Chemistry, 07.03.2020 04:59, rileyeddins1010

2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electrons are transferred in the reaction

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answered: joseestrada27

Answer :

The oxidation state of tin changes from (+2) to (+4).

The number electrons transferred in the reaction are, 10 electron.

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 basic 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 more number of oxygen are present.

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

Now balance the charge.

The given chemical reaction is,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The oxidation-reduction half reaction will be :

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $BrO_3^-\rightarrow Br_2$

First balance the main element in the reaction.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $2BrO_3^-\rightarrow Br_2$

Now balance oxygen atom on both side.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+H_2O\rightarrow Br_2$

Now balance hydrogen atom on both side.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+3H_2O\rightarrow Br_2+12OH^-$

Now balance the charge.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O+2e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The charges are not balance on both side of the reaction. We are multiplying oxidation reaction by 5 and then added both equation, we get the balanced redox reaction.

Oxidation : $5SnO_2^{2-}+5OH^-\rightarrow 5SnO_3^{2-}+5H_2O+10e^-$