sn^{2+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sn^{4+}(aq)$

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